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Consider a continuous function \( f : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{R} \). Then the image of \( f \) is of the form \( \left\lbrack {a, b}\right\rbrack \) for some real numbers \( a \leq b \). | The point is that the image of \( f \) is compact in \( \mathbb{R} \), and hence closed and bounded. You can convince yourself that the closed sets are just unions of closed intervals. That implies the extreme value theorem. When \( X = \left\lbrack {0,1}\right\rbrack \), the image is also connected, so there should on... | No |
Proposition 8.4.9 (Continuous on compact \( \Rightarrow \) uniformly continuous)\n\nIf \( M \) is compact and \( f : M \rightarrow N \) is continuous, then \( f \) is uniformly continuous. | Proof using sequences. Fix \( \varepsilon > 0 \), and assume for contradiction that for every \( \delta = 1/k \) there exists points \( {x}_{k} \) and \( {y}_{k} \) within \( \delta \) of each other but with images \( \varepsilon > 0 \) apart. By compactness, take a convergent subsequence \( {x}_{{i}_{k}} \rightarrow p... | Yes |
Theorem 8.5.1 (Heine-Borel for general metric spaces)\n\nFor a metric space \( M \), the following are equivalent:\n\n(i) Every sequence has a convergent subsequence,\n\n(ii) The space \( M \) is complete and totally bounded, and\n\n(iii) Every open cover has a finite subcover. | We leave the proof that (i) \( \Leftrightarrow \) (ii) as Problem \( 8{\mathrm{\;F}}^{ \dagger } \) ; the idea of the proof is much in the spirit of Theorem 8.2.2.\n\nProof that (i) and (ii) \( \Rightarrow \) (iii). We prove the following lemma, which is interesting in its own right.\n\nLemma 8.5.2 (Lebesgue n | No |
Lemma 8.5.2 (Lebesgue number lemma)\n\nLet \( M \) be a compact metric space and \( \left\{ {U}_{\alpha }\right\} \) an open cover. Then there exists a real number \( \delta > 0 \), called a Lebesgue number for that covering, such that the \( \delta \) -neighborhood of any point \( p \) lies entirely in some \( {U}_{\a... | Proof of lemma. Assume for contradiction that for every \( \delta = 1/k \) there is a point \( {x}_{k} \in M \) such that its \( 1/k \) -neighborhood isn’t contained in any \( {U}_{\alpha } \) . In this way we construct a sequence \( {x}_{1},{x}_{2},\ldots \) ; thus we’re allowed to take a subsequence which converges t... | Yes |
In fact, any abelian group \( G = \left( {G, + }\right) \) is a \( \mathbb{Z} \) -module. The multiplication can be defined by\n\n\[ n \cdot g = \underset{n\text{ times }}{\underbrace{g + \cdots + g}}\;\left( {-n}\right) \cdot g = n \cdot \left( {-g}\right) \] | for \( n \geq 0 \) . (Here \( - g \) is the additive inverse of \( g \) .) | No |
Example 9.3.2 (Euclidean plane)\n\nTake the vector space \( {\mathbb{R}}^{2} = \{ \left( {x, y}\right) \mid x \in \mathbb{R}, y \in \mathbb{R}\} \) . We can consider it as a direct sum of its \( x \) -axis and \( y \) -axis:\n\n\[ \nX = \{ \left( {x,0}\right) \mid x \in \mathbb{R}\} \text{ and }Y = \{ \left( {0, y}\rig... | This gives us a \ | No |
Theorem 9.4.5 (Maximality and minimality of bases)\n\nLet \( V \) be a vector space over some field \( k \) and take \( {e}_{1},\ldots ,{e}_{n} \in V \) . The following are equivalent:\n\n(a) The \( {e}_{i} \) form a basis.\n\n(b) The \( {e}_{i} \) are spanning, but no proper subset is spanning.\n\n(c) The \( {e}_{i} \... | Proof. Straightforward, do it yourself if you like. The key point to notice is that you need to divide by scalars for the converse direction, hence \( V \) is required to be a vector space instead of just a module for the implications (b) \( \Rightarrow \) (a) and (c) \( \Rightarrow \) (a). | No |
Theorem 9.4.7 (Dimension theorem for vector spaces)\n\nIf a vector space \( V \) has a finite basis, then every other basis has the same number of elements. | Proof. We prove something stronger: Assume \( {v}_{1},\ldots ,{v}_{n} \) is a spanning set while \( {w}_{1},\ldots ,{w}_{m} \) is linearly independent. We claim that \( n \geq m \ ).\n\nLet \( {A}_{0} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be the spanning set. Throw in \( {w}_{1} \) : by the spanning condition,... | Yes |
Let \( V = \left\{ {a{x}^{2} + {bx} + c \mid a, b, c \in \mathbb{R}}\right\} \) . Then \( T\left( {a{x}^{2} + {bx} + c}\right) = {aT}\left( {x}^{2}\right) + {bT}\left( x\right) + {cT}\left( 1\right) \) . | Now I can even be more concrete. I could tell you what \( T\left( {e}_{1}\right) \) is, but seeing as I have a basis of \( W \), I can actually just tell you what \( T\left( {e}_{1}\right) \) is in terms of this basis. Specifically, there are unique \( {a}_{11},{a}_{21},\ldots ,{a}_{n1} \in k \) such that\n\n\[ T\left(... | Yes |
Theorem 9.7.5 (Basis completion)\n\nLet \( V \) be an \( n \) -dimensional space, and \( {V}^{\prime } \) a subspace of \( V \) . Then\n\n(a) \( {V}^{\prime } \) is also finite-dimensional.\n\n(b) If \( {e}_{1},\ldots ,{e}_{m} \) is a basis of \( {V}^{\prime } \), then there exist \( {e}_{m + 1},\ldots ,{e}_{n} \) in \... | Proof. Omitted, since it is intuitive and the proof is not that enlightening. (However, we will use this result repeatedly later on, so do take the time to internalize it now.) | No |
Theorem 9.7.6 (Picking a basis for linear maps)\n\nLet \( T : V \rightarrow W \) be a map of finite-dimensional vector spaces, with \( n = \dim V \) , \( m = \dim W \) . Then there exists a basis \( {v}_{1},\ldots ,{v}_{n} \) of \( V \) and a basis \( {w}_{1},\ldots ,{w}_{m} \) of \( W \) , as well as a nonnegative int... | Sketch of Proof. You might like to try this one yourself before reading on: it's a repeated application of Theorem 9.7.5.\n\nLet \( \ker T \) have dimension \( n - k \) . We can pick \( {v}_{k + 1},\ldots ,{v}_{n} \) a basis of \( \ker T \) . Then extend it to a basis \( {v}_{1},\ldots ,{v}_{n} \) of \( V \) . The map ... | No |
Theorem 9.8.1 (Lagrange interpolation)\n\nLet \( {x}_{1},\ldots ,{x}_{n + 1} \) be distinct real numbers and \( {y}_{1},\ldots ,{y}_{n + 1} \) any real numbers. Then there exists a unique polynomial \( P \) of degree at most \( n \) such that\n\n\[ P\left( {x}_{i}\right) = {y}_{i} \]\n\nfor every \( i \) . | Proof. The idea is to consider the vector space \( V \) of polynomials with degree at most \( n \) , as well as the vector space \( W = {\mathbb{R}}^{n + 1} \).\n\n---\n\n\( {}^{2} \) Source: Communicated to me by Joe Harris at the first Harvard-MIT Undergraduate Math Symposium.\n\n---\n\n\n\nQuestion 9.8.2. Check that... | No |
Theorem 10.3.6 (Eigenvalues always exist over algebraically closed fields)\n\nSuppose \( k \) is an algebraically closed field. Let \( V \) be a finite dimensional \( k \) -vector space. Then if \( T : V \rightarrow V \) is a linear map, there exists an eigenvalue \( \lambda \in k \) . | Proof. (From [Ax97]) The idea behind this proof is to consider \ | No |
Let \( T : {k}^{6} \rightarrow {k}^{6} \) and suppose \( T \) is given by the matrix \[ T = \left\lbrack \begin{array}{llllll} 5 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 7 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 \end{array}\right\rbrack \] | Reading the matrix, we can compute all the eigenvectors and eigenvalues: for any constants \( a, b \in k \) we have \[ T\left( {a \cdot {e}_{1}}\right) = {5a} \cdot {e}_{1} \] \[ T\left( {a \cdot {e}_{2}}\right) = {2a} \cdot {e}_{2} \] \[ T\left( {a \cdot {e}_{4}}\right) = {7a} \cdot {e}_{4} \] \[ T\left( {a \cdot {e}_... | Yes |
Example 10.5.4 (Double staircase)\n\nLet \( V = {k}^{\oplus 5} \) have basis \( {e}_{1},{e}_{2},{e}_{3},{e}_{4},{e}_{5} \) . Then the map\n\n\[ \n{e}_{3} \mapsto {e}_{2} \mapsto {e}_{1} \mapsto 0\text{ and }{e}_{5} \mapsto {e}_{4} \mapsto 0\n\]\n\nis nilpotent. | Picture, with some zeros omitted for emphasis:\n\n\[ \nT = \left\lbrack \begin{array}{lllll} 0 & 1 & 0 & & \\ 0 & 0 & 1 & & \\ 0 & 0 & 0 & & \\ & & & 0 & 1 \\ & & & 0 & 0 \end{array}\right\rbrack \n\] | Yes |
Proposition 10.6.3 (Invariant subspace decomposition)\n\nLet \( V \) be a finite-dimensional vector space. Given any map \( T : V \rightarrow V \), we can write\n\n\[ V = {V}_{1} \oplus {V}_{2} \oplus \cdots \oplus {V}_{m} \]\n\nwhere each \( {V}_{i} \) is \( T \) -invariant, and for any \( i \) the map \( T : {V}_{i} ... | Proof. Same as the proof that every integer is the product of primes. If \( V \) is not decomposable, we are done. Otherwise, by definition write \( V = {W}_{1} \oplus {W}_{2} \) and then repeat on each of \( {W}_{1} \) and \( {W}_{2} \).\n\nIncredibly, with just that we're almost done! Consider a decomposition as abov... | No |
\[ \mathbb{R}\left\lbrack x\right\rbrack { \otimes }_{\mathbb{R}}\mathbb{R}\left\lbrack y\right\rbrack = \mathbb{R}\left\lbrack {x, y}\right\rbrack \] | That is, the tensor product of polynomials in \( x \) with real polynomials in \( y \) turns out to just be two-variable polynomials \( \mathbb{R}\left\lbrack {x, y}\right\rbrack \) . | Yes |
Proposition 11.1.4 (Basis of \( V \otimes W \) )\n\nLet \( V \) and \( W \) be finite-dimensional \( k \) -vector spaces. If \( {e}_{1},\ldots ,{e}_{m} \) is a basis of \( V \) and \( {f}_{1},\ldots ,{f}_{n} \) is a basis of \( W \), then the basis of \( V{ \otimes }_{k}W \) is precisely \( {e}_{i} \otimes {f}_{j} \), ... | Proof. Omitted; it's easy at least to see that this basis is spanning. | No |
The set of real functions \( f\left( {x, y, z}\right) \) is an infinite-dimensional real vector space. | Indeed, we can add two functions to get \( f + g \), and we can think of functions like \( {2f} \). | No |
Theorem 11.3.3 \( \left( {{V}^{ \vee } \otimes W \Leftrightarrow \text{linear maps}V \rightarrow W}\right) \) | Let \( V \) and \( W \) be finite-dimensional vector spaces. We described a map\n\n\[ \Psi : {V}^{ \vee } \otimes W \rightarrow \operatorname{Hom}\left( {V, W}\right) \]\n\nby sending \( {\xi }_{1} \otimes {w}_{1} + \cdots + {\xi }_{m} \otimes {w}_{m} \) to the linear map\n\n\[ v \mapsto {\xi }_{1}\left( v\right) {w}_{... | Yes |
Let \( V = {\mathbb{R}}^{2} \) and take a basis \( {e}_{1},{e}_{2} \) of \( V \) . Then define \( T : V \rightarrow V \) by\n\n\[ T = \left\lbrack \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right\rbrack \]\n\nThen we have\n\n\[ \Psi \left( {{e}_{1}^{ \vee } \otimes {e}_{1} + 2{e}_{2}^{ \vee } \otimes {e}_{1} + 3{e}_{... | Proof of Theorem 11.3.3. This looks intimidating, but it's actually not difficult. We proceed in two steps:\n\n1. First, we check that \( \Psi \) is surjective; every linear map has at least one representation in \( {V}^{ \vee } \otimes W \) . To see this, take any \( T : V \rightarrow W \) . Suppose \( V \) has basis ... | No |
Let \( V = {\mathbb{R}}^{2} \), and let \( v = a{e}_{1} + b{e}_{2}, w = c{e}_{1} + d{e}_{2} \). Now let’s compute \( v \land w \) in \( {\Lambda }^{2}\left( V\right) \). | \[ v \land w = \left( {a{e}_{1} + b{e}_{2}}\right) \land \left( {c{e}_{1} + d{e}_{2}}\right) \] \[ = {ac}\left( {{e}_{1} \land {e}_{1}}\right) + {bd}\left( {{e}_{2} \land {e}_{2}}\right) + {ad}\left( {{e}_{1} \land {e}_{2}}\right) + {bc}\left( {{e}_{2} \land {e}_{1}}\right) \] \[ = \operatorname{ad}\left( {{e}_{1} \lan... | Yes |
Proposition 12.1.4 (Basis of \( {\Lambda }^{2}\left( V\right) \) )\n\nLet \( V \) be a vector space with basis \( {e}_{1},\ldots ,{e}_{n} \) . Then a basis of \( {\Lambda }^{2}\left( V\right) \) is\n\n\[ {e}_{i} \land {e}_{j} \]\n\nwhere \( i < j \) . Hence \( {\Lambda }^{2}\left( V\right) \) has dimension \( \left( \b... | Proof. Surprisingly slippery, and also omitted. (You can derive it from the corresponding theorem on tensor products.) | No |
Proposition 12.1.6 (Basis of the wedge product)\n\nLet \( V \) be a vector space with basis \( {e}_{1},\ldots ,{e}_{n} \) . A basis for \( {\Lambda }^{m}\left( V\right) \) consists of the elements\n\n\[ {e}_{{i}_{1}} \land {e}_{{i}_{2}} \land \cdots \land {e}_{{i}_{m}} \]\n\nwhere\n\n\[ 1 \leq {i}_{1} < {i}_{2} < \cdot... | Sketch of proof. We knew earlier that \( {e}_{{i}_{1}} \otimes \cdots \otimes {e}_{{i}_{m}} \) was a basis for the tensor product. Here we have the additional property that (a) if two basis elements re-appear then the whole thing becomes zero, thus we should assume the \( i \) ’s are all distinct; and (b) we can shuffl... | No |
In \( V = {\mathbb{R}}^{4} \) with standard basis \( {e}_{1},{e}_{2},{e}_{3},{e}_{4} \), let \( T\left( {e}_{1}\right) = {e}_{2}, T\left( {e}_{2}\right) = 2{e}_{3}, T\left( {e}_{3}\right) = {e}_{3} \) and \( T\left( {e}_{4}\right) = 2{e}_{2} + {e}_{3} \). Then, for example, \( {\Lambda }^{2}\left( T\right) \) sends | \n\[{e}_{1} \land {e}_{2} + {e}_{3} \land {e}_{4} \mapsto T\left( {e}_{1}\right) \land T\left( {e}_{2}\right) + T\left( {e}_{3}\right) \land T\left( {e}_{4}\right)\]\n\[= {e}_{2} \land 2{e}_{3} + {e}_{3} \land \left( {2{e}_{2} + {e}_{3}}\right)\]\n\[= 2\left( {{e}_{2} \land {e}_{3} + {e}_{3} \land {e}_{2}}\right)\]\n\[... | Yes |
Example 12.2.3 (The determinant of a \( 2 \times 2 \) matrix)\n\nLet \( V = {\mathbb{R}}^{2} \) again with basis \( {e}_{1} \) and \( {e}_{2} \) . Let\n\n\[ T = \left\lbrack \begin{array}{ll} a & c \\ b & d \end{array}\right\rbrack \]\n\nIn other words, \( T\left( {e}_{1}\right) = a{e}_{1} + b{e}_{2} \) and \( T\left( ... | Now let’s consider \( {\Lambda }^{2}\left( V\right) \) . It has a basis \( {e}_{1} \land {e}_{2} \) . Now \( {\Lambda }^{2}\left( T\right) \) sends it to\n\n\[ {e}_{1} \land {e}_{2}\xrightarrow[]{{\Lambda }^{2}\left( T\right) }T\left( {e}_{1}\right) \land T\left( {e}_{2}\right) = \left( {a{e}_{1} + b{e}_{2}}\right) \la... | Yes |
Example 12.3.2 (Example of Cayley-Hamilton using determinant definition)\n\nSuppose \( T = \left\lbrack \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right\rbrack \) . Using the determinant definition of characteristic polynomial, we find that \( {p}_{T}\left( X\right) = \left( {X - 1}\right) \left( {X - 4}\right) - \le... | \[ {T}^{2} - {5T} - 2 = \left\lbrack \begin{matrix} 7 & {10} \\ {15} & {22} \end{matrix}\right\rbrack - 5 \cdot \left\lbrack \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right\rbrack - 2 \cdot \left\lbrack \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right\rbrack = \left\lbrack \begin{array}{ll} 0 & 0 \\ 0 & 0 \end{arr... | Yes |
Lemma 13.2.4 (Cauchy-Schwarz)\n\nLet \( V \) be an inner product space. For any \( v, w \in V \) we have\n\n\[ \left| {\langle v, w\rangle }\right| \leq \parallel v\parallel \parallel w\parallel \]\n\nwith equality if and only if \( v \) and \( w \) are linearly dependent. | Proof. The theorem is immediate if \( \langle v, w\rangle = 0 \) . It is also immediate if \( \parallel v\parallel \parallel w\parallel = 0 \), since then one of \( v \) or \( w \) is the zero vector. So henceforth we assume all these quantities are nonzero (as we need to divide by them later).\n\nThe key to the proof ... | Yes |
Theorem 13.2.5 (Triangle inequality)\n\nWe always have\n\n\[ \parallel v\parallel + \parallel w\parallel \geq \parallel v + w\parallel \]\n\nwith equality if and only if \( v \) and \( w \) are linearly dependent. | Exercise 13.2.6. Prove this by squaring both sides, and applying Cauchy-Schwarz. | No |
Lemma 13.3.2 (Orthogonal vectors are independent)\n\nAny set of pairwise orthogonal vectors \( {v}_{1},{v}_{2},\ldots ,{v}_{n} \), with \( \begin{Vmatrix}{v}_{i}\end{Vmatrix} \neq 0 \) for each \( i \), is linearly independent. | Proof. Consider a dependence\n\n\[ {a}_{1}{v}_{1} + \cdots + {a}_{n}{v}_{n} = 0 \]\n\nfor \( {a}_{i} \) in \( \mathbb{R} \) or \( \mathbb{C} \) . Then\n\n\[ {0}_{V} = \left\langle {{v}_{1},\sum {a}_{i}{v}_{i}}\right\rangle = \overline{{a}_{1}}{\begin{Vmatrix}{v}_{1}\end{Vmatrix}}^{2}. \]\n\nHence \( {a}_{1} = 0 \), sin... | Yes |
Theorem 13.3.5 (Gram-Schmidt)\n\nLet \( V \) be a finite-dimensional inner product space. Then it has an orthonormal basis. | Sketch of Proof. One constructs the orthonormal basis explicitly from any basis \( {e}_{1},\ldots \) , \( {e}_{n} \) of \( V \) . Define \( {\operatorname{proj}}_{u}\left( v\right) = \frac{\langle v, u\rangle }{\langle u, u\rangle }u \) . Then recursively define\n\n\[ \n{u}_{1} = {e}_{1} \n\]\n\n\[ \n{u}_{2} = {e}_{2} ... | No |
Proposition 13.4.2 (Convergence criteria in a Hilbert space)\n\nThe sequence \( \left( {v}_{i}\right) \) defined above converges if and only if \( \sum {\left| {c}_{i}\right| }^{2} < \infty \) . | Proof. This will make more sense if you read Chapter 26, so you could skip this proof if you haven’t read the chapter. The sequence \( {v}_{i} \) converges if and only if it is Cauchy, meaning that when \( i < j \), \n\n\[ \n{\begin{Vmatrix}{v}_{j} - {v}_{i}\end{Vmatrix}}^{2} = {\left| {c}_{i + 1}\right| }^{2} + \cdots... | No |
Proposition 14.2.1 (Facts about orthonormal bases)\n\nLet \( V \) be a complex Hilbert space with inner form \( \langle - , - \rangle \) and suppose \( x = \mathop{\sum }\limits_{\xi }{a}_{\xi }{e}_{\xi } \) and \( y = \mathop{\sum }\limits_{\xi }{b}_{\xi }{e}_{\xi } \) where \( {e}_{\xi } \) are an orthonormal basis. ... | Exercise 14.2.2. Prove all of these. (You don't need any of the preceding section, it's only there to motivate the notation with lots of scary \( \xi \) ’s.) | No |
Let \( n = 2 \) . Then binary functions \( \{ \pm 1{\} }^{2} \rightarrow \mathbb{C} \) have a basis given by the four polynomials\n\n\[1,\;{x}_{1},\;{x}_{2},\;{x}_{1}{x}_{2}.\]\n\nFor example, consider the function \( f \) which is 1 at \( \left( {1,1}\right) \) and 0 elsewhere. Then we can put\n\n\[f\left( {{x}_{1},{x... | So the Fourier coefficients are \( \widehat{f}\left( S\right) = \frac{1}{4} \) for each of the four \( S \) ’s. | Yes |
Proposition 14.3.6 \( \left( {e}_{\xi }\right. \) are orthonormal) | Proof. I recommend skipping this one, but it is:\n\n\[ \left\langle {{e}_{\xi },{e}_{{\xi }^{\prime }}}\right\rangle = \frac{1}{\left| Z\right| }\mathop{\sum }\limits_{{x \in Z}}e\left( {\xi \cdot x}\right) \overline{e\left( {{\xi }^{\prime } \cdot x}\right) }\n\]\n\n\[ = \frac{1}{\left| Z\right| }\mathop{\sum }\limits... | Yes |
Example 14.3.7 (Cube roots of unity filter)\n\nSuppose \( Z = \mathbb{Z}/3\mathbb{Z} \), with the inner form given by \( \xi \cdot x = \left( {\xi x}\right) /3 \) . Let \( \omega = \exp \left( {\frac{2}{3}{\pi i}}\right) \) be a primitive cube root of unity. Note that\n\n\[ \n{e}_{\xi }\left( x\right) = \left\{ \begin{... | In this way we derive that the transforms are\n\n\[ \n\widehat{f}\left( 0\right) = \frac{a + b + c}{3}\n\]\n\n\[ \n\widehat{f}\left( 1\right) = \frac{a + {\omega }^{2}b + {\omega c}}{3}\n\]\n\n\[ \n\widehat{f}\left( 2\right) = \frac{a + {\omega b} + {\omega }^{2}c}{3}.\n\] | Yes |
Theorem 14.3.11 (The classical Fourier basis)\n\nFor each integer \( n \), define\n\n\[ \n{e}_{n}\left( x\right) = \exp \left( {inx}\right) \n\]\n\nThen \( {e}_{n} \) form an orthonormal basis of the Hilbert space \( {L}^{2}\left( \left\lbrack {-\pi ,\pi }\right\rbrack \right) \) . | Thus this time the frequency set \( \mathbb{Z} \) is infinite, and we have\n\n\[ \nf\left( x\right) = \mathop{\sum }\limits_{n}\widehat{f}\left( n\right) \exp \left( {inx}\right) \;\text{ almost everywhere } \n\]\n\nfor coefficients \( \widehat{f}\left( n\right) \) with \( \mathop{\sum }\limits_{n}{\left| \widehat{f}\l... | No |
Corollary 14.5.1 (Parseval theorem)\n\nLet \( f : Z \rightarrow \mathbb{C} \), where \( Z \) is a finite abelian group. Then\n\n\[ \mathop{\sum }\limits_{\xi }{\left| \widehat{f}\left( \xi \right) \right| }^{2} = \frac{1}{\left| Z\right| }\mathop{\sum }\limits_{{x \in Z}}{\left| f\left( x\right) \right| }^{2} \]\n\nSim... | Proof. Recall that \( \langle f, f\rangle \) is equal to the square sum of the coefficients. | No |
Corollary 14.5.2 (Fourier inversion formula)\n\nLet \( f : Z \rightarrow \mathbb{C} \), where \( Z \) is a finite abelian group. Then\n\n\[ \widehat{f}\left( \xi \right) = \frac{1}{\left| Z\right| }\mathop{\sum }\limits_{{x \in Z}}f\left( x\right) \overline{{e}_{\xi }\left( x\right) }.\]\n\nSimilarly, if \( f : \left\l... | Proof. Recall that in an orthonormal basis \( {\left( {e}_{\xi }\right) }_{\xi } \), the coefficient of \( {e}_{\xi } \) in \( f \) is \( \left\langle {f,{e}_{\xi }}\right\rangle \) . | No |
Theorem 14.6.1 (Basel problem)\n\nWe have\n\n\[ \mathop{\sum }\limits_{{n \geq 1}}\frac{1}{{n}^{2}} = \frac{{\pi }^{2}}{6} \] | The proof is to consider the identity function \( f\left( x\right) = x \), which is certainly square-integrable.\n\nThen by Parseval, we have\n\n\[ \mathop{\sum }\limits_{{n \in \mathbb{Z}}}{\left| \widehat{f}\left( n\right) \right| }^{2} = \langle f, f\rangle = \frac{1}{2\pi }{\int }_{\left\lbrack -\pi ,\pi \right\rbr... | No |
Example 15.1.2 (Example of a dual map)\n\nWork over \( \mathbb{R} \) . Let’s consider \( V \) with basis \( {e}_{1},{e}_{2},{e}_{3} \) and \( W \) with basis \( {f}_{1},{f}_{2} \) . Suppose that\n\n\[ T\left( {e}_{1}\right) = {f}_{1} + 2{f}_{2} \]\n\n\[ T\left( {e}_{2}\right) = 3{f}_{1} + 4{f}_{2} \]\n\n\[ T\left( {e}_... | If we write the matrices for \( T \) and \( {T}^{ \vee } \) in terms of our basis, we now see that\n\n\[ T = \left\lbrack \begin{array}{lll} 1 & 3 & 5 \\ 2 & 4 & 6 \end{array}\right\rbrack \;\text{ and }\;{T}^{ \vee } = \left\lbrack \begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array}\right\rbrack \]\nSo in our select... | Yes |
Theorem 15.1.3 (Transpose interpretation of \( {T}^{ \vee } \) )\n\nLet \( V \) and \( W \) be finite-dimensional \( k \) -vector spaces. Then, for any \( T : V \rightarrow W \), the following two matrices are transposes:\n\n- The matrix for \( T : V \rightarrow W \) expressed in the basis \( \left( {e}_{i}\right) ,\le... | Proof. The \( \left( {i, j}\right) \) th entry of the matrix \( T \) corresponds to the coefficient of \( {f}_{j} \) in \( T\left( {e}_{i}\right) \) , which corresponds to the coefficient of \( {e}_{i}^{ \vee } \) in \( {f}_{j}^{ \vee } \circ T \) . | Yes |
Theorem 15.2.2 \( \left( {V \cong {V}^{ \vee }}\right. \) for real inner form) | Proof. It suffices to show that the map is injective and surjective.\n\n- Injective: suppose \( \left\langle {{v}_{1}, v}\right\rangle = \left\langle {{v}_{2}, v}\right\rangle \) for every vector \( v \in V \) . This means \( \left\langle {{v}_{1} - {v}_{2}, v}\right\rangle = \) 0 for every vector \( v \in V \) . This ... | Yes |
Theorem 15.2.3 ( \( V \) versus \( {V}^{ \vee } \) for complex inner forms)\n\nLet \( V \) be a finite-dimensional complex inner product space and \( {V}^{ \vee } \) its dual. Then the map \( V \rightarrow {V}^{ \vee } \) by\n\n\[ v \mapsto \langle -, v\rangle \in {V}^{ \vee } \]\n\nis a bijection of sets. | Wait, what? Well, the proof above shows that it is both injective and surjective, but why is it not an isomorphism? The answer is that it is not a linear map: since the form is sesquilinear we have for example\n\n\[ {iv} \mapsto \langle - ,{iv}\rangle = - i\langle -, v\rangle \]\n\nwhich has introduced a minus sign! In... | No |
We compute \( {T}^{ \dagger }\left( {f}_{1}\right) \) . It is the unique vector \( x \in V \) such that\n\n\[ \langle v, x{\rangle }_{V} = {\left\langle T\left( v\right) ,{f}_{1}\right\rangle }_{W} \]\n\nfor any \( v \in V \) . | If we expand \( v = a{e}_{1} + b{e}_{2} + c{e}_{3} \) the above equality becomes\n\n\[ {\left\langle a{e}_{1} + b{e}_{2} + c{e}_{3}, x\right\rangle }_{V} = {\left\langle T\left( a{e}_{1} + b{e}_{2} + c{e}_{3}\right) ,{f}_{1}\right\rangle }_{W} \]\n\n\[ = {ia} + {3b} + {5c}\text{.} \]\n\nHowever, since \( x \) is in the... | Yes |
Theorem 15.3.3 (Adjoints are conjugate transposes)\n\nFix an orthonormal basis of a finite-dimensional inner product space \( V \) . Let \( T : V \rightarrow \) \( V \) be a linear map. If we write \( T \) as a matrix in this basis, then the matrix \( {T}^{ \dagger } \) (in the same basis) is the conjugate transpose of... | Proof. One-line version: take \( v \) and \( w \) to be basis elements, and this falls right out.\n\nFull proof: let\n\n\[ T = \left\lbrack \begin{matrix} {a}_{11} & \ldots & {a}_{1n} \\ \vdots & \ddots & \vdots \\ {a}_{n1} & \ldots & {a}_{nn} \end{matrix}\right\rbrack \]\n\nin this basis \( {e}_{1},\ldots ,{e}_{n} \) ... | Yes |
Theorem 15.4.6 (Hermitian matrices have real eigenvalues)\n\nA Hermitian matrix \( T \) is diagonalizable, and all its eigenvalues are real. | Proof. Obviously Hermitian \( \Rightarrow \) normal, so write it in the orthonormal basis of eigenvectors. To see that the eigenvalues are real, note that \( T = {T}^{ \dagger } \) means \( {\lambda }_{i} = \overline{{\lambda }_{i}} \) for every \( i \) . | No |
Theorem 16.2.5 (Orbit-stabilizer theorem)\n\nLet \( \mathcal{O} \) be an orbit, and pick any \( x \in \mathcal{O} \) . Let \( S = {\operatorname{Stab}}_{G}\left( x\right) \) be a subgroup of \( G \) . There is a natural bijection between \( \mathcal{O} \) and left cosets. In particular,\n\n\[ \left| \mathcal{O}\right| ... | Proof. The point is that every coset \( {gS} \) just specifies an element of \( \mathcal{O} \), namely \( g \cdot x \) . The fact that \( S \) is a stabilizer implies that it is irrelevant which representative we pick.\n\nSince the \( \left| \mathcal{O}\right| \) cosets partition \( G \), each of size \( \left| S\right... | Yes |
Example 16.4.3 (Conjugacy classes of \( {S}_{n} \) correspond to cycle types) | Intuitively, the discussion above says that two elements of \( {S}_{n} \) should be conjugate if they have the same \ | No |
Proposition 17.1.2 (Triple product of primes)\n\nIf \( \\left| G\\right| = {pqr} \) is the product of distinct primes, then \( G \) must have a normal Sylow subgroup. | Proof. WLOG, assume \( p < q < r \) . Notice that \( {n}_{p} \equiv 1\\left( {\\;\\operatorname{mod}\\;p}\\right) ,{n}_{p} \mid {qr} \) and cyclically, and assume for contradiction that \( {n}_{p},{n}_{q},{n}_{r} > 1 \) .\n\nSince \( {n}_{r} \mid {pq} \), we have \( {n}_{r} = {pq} \) since \( {n}_{r} \) divides neither... | Yes |
Example 17.3.5 (Fundamental theorem of arithmetic when \( n = {12} \) ) | Let’s consider the group \( \mathbb{Z}/{12}\mathbb{Z} \) . It’s not hard to check that the possible composition series are\n\n\( \{ 1\} \trianglelefteq \mathbb{Z}/2\mathbb{Z} \trianglelefteq \mathbb{Z}/4\mathbb{Z} \trianglelefteq \mathbb{Z}/{12}\mathbb{Z} \) with factors \( \mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z}... | Yes |
Proposition 18.2.5 (PID's are Noetherian UFD's)\n\nIf \( R \) is a PID, then it is Noetherian and also a UFD. | Proof. The fact that \( R \) is Noetherian is obvious. For \( R \) to be a UFD we essentially repeat the proof for \( \mathbb{Z} \), using the fact that \( \left( {a, b}\right) \) is principal in order to extract \( \gcd \left( {a, b}\right) \) . | No |
Theorem 18.2.6 (Chinese remainder theorem for rings)\n\nLet \( m \) and \( n \) be relatively prime elements, meaning \( \left( m\right) + \left( n\right) = \left( 1\right) \) . Then\n\n\[ R/\left( {mn}\right) \cong R/\left( m\right) \times R/\left( n\right) . \] | Proof. This is the same as the proof of the usual Chinese remainder theorem. First, since \( \left( {m, n}\right) = \left( 1\right) \) we have \( {am} + {bn} = 1 \) for some \( a \) and \( b \) . Then we have a map\n\n\[ R/\left( m\right) \times R/\left( n\right) \rightarrow R/\left( {mn}\right) \;\text{ by }\;\left( {... | No |
Corollary 18.3.2 (Structure theorem, primary form)\n\nLet \( R \) be a PID and let \( M \) be any finitely generated \( R \) -module. Then\n\n\[ M \cong {R}^{\oplus r} \oplus R/\left( {q}_{1}\right) \oplus R/\left( {q}_{2}\right) \oplus \cdots \oplus R/\left( {q}_{m}\right) \]\n\nwhere \( {q}_{i} = {p}_{i}^{{e}_{i}} \)... | Proof of corollary. Factor each \( {s}_{i} \) into prime factors (since \( R \) is a UFD), then use the Chinese remainder theorem. | No |
Lemma 18.4.2 (Direct sum of Noetherian modules is Noetherian)\n\nLet \( M \) and \( N \) be two Noetherian \( R \) -modules. Then the direct sum \( M \oplus N \) is also a Noetherian \( R \) -module. | Proof. It suffices to show that if \( L \subseteq M \oplus N \), then \( L \) is finitely generated. One guess is that \( L = P \oplus Q \), where \( P \) and \( Q \) are the projections of \( L \) onto \( M \) and \( N \) . Unfortunately this is false (take \( M = N = \mathbb{Z} \) and \( L = \{ \left( {n, n}\right) \... | No |
Example 18.5.4 (Example of Smith normal form)\n\nTo give a flavor of the idea of the proof, let's work through a concrete example with\n\nthe \( \\mathbb{Z} \) -matrix\n\[ \n\\left\\lbrack \\begin{array}{lll} {18} & {38} & {48} \\ {14} & {30} & {32} \\end{array}\\right\\rbrack \n\] | The GCD of all the entries is 2, and so motivated by this, we perform the Euclidean algorithm on the left column: subtract the second row from the first row, then three times the first row from the second:\n\n\[ \n\\left\\lbrack \\begin{array}{lll} {18} & {38} & {48} \\ {14} & {30} & {32} \\end{array}\\right\\rbrack \\... | Yes |
Let \( A = \mathbb{R}\left\lbrack {S}_{3}\right\rbrack \) . Then let\n\n\[ V = {\mathbb{R}}^{\oplus 3} = \{ \left( {x, y, z}\right) \mid x, y, z \in \mathbb{R}\} . \]\n\nWe can let \( A \) act on \( V \) as follows: given a permutation \( \pi \in {S}_{3} \), we permute the corresponding coordinates in \( V \) . So for ... | From the matrix perspective, what we are doing is representing the permutations in \( {S}_{3} \) as permutation matrices on \( {k}^{\oplus 3} \), like\n\n\[ \left( \begin{array}{ll} 1 & 2 \end{array}\right) \mapsto \left\lbrack \begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \] | No |
Proposition 19.3.4 (Representations of \( A \oplus B \) are \( {V}_{A} \oplus {V}_{B} \) )\n\nLet \( A \) and \( B \) be \( k \) -algebras. Then every representation of \( A \oplus B \) is of the form\n\n\[ \n{V}_{A} \oplus {V}_{B} \n\]\n\nwhere \( {V}_{A} \) and \( {V}_{B} \) are representations of \( A \) and \( B \)... | Sketch of Proof. Let \( \left( {V,\rho }\right) \) be a representation of \( A \oplus B \) . For any \( v \in V,\rho \left( {{1}_{A} + {1}_{B}}\right) v = \) \( \rho \left( {1}_{A}\right) v + \rho \left( {1}_{B}\right) v \) . One can then set \( {V}_{A} = \left\{ {\rho \left( {1}_{A}\right) v \mid v \in V}\right\} \) a... | No |
Example 19.4.4 (Representation of \( {S}_{n} \) decomposes)\n\nLet \( A = \mathbb{R}\left\lbrack {S}_{3}\right\rbrack \) again, acting via permutation of coordinates on\n\n\[ V = {\mathbb{R}}^{\oplus 3} = \{ \left( {x, y, z}\right) \mid x, y, z \in \mathbb{R}\} . \]\n\nConsider again the two subspaces\n\n\[ {W}_{1} = \... | - For \( {W}_{1} \) it’s obvious, since \( {W}_{1} \) is one-dimensional.\n\n- For \( {W}_{2} \), consider any vector \( w = \left( {a, b, c}\right) \) with \( a + b + c = 0 \) and not all zero. Then WLOG we can assume \( a \neq b \) (since not all three coordinates are equal). In that case,(12) sends \( w \) to \( {w}... | Yes |
Theorem 19.5.7 (Schur's lemma for algebraically closed fields)\n\nLet \( k \) be an algebraically closed field. Let \( V \) be an irrep of a \( k \) -algebra \( A \) . Then any intertwining operator \( T : V \rightarrow V \) is multiplication by a scalar. | Exercise 19.5.8. Use the fact that \( T \) has an eigenvalue \( \lambda \) to deduce this from Schur’s lemma. (Consider \( T - \lambda \cdot {\mathrm{{id}}}_{V} \), and use Schur to deduce it’s zero.) | No |
Corollary 20.1.3 (Subrepresentations of completely reducible representations)\n\nLet \( V = \bigoplus {V}_{i}^{\oplus {n}_{i}} \) be completely reducible. Then any subrepresentation \( W \) of \( V \) is isomorphic to \( \bigoplus {V}_{i}^{\oplus {m}_{i}} \) where \( {m}_{i} \leq {n}_{i} \) for each \( i \), and the in... | Proof. Apply Schur’s lemma to the inclusion \( W \hookrightarrow V \) . | No |
Any finite-dimensional algebra \( A \) has at most \( \dim A \) irreps. | If \( {V}_{i} \) are such irreps then \( A \rightarrow {\bigoplus }_{i}{V}_{i}^{\oplus \dim {V}_{i}} \), hence we have the inequality \( \sum {\left( \dim {V}_{i}\right) }^{2} \leq \dim A \) | Yes |
Theorem 20.3.2 (Semisimple algebras)\n\nLet \( A \) be a finite-dimensional algebra. Then the following are equivalent:\n\n(i) \( A \cong {\bigoplus }_{i}{\operatorname{Mat}}_{{d}_{i}}\left( k\right) \) for some \( {d}_{i} \).\n\n(ii) \( A \) is semisimple.\n\n(iii) \( \operatorname{Reg}\left( A\right) \) is completely... | Proof. (i) \( \Rightarrow \) (ii) follows from Theorem 19.6.1 and Proposition 19.3.4. (ii) \( \Rightarrow \) (iii) is tautological.\n\nTo see (iii) \( \Rightarrow \) (i), we use the following clever trick. Consider\n\n\[{\operatorname{Hom}}_{\text{rep }}\left( {\operatorname{Reg}\left( A\right) ,\operatorname{Reg}\left... | No |
Theorem 20.3.3 (Sum of squares formula)\n\nFor a finite-dimensional algebra \( A \) we have\n\n\[ \mathop{\sum }\limits_{i}\dim {\left( {V}_{i}\right) }^{2} \leq \dim A \]\n\nwhere the \( {V}_{i} \) are the irreps of \( A \) ; equality holds exactly when \( A \) is semisimple, in which case\n\n\[ \operatorname{Reg}\lef... | Proof. The inequality was already mentioned in Corollary 20.2.2. It is equality if and only if the map \( \rho : A \rightarrow {\bigoplus }_{i}\operatorname{Mat}\left( {V}_{i}\right) \) is an isomorphism; this means all \( {V}_{i} \) are present. | Yes |
Theorem 20.4.1 (Maschke's theorem)\n\nLet \( G \) be a finite group, and \( k \) an algebraically closed field whose characteristic does not divide \( \left| G\right| \) . Then \( k\left\lbrack G\right\rbrack \) is semisimple. | Proof. Consider any finite-dimensional representation \( \left( {V,\rho }\right) \) of \( k\left\lbrack G\right\rbrack \) . Given a proper subrepresentation \( W \subseteq V \), our goal is to construct a supplementary \( G \) -invariant subspace \( {W}^{\prime } \) which satisfies\n\n\[ V = W \oplus {W}^{\prime } \]\n... | No |
Theorem 21.2.4 (Character of representations of algebras)\n\nLet \( A \) be an algebra over an algebraically closed field. Then\n\n(a) Characters of pairwise non-isomorphic irreps are linearly independent as elements of \( {A}^{\text{ab }} \) .\n\n(b) If \( A \) is finite-dimensional and semisimple, then the characters... | Proof. Part (a) is more or less obvious by the density theorem. Suppose there is a linear dependence, so that for every \( a \) we have\n\n\[ {c}_{1}{\chi }_{{V}_{1}}\left( a\right) + {c}_{2}{\chi }_{{V}_{2}}\left( a\right) + \cdots + {c}_{r}{\chi }_{{V}_{r}}\left( a\right) = 0 \]\n\nfor some integer \( r \) . | No |
Example 21.4.1 (Dihedral group on 10 elements)\n\nLet \( {D}_{10} = \left\langle {r, s \mid {r}^{5} = {s}^{2} = 1,{rs} = s{r}^{-1}}\right\rangle \) . Let \( \omega = \exp \left( \frac{2\pi i}{5}\right) \) . We write four representations of \( {D}_{10} \) :\n\n- \( {\mathbb{C}}_{\text{triv }} \), all elements of \( {D}_... | We do so by writing the character table:\n\n<table><thead><tr><th>\( {D}_{10} \)</th><th>1</th><th>\( r,{r}^{4} \)</th><th>\( {r}^{2},{r}^{3} \)</th><th>\( s{r}^{k} \)</th></tr></thead><tr><td>\( {\mathbb{C}}_{\text{triv }} \)</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>\( {\mathbb{C}}_{\text{sign }} \)</t... | Yes |
Theorem 22.1.1 (Frobenius divisibility)\n\nLet \( V \) be a complex irrep of a finite group \( G \) . Then \( \dim V \) divides \( \left| G\right| \) . | The proof of this will require algebraic integers (developed in the algebraic number theory chapter). Recall that an algebraic integer is a complex number which is the root of a polynomial with integer coefficients, and that these algebraic integers form a ring \( \overline{\mathbb{Z}} \) under addition and multiplicat... | No |
Lemma 22.1.2 (Elements of \( \mathbb{Z}\left\lbrack G\right\rbrack \) are integral)\n\nLet \( \alpha \in \mathbb{Z}\left\lbrack G\right\rbrack \) . Then there exists a monic polynomial \( P \) with integer coefficients such that \( P\left( \alpha \right) = 0 \) . | Proof. Let \( {A}_{k} \) be the \( \mathbb{Z} \) -span of \( 1,{\alpha }^{1},\ldots ,{\alpha }^{k} \) . Since \( \mathbb{Z}\left\lbrack G\right\rbrack \) is Noetherian, the inclusions \( {A}_{0} \subseteq {A}_{1} \subseteq {A}_{2} \subseteq \ldots \) cannot all be strict, hence \( {A}_{k} = {A}_{k + 1} \) for some \( k... | No |
Lemma 22.2.2 (On \( \gcd \left( {\left| C\right| ,\dim V}\right) = 1 \) )\n\nLet \( V = \left( {V,\rho }\right) \) be an complex irrep of \( G \) . Assume \( C \) is a conjugacy class of \( G \) with \( \gcd \left( {\left| C\right| ,\dim V}\right) = 1 \) . Then for any \( g \in C \), either\n\n- \( \rho \left( g\right)... | Proof. If \( {\varepsilon }_{i} \) are the \( n \) eigenvalues of \( \rho \left( g\right) \) (which are roots of unity), then from the proof of Frobenius divisibility we know \( \frac{\left| C\right| }{n}{\chi }_{V}\left( g\right) \in \overline{\mathbb{Z}} \), thus from \( \gcd \left( {\left| C\right|, n}\right) = 1 \)... | Yes |
Lemma 22.2.3 (Simple groups don't have prime power conjugacy classes)\n\nLet \( G \) be a finite simple group. Then \( G \) cannot have a conjugacy class of order \( {p}^{k} \) (where \( k > 0 \) ). | Proof. By contradiction. Assume \( C \) is such a conjugacy class, and fix any \( g \in C \) . By the second orthogonality formula (Problem \( {21}{\mathrm{E}}^{ \star } \) ) applied \( g \) and \( {1}_{G} \) (which are not conjugate since \( \left. {g \neq {1}_{G}}\right) \) we have\n\n\[
\mathop{\sum }\limits_{{i = 1... | No |
Example 22.3.1 (Frobenius determinants)\n\n(a) If \( G = \mathbb{Z}/2\mathbb{Z} = \left\langle {T \mid {T}^{2} = 1}\right\rangle \) then the matrix would be\n\n\[ \n{M}_{G} = \left\lbrack \begin{array}{ll} {x}_{\mathrm{{id}}} & {x}_{T} \\ {x}_{T} & {x}_{\mathrm{{id}}} \end{array}\right\rbrack \n\]\n\nThen \( \det {M}_{... | Then \( \det {M}_{G} = \left( {{x}_{\mathrm{{id}}} - {x}_{T}}\right) \left( {{x}_{\mathrm{{id}}} + {x}_{T}}\right) \). | Yes |
Theorem 22.3.2 (Frobenius determinant)\n\nThe polynomial \( \det {M}_{G} \) (in \( \left| G\right| \) variables) factors into a product of irreducible polynomials such that\n\n(i) The number of polynomials equals the number of conjugacy classes of \( G \), and\n\n(ii) The multiplicity of each polynomial equals its degr... | Proof. Let \( V = \left( {V,\rho }\right) = \operatorname{Reg}\left( {\mathbb{C}\left\lbrack G\right\rbrack }\right) \) and let \( {V}_{1},\ldots ,{V}_{r} \) be the irreps of \( G \) . Let’s consider the map \( T : \mathbb{C}\left\lbrack G\right\rbrack \rightarrow \mathbb{C}\left\lbrack G\right\rbrack \) which has matr... | Yes |
Example 23.3.4 (Quantum measurement of a state \( |\psi \rangle \) ) Let \( H = {\mathbb{C}}^{\oplus 2} \) with orthonormal basis \( \left| {0\rangle \text{and}}\right| 1\rangle \) and consider the state \[ \left| {\psi \rangle = \frac{i}{\sqrt{5}}}\right| 0\rangle + \frac{2}{\sqrt{5}}|1\rangle = \left\lbrack \begin{ar... | (b) Now consider \( T = \mathrm{{id}} \), and arbitrarily pick two orthonormal eigenvectors \( \left| {0{\rangle }_{T},}\right| 1{\rangle }_{T} \) ; thus \( \psi = {c}_{0}\left| {0{\rangle }_{T} + {c}_{1}}\right| 1{\rangle }_{T} \) . Since all eigenvalues of \( T \) are +1, our measurement will always be +1 no matter w... | Yes |
Example 23.4.1 (Two non-entangled qubits)\n\nSuppose we have qubit \( A \) in the state \( \frac{i}{\sqrt{5}}\left| {0{\rangle }_{A} + \frac{2}{\sqrt{5}}}\right| 1{\rangle }_{A} \) and qubit \( B \) in the state \( \frac{1}{\sqrt{2}}\left| {0{\rangle }_{B} + \frac{1}{\sqrt{2}}}\right| 1{\rangle }_{B} \) . So, the two q... | (We could have used other bases, like \( \left| { \rightarrow {\rangle }_{A} \otimes }\right| 0{\rangle }_{B} \) and \( \left| { \leftarrow {\rangle }_{A} \otimes }\right| 0{\rangle }_{B} \) for the first eigenspace, but it doesn’t matter.) Expanding \( |\psi \rangle \) in the four-element basis, we find that we'll get... | Yes |
Example 23.4.4 (Simultaneously measuring a general 2-Qubit state)\n\nConsider a normalized state \( |\psi \rangle \) in \( H = {\mathbb{C}}^{\oplus 2} \otimes {\mathbb{C}}^{\oplus 2} \), say\n\n\[ \left| {\psi \rangle = \alpha }\right| {00}\rangle + \beta \left| {{01}\rangle + \gamma }\right| {10}\rangle + \delta |{11}... | Thus we get each of the eigenvalues \( 0,1,2,3 \) with probability \( {\left| \alpha \right| }^{2},{\left| \beta \right| }^{2},{\left| \gamma \right| }^{2},{\left| \delta \right| }^{2} \) . So if we like we can make \ | No |
Consider the state \[ \left| {\Psi }_{ - }\right\rangle = \frac{1}{\sqrt{2}}\left| {{01}\rangle - \frac{1}{\sqrt{2}}}\right| {10}\rangle \] which is called the singlet state. One can see that \( \left| {\Psi }_{ - }\right\rangle \) is not a simple tensor, which means that it doesn’t just consist of two qubits side by s... | The eigenspace decomposition of \( T \) can be described as: - The span of \( \left| {{00}\rangle \text{and}}\right| {01}\rangle \), with eigenvalue +1 . - The span of \( \left| {{10}\rangle \text{and}}\right| {11}\rangle \), with eigenvalue -1 . So one of two things will happen: - With probability \( \frac{1}{2} \), w... | Yes |
Theorem 24.1.1 (AND, OR, NOT, COPY are universal)\n\nThe set of four gates AND, OR, NOT, COPY is universal in the sense that any boolean function \( f : \{ 0,1{\} }^{n} \rightarrow \{ 0,1\} \) can be implemented as a circuit using only these gates. | Proof. Somewhat silly: we essentially write down a circuit that OR's across all input strings in \( {f}^{\text{pre }}\left( 1\right) \) . For example, suppose we have \( n = 3 \) and want to simulate the function \( f\left( {abc}\right) \) with \( f\left( {011}\right) = f\left( {110}\right) = 1 \) and 0 otherwise. Then... | Yes |
Theorem 24.2.6 (Toffoli gate is universal)\n\nThe Toffoli gate is universal. | Proof. We will show it can reversibly simulate AND, NOT, hence OR, which we know is enough to show universality. (We don't need COPY because of reversibility.)\n\nFor the AND gate, we draw the circuit\n\n\n\nwith one... | Yes |
Example 25.1.2 (Example of discrete inverse Fourier transform)\n\nLet \( N = 6,\omega = {\omega }_{6} = \exp \left( \frac{2\pi i}{6}\right) \) and suppose \( \left( {{x}_{0},{x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}}\right) = \left( {0,1,0,1,0,1}\right) \) (hence \( {x}_{i} \) is periodic modulo 2). | Thus,\n\n\[ {y}_{0} = \frac{1}{6}\left( {{\omega }^{0} + {\omega }^{0} + {\omega }^{0}}\right) = 1/2 \]\n\n\[ {y}_{1} = \frac{1}{6}\left( {{\omega }^{1} + {\omega }^{3} + {\omega }^{5}}\right) = 0 \]\n\n\[ {y}_{2} = \frac{1}{6}\left( {{\omega }^{2} + {\omega }^{6} + {\omega }^{10}}\right) = 0 \]\n\n\[ {y}_{3} = \frac{1... | Yes |
Proposition 25.2.3 (Tensor representation)\n\nLet \( \\left| {x\\rangle = \\left| {{x}_{n}{x}_{n - 1}\\ldots {x}_{1}}\\right\\rangle \\text{. Then}}\\right| \)\n\n\[ \n{U}_{\\mathrm{{QFT}}}\\left( \\left| {{x}_{n}{x}_{n - 1}\\ldots {x}_{1}}\\right\\rangle \\right) = \\frac{1}{\\sqrt{N}}\\left( {\\left| {0\\rangle + \\e... | Proof. Direct (and quite annoying) computation. In short, expand everything. | No |
Example 25.3.2 (Factoring 77: generating the periodic state)\n\nLet’s say we’re trying to factor \( M = {77} \), and we randomly select \( x = 2 \), and want to find its order \( r \) . Let \( n = {13} \) and \( N = {2}^{13} \), and start by initializing the state\n\n\[ \left| {\psi \rangle = \frac{1}{\sqrt{N}}\mathop{... | In general, the operation is:\n\n- Pick a sufficiently large \( N = {2}^{n} \) (say, \( N \geq {M}^{2} \) ).\n\n- Generate \( \left| {\psi \rangle = \mathop{\sum }\limits_{{k = 0}}^{{{2}^{n} - 1}}}\right| k\rangle \) .\n\n- Build a circuit \( {U}_{x} \) which computes \( \left| {{x}^{k}{\;\operatorname{mod}\;M}}\right\... | Yes |
Example 25.3.3 (Finishing the factoring of \( M = {77} \) ) | As before, we made an observation to the second qubit, and thus the first qubit collapses to the state \( \left| {\phi \rangle = }\right| 7\rangle + |7 + r\rangle + \ldots \) . Now we make a measurement and obtain \( j = {4642} \), which means that for some integer \( s \) we have\n\n\[\n\frac{4642r}{{2}^{13}} \approx ... | Yes |
Consider the sequence defined by\n\n\[ \n{a}_{1} = {1.2} \]\n\n\[ \n{a}_{2} = {1.24} \]\n\n\[ \n{a}_{3} = {1.248} \]\n\n\[ \n{a}_{4} = {1.24816} \]\n\n\[ \n{a}_{5} = {1.2481632} \]\n\n\[ \n\\vdots \]\n\nand so on, where in general we stuck on the decimal representation of the next power of 2. This will converge to some... | In general, \ | No |
Example 26.3.5 \( \left( {{0.9999}\cdots = 1}\right) \) | We simply define a repeating decimal to be the limit of the sequence \( {0.9},{0.99} \) , \( {0.999}\ldots \) And it is obvious that the limit of this sequence is 1. | Yes |
We can now prove the classic telescoping series\n\n\\[ \n\\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }\\frac{1}{k\\left( {k + 1}\\right) } \n\\] | Note that the \( k \) th partial sum is\n\n\\[ \n\\mathop{\\sum }\\limits_{{k = 1}}^{n}\\frac{1}{k\\left( {k + 1}\\right) } = \\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\cdots + \\frac{1}{n\\left( {n + 1}\\right) } \n\\]\n\n\\[ \n= \\left( {\\frac{1}{1} - \\frac{1}{2}}\\right) + \\cdots + \\left( {\\frac{1}{n} - ... | Yes |
Example 26.4.4 (Harmonic series diverges)\n\nWe can also make sense of the statement that \( \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{k} = \infty \) (i.e. it diverges). | We may bound the \( {2}^{n} \) th partial sums from below:\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{{2}^{n}}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{{2}^{n}} \]\n\n\[ \geq \frac{1}{1} + \frac{1}{2} + \left( {\frac{1}{4} + \frac{1}{4}}\right) + \left( {\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{... | Yes |
Proposition 26.4.5 (Partial sums of nonnegatives bounded implies convergent)\n\nLet \( \mathop{\sum }\limits_{k}{a}_{k} \) be a series of nonnegative real numbers. Then \( \mathop{\sum }\limits_{k}{a}_{k} \) converges to some limit if and only if there is a constant \( M \) such that\n\n\[{a}_{1} + \cdots + {a}_{n} < M... | Proof. This is actually just Theorem 26.3.3 in disguise, but since we left the proof as an exercise back then, we'll write it out this time.\n\nObviously if no such \( M \) exists then convergence will not happen, since this means the sequence \( {s}_{n} \) of partial sums is unbounded.\n\nConversely, if such \( M \) e... | Yes |
Proposition 26.5.2 (Absolute convergence \( \Rightarrow \) convergence)\n\nIf a series \( \mathop{\sum }\limits_{k}{a}_{k} \) of real numbers converges absolutely, then it converges in the usual sense. | Exercise 26.5.3 (Great exercise). Prove this by using the Cauchy criteria: show that if the partial sums of \( \mathop{\sum }\limits_{k}\left| {a}_{k}\right| \) are Cauchy, then so are the partial sums of \( \mathop{\sum }\limits_{k}{a}_{k} \). | No |
Theorem 26.5.4 (Permutation of terms okay for absolute convergence)\n\nConsider a series \( \mathop{\sum }\limits_{k}{a}_{k} \) which is absolutely convergent and has limit \( L \) . Then any permutation of the terms will also converge to \( L \) . | Proof. Suppose \( \mathop{\sum }\limits_{k}{a}_{k} \) converges to \( L \), and \( {b}_{n} \) is a rearrangement. Let \( \varepsilon > 0 \) . We will show that the partial sums of \( {b}_{n} \) are eventually within \( \varepsilon \) of \( L \) .\n\nThe hypothesis means that there is a large \( N \) in terms of \( \var... | Yes |
Define the function \( f : \mathbb{R} \rightarrow \mathbb{R} \) as follows:\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} 1 & \text{ if }x = 0 \\ \frac{1}{q} & \text{ if }x = \frac{p}{q}\text{ where }q > 0\text{ and }\gcd \left( {p, q}\right) = 1 \\ 0 & \text{ if }x \notin \mathbb{Q}. \end{array}\right. \] | Then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}f\left( x\right) = 0 \] | No |
For a prime \( p \), show the value of\n\n\[ \n{f}_{p}\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{{p - 1}}\frac{1}{{\left( px + k\right) }^{2}}\;\left( {\;\operatorname{mod}\;{p}^{3}}\right) \n\]\n\ndoes not depend on \( x \) . | However, with \( p \) -adic analysis we’re going to be able to overcome these limitations and give a \ | No |
Proposition 27.2.4 (Non-multiples of \( p \) are all invertible)\n\nThe number \( x \in {\mathbb{Z}}_{p} \) is invertible if and only if \( {x}_{1} \neq 0 \) . In symbols,\n\n\[ x \in {\mathbb{Z}}_{p}^{ \times } \Leftrightarrow x ≢ 0\;\left( {\;\operatorname{mod}\;p}\right) \] | Proof. If \( x \equiv 0\left( {\;\operatorname{mod}\;p}\right) \) then \( {x}_{1} = 0 \), so clearly not invertible. Otherwise, \( {x}_{e} ≢ 0 \) \( \left( {\;\operatorname{mod}\;p}\right) \) for all \( e \), so we can take an inverse \( {y}_{e} \) modulo \( {p}^{e} \), with \( {x}_{e}{y}_{e} \equiv 1\left( {\;\operato... | Yes |
Example 27.2.5 (We have \( - \frac{1}{2} = \ldots {1111}_{3} \in {\mathbb{Z}}_{3} \) ) | We claim the earlier example is actually\n\n\[ \n- \frac{1}{2} = \left( {1{\;\operatorname{mod}\;3},4{\;\operatorname{mod}\;9},{13}{\;\operatorname{mod}\;{27}},{40}{\;\operatorname{mod}\;{81}},\ldots }\right) = 1 + 3 + {3}^{2} + \ldots \n\] \n\n\[ \n= \overline{\ldots }{\overline{1111}}_{3} \n\] \n\nIndeed, multiplying... | Yes |
Proposition 27.3.3 \( \left( {\left| \bullet \right| }_{p}\right. \) is an ultrametric) | For any \( x, y \in {\mathbb{Z}}_{p} \), we have the strong triangle inequality\n\n\[ \n{\left| x + y\right| }_{p} \leq \max \left\{ {{\left| x\right| }_{p},{\left| y\right| }_{p}}\right\} \n\]\n\nEquality holds if (but not only if) \( {\left| x\right| }_{p} \neq {\left| y\right| }_{p} \) . | No |
Proposition 27.3.5 \( \left( {{\left| {x}_{k}\right| }_{p} \rightarrow 0}\right. \) iff convergence of series) | Proof. By multiplying by a large enough power of \( p \), we may assume \( {x}_{k} \in {\mathbb{Z}}_{p} \) . (This isn’t actually necessary, but makes the notation nicer.)\n\nObserve that \( {x}_{k}\left( {\;\operatorname{mod}\;p}\right) \) must eventually stabilize, since for large enough \( n \) we have \( {\left| {x... | Yes |
Proposition 27.3.6 (Geometric series)\n\nLet \( x \in {\mathbb{Z}}_{p} \) with \( {\left| x\right| }_{p} < 1 \) . Then\n\n\[\n\frac{1}{1 - x} = 1 + x + {x}^{2} + {x}^{3} + \ldots\n\] | Proof. Note that the partial sums satisfy \( 1 + x + {x}^{2} + \cdots + {x}^{n} = \frac{1 - {x}^{n}}{1 - x} \), and \( {x}^{n} \rightarrow 0 \) as \( n \rightarrow \infty \) since \( {\left| x\right| }_{p} < 1 \) . | Yes |
Theorem 27.3.8 \( \left( {\mathbb{Q}}_{p}\right. \) is complete) | The space \( {\mathbb{Q}}_{p} \) is the completion of \( \mathbb{Q} \) with respect to \( {\left| \bullet \right| }_{p} \) . | Yes |
The function \( f\left( x\right) = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}\left( \begin{array}{l} x \\ n \end{array}\right) \) is analytic if and only if | \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{n}}{n!} = 0 \] | No |
Theorem 27.4.6 (Skolem-Mahler-Lech)\n\nLet \( {\left( {x}_{i}\right) }_{i \geq 0} \) be an integral linear recurrence, meaning \( {\left( {x}_{i}\right) }_{i \geq 0} \) is a sequence of integers\n\n\[ \n{x}_{n} = {c}_{1}{x}_{n - 1} + {c}_{2}{x}_{n - 2} + \cdots + {c}_{k}{x}_{n - k}\;n = 1,2,\ldots \n\]\n\nholds for som... | Proof. According to the theory of linear recurrences, there exists a matrix \( A \) such that we can write \( {x}_{i} \) as a dot product\n\n\[ \n{x}_{i} = \left\langle {{A}^{i}u, v}\right\rangle \n\]\nLet \( p \) be a prime not dividing \( \det A \) . Let \( T \) be an integer such that \( {A}^{T} \equiv \operatorname... | Yes |
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