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T , and therefore x = 1 det A C11 C21 · · · Cn1 C12 C22 · · · Cn2 ... ... . . . ... C1n C2n · · · Cnn b1 b2 ... bn . Consider the first component x1 of x: x1 = 1 det A (b1C11 + b2C21 + · · · + bnCn1). The expression b1C11 + b2C21 + · · · + bnCn1 is the expansion of the determinant along the first... | {
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the number x. Let A be an invertible matrix and let w1 = Av 1, w2 = Av 2, w3 = Av 3. How are Vol( v1, v2, v2) and Vol( w1, w2, w2) related? Compute: Vol( w1, w2, w3) = abs(det [w1 w2 w3 ])= abs (det [Av 1 Av 2 Av 3 ]) = abs (det( A [v1 v2 v3 ])) = abs (det A · det [v1 v2 v3 ]) = abs(det A) · Vol( v1, v2, v3). Therefore... | {
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interpretation of the determinant (volume) > 108 # Lecture 14 Vector Spaces # 14.1 Vector Spaces When you read/hear the word vector you may immediately think of two points in R2 (or R3) connected by an arrow. Mathematically speaking, a vector is just an element of a vector space . This then begs the question: What is a... | {
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α(u + v) = αu + αv (8) (α + β)v = αv + βv (9) α(βv) = ( αβ )v (10) 1v = v It can be shown that 0 · v = 0 for any vector v in V. To better understand the definition of a vector space, we first consider a few elementary examples. Example 14.2. Let V be the unit disc in R2: V = {(x, y ) ∈ R2 | x2 + y2 ≤ 1} Is V a vector s... | {
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Is V a vector space? Solution. We will show that V is a vector space. First, we verify that V is closed under addition. We first note that an arbitrary point in V can be written as u = ( x, 2x). Let then u = ( a, 2a) and v = ( b, 2b) be points in V. Then u + v = ( a + b, 2a + 2 b) = ( a + b, 2( a + b)) . Therefore V is... | {
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u(t) = u0 + u1t + · · · + untn and let v(t) = v0 + v1t + · · · + vntn be polynomials in V. We define the addition of u and v as the new polynomial ( u + v) as follows: (u + v)( t) = u(t) + v(t) = ( u0 + v0) + ( u1 + v1)t + · · · + ( un + vn)tn. > 111 Vector Spaces Then u + v is a polynomial of degree at most n and thus... | {
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be verified that all other properties of the definition of a vector space also hold. Thus, the set Mm×n is a vector space. Example 14.7. The n-dimensional Euclidean space V = Rn under the usual operations of addition and scalar multiplication is vector space. Example 14.8. Let V = C[a, b ] denote the set of functions w... | {
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︷︷ ︸ > x , 2 ( a + b) ︸ ︷︷ ︸ > x ). Because the x and y components of u + v satisfy y = 2 x then u + v is inside in W. Thus, W is closed under addition. Let α be any scalar and let u = ( a, 2a) be an element of W. Then αu = ( αa, α 2a) = ( αa ︸︷︷︸ > x , 2 ( αa ) ︸︷︷ ︸ > x )Because the x and y components of αu satisfy y... | {
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14.12. Let V = Mn×n be the vector space of all n × n matrices. We define the trace of a matrix A ∈ Mn×n as the sum of its diagonal entries: tr( A) = a11 + a22 + · · · + ann . Let W be the set of all n × n matrices whose trace is zero: W = {A ∈ Mn×n | tr( A) = 0 }. Is W a subspace of V? Solution. If 0 is the n × n zero ... | {
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} In other words, W consists of polynomials of degree n in the variable t whose derivative at t = 1 is zero. Is W a subspace of V? Solution. The zero polynomial 0(t) = 0 clearly has derivative at t = 1 equal to zero, that is, 0′(1) = 0, and thus the zero polynomial is in W. Now suppose that u(t) and v(t) are two polyno... | {
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does not equal −1 at t = 2. Therefore, W does not contain the zero polynomial and, because all three conditions of a subspace must be satisfied for W to be a subspace, then W is not a subspace of Pn[t]. As an exercise, you may want to investigate whether or not W is closed under addition and scalar multiplication. > 11... | {
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span of a set of vectors in V is a subspace of V. Theorem 14.18: If v1, v2, . . . , vp are vectors in V then span {v1, . . . , vp} is a subspace of V. Solution. Let u = t1v1+· · · +tpvp and w = s1v1+· · · +spvp be two vectors in span {v1, v2, . . . , vp}.Then u + w = ( t1v1 + · · · + tpvp) + ( s1v1 + · · · + spvp) = ( ... | {
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• that the span of a set of vectors in V is a subspace of V > 116 Lecture 15 # Lecture 15 Linear Maps Before we begin this Lecture, we review subspaces. Recall that W is a subspace of a vector space V if W is a subset of V and 1. the zero vector 0 in V is also in W,2. for any vectors u, v in W the sum u + v is also in ... | {
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the vector space of n × n matrices and let T : V → V be the mapping T(A) = A + AT . > 117 Linear Maps Is T is a linear mapping? Solution. Let A and B be matrices in V. Then using the properties of the transpose and regrouping we obtain: T(A + B) = ( A + B) + ( A + B)T = A + B + AT + BT = ( A + AT ) + ( B + BT )= T(A) +... | {
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therefore det( A) + det( B) = −1. On the other hand, A + B = [1 10 4 ] and thus det( A + B) = 4. Thus, det( A + B) 6 = det( A) + det( B). Example 15.4. Let V = Pn[t] be the vector space of polynomials in the variable t of degree no more than n ≥ 1. Consider the mapping T : V → V define as T(f (t)) = 2 f (t) + f ′(t). >... | {
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the zero vector, that is, T(v) = 0. We denote the kernel of T by ker( T): ker( T) = {v ∈ V | T(v) = 0}. 2. The range of T is the set of vectors b in the codomain U for which there exists at least one v in V such that T(v) = b. We denote the range of T by Range( T): Range( T) = {b ∈ U | there exists some v ∈ U such that... | {
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0. Therefore, the zero vector 0 is in ker( T). This proves that ker( T) is a subspace of V. The proof that Range( T) is a subspace of U is left as an exercise. Example 15.7. Let V = Mn×n be the vector space of n × n matrices and let T : V → V be the mapping T(A) = A + AT . Describe the kernel of T. Solution. A matrix A... | {
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mapping T : V → V defined as T(f (x)) = f (x) + f ′(x). Solution. A function f is in the kernel of T if T(f (x)) = 0, that is, if f (x) + f ′(x) = 0. Equivalently, if f ′(x) = −f (x). What functions f do you know of satisfy f ′(x) = −f (x)? How about f (x) = e−x? It is clear that f ′(x) = −e−x = −f (x) and thus f (x) =... | {
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= 0}. Hence, the following holds ker( TA) = Null( A). Because the kernel of a linear mapping is a subspace we obtain the following. Theorem 15.10: If A ∈ Mm×n then Null( A) is a subspace of Rn.Hence, by Theorem 15.10 , if u and v are two solutions to the linear system Ax = 0 then αu + βv is also a solution: A(αu + βv) ... | {
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Null( A) = span {v1, v2, . . . , vd}. Hence, the vectors v1, v2, . . . , vn form a spanning set for Null( A). Example 15.12. Find a spanning set for the null space of the matrix A = −3 6 −1 1 −71 −2 2 3 −12 −4 5 8 −4 . Solution. The null space of A is the solution set of the homogeneous system Ax = 0.Performing e... | {
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vn ] and x = ( x1, x 2, . . . , x n) then recall that Ax = x1v1 + x2v2 + · · · + xnvn and thus Ax = x1v1 + x2v2 + · · · + xnvn = b. Thus, a vector b is in the range of A if it can be written as a linear combination of the columns v1, v2, . . . , vn of A. This motivates the following definition. Definition 15.13: Let A ... | {
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should know the following: • what the null space of a matrix is and how to compute it • what the column space of a matrix is and how to determine if a given vector is in the column space • what the range and kernel of a linear mapping is > 124 # Lecture 16 Linear Independence, Bases, and Dimension # 16.1 Linear Indepen... | {
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example, suppose that {v1, v2, v3, v4} are linearly dependent. Then there are scalars c1, c 2, c 3, c 4, not all of them zero, such that equation ( ⋆) holds. Suppose, for the sake of argument, that c3 6 = 0. Then, v3 = −c1 c3 v1 − c2 c3 v2 − c4 c3 v4.Linear Independence, Bases, and Dimension Therefore, when a set of ve... | {
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2A2 = [3 60 −3 ] − [−2 62 0 ] = [ 5 0 −2 −3 ] = A3 Therefore, 3 A1 − 2A2 − A3 = 0 and thus we have found scalars c1, c 2, c 3 not all zero such that c1A1 + c2A2 + c3A3 = 0. # 16.2 Bases We now introduce the important concept of a basis. Given a set of vectors {v1, . . . , vp−1, vp} in V, we showed that W = span {v1, v2... | {
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{v1, . . . , vp} is a basis for W and we remove say vp, then ˜B = {v1, . . . , vp−1} cannot be a basis for W.Why? If B = {v1, . . . , vp} is a basis then it is linearly independent and therefore vp cannot be written as a linear combination of the others. In other words, vp ∈ W is not in the span of ˜B = {v1, . . . , vp... | {
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c 3 such that c1e1 + c2e2 + c3e3 = 0 then clearly they must all be zero, c1 = c2 = c3 = 0. Therefore, by definition, B = {e1, e2, e3} is a basis for R3. This basis is called the standard basis for R3. Analogous arguments hold for {e1, e2, . . . , en} in Rn. Example 16.6. Is B = {v1, v2, v3} a basis for R3? v1 = 20 −... | {
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Find a basis for the vector space of 2 × 2 matrices. Example 16.9. Recall that a n × n is skew-symmetric A if AT = −A. We proved that the set of n × n matrices is a subspace. Find a basis for the set of 3 × 3 skew-symmetric matrices. # 16.3 Dimension of a Vector Space The following theorem will lead to the definition o... | {
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Rn. Thus, in either case ( p > n or p < n ), the set {u1, u2, . . . , up} cannot be a basis for Rn. Hence, any basis in Rn must contain n vectors. The previous theorem does not say that every set {v1, v2, . . . , vn} of nonzero vectors in Rn containing n vectors is automatically a basis for Rn. For example, v1 = 1... | {
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basis of Rn, the set B must be linearly independent and span B = Rn. In fact, it can be shown that if B is linearly independent then the spanning condition span B = Rn is automatically satisfied, and vice-versa. For example, say the vec-tors {v1, v2, . . . , vn} in Rn are linearly independent, and put A = [ v1 v2 · · ·... | {
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basis for W and in this case the dimension of W is k.Since an n-dimensional vector space V requires exactly n vectors in any basis, then if W is a strict subspace of V then dim W 130 Lecture 16 The general solution to Ax = 0 in parametric form is x = t −5 −3/201 + s −6 −5/210 = tv1 + sv2 By constru... | {
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= d = n − rank( A) Example 16.15. Find a basis for Col( A) and the dim Col( A) if A = 1 2 3 −4 81 2 0 2 82 4 −3 10 93 6 0 6 9 . Solution. By definition, the column space of A is the span of the columns of A, which we denote by A = [ v1 v2 v3 v4 v5]. Thus, to find a basis for Col( A), by trial and error we... | {
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2 v1 − 2v3. Thus, because b1, b3, b5 are linearly inde-pendent columns of B =rref (A), then v1, v3, v5 are linearly independent columns of A.Therefore, we have Col( A) = span {v1, v3, v5} = span 1123 , 30 −30 , 8899 and consequently dim Col( A) = 3. This p... | {
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a matrix. Definition 17.2: The nullity of a matrix A is the dimension of its nullspace Null( A). We will use nullity( A) to denote the nullity of A.Recall that ( A) = ker( TA), and thus the nullity of A is the dimension of the kernel of the linear mapping TA.The rank and nullity of a matrix are connected via the follow... | {
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= 2. Example 17.5. Find the rank and nullity of the matrix A = 1 −3 −1 −1 4 2 −1 3 0 . Solution. Row reduce far enough to identify where the leading entries are: A R1+R2,R 1+R3 −−−−−−−−→ 1 −3 −10 1 10 0 −1 There are r = 3 leading entries and therefore rank( A) = 3. The nullity is therefore nullity( A) = 3 −... | {
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as a linear combination of B: x∗ = c1v1 + c2v2 + · · · + cnvn. Moreover, from the definition of linear independence given in Definition 6.1 , any vector x ∈ span( B) can be written in only one way as a linear combination of v1, . . . , vn. In other words, for the x∗ above, there does not exist other scalars t1, . . . ,... | {
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for V and let x ∈ V. The coordinates of x relative to the basis B are the unique scalars c1, c 2, . . . , c n such that x = c1v1 + c2v2 + · · · + cnvn. In vector notation, the B-coordinates of x will be denoted by [x]B = c1 c2 ... cn and we will call [ x]B the coordinate vector of x relative to B.The notati... | {
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].Then the B-coordinates of v is the unique column vector [ v]B solving the linear system Px = v > 138 Lecture 18 that is, x = [ v]B is the unique solution to Px = v. Because v1, v2, . . . , vn are linearly independent, the solution to Px = v is [v]B = P−1v. We remark that if an inconsistent row arises when you row red... | {
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139 Coordinate Systems Therefore, the coordinate vector of v relative to {e1, e2, e3} is [v]E = 311 −7 Example 18.5. Let P3[t] be the vector space of polynomials of degree at most 3. (i) Show that B = {1, t, t 2, t 3} is a basis for P3[t]. (ii) Find the coordinates of v(t) = 3 − t2 − 7t3 relative to B. Solution. ... | {
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0 ] + m22 [0 00 1 ] If c1 [1 00 0 ] + c2 [0 10 0 ] + c3 [0 01 0 ] + c4 [0 00 1 ] = [c1 c2 c3 c4 ] = [0 00 0 ] > 140 Lecture 18 then clearly c1 = c2 = c3 = c4 = 0. Therefore, B is linearly independent, and consequently a basis for M2×2. The coordinates of A = [ 3 0 −4 −1 ] in the basis B = {[ 1 00 0 ] , [0 10 0 ] , [0 0... | {
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[ x]B = (1 , 0, −1). (b) Find the B-coordinates of v = (2 , −1, 0). Solution. The matrix P maps B-coordinates to standard coordinates in R3. Therefore, x = P[x]B = −211 > 141 Coordinate Systems On the other hand, the inverse matrix P−1 maps standard coordinates in R3 to B-coordinates. One can verify that P−1 = ... | {
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∈ V and α ∈ R. Let B = {v1, v2, . . . , vn} be a basis of V and let γ = {w1, w2, . . . , wm} be a basis of W. Then for any v ∈ V there exists scalars c1, c 2, . . . , c n such that v = c1v1 + c2v2 + · · · + cnvn > 142 Lecture 18 and thus [ v]B = ( c1, c 2, . . . , c n) are the coordinates of v in the basis B By lineari... | {
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V = P2[t] of polynomial of degree no more than two and let T : V → V be defined by T(v(t)) = 4 v′(t) − 2v(t)It is straightforward to verify that T is a linear mapping. Let B = {v1, v2, v3} = {t − 1, 3 + 2 t, t 2 + 1 }. (a) Verify that B is a basis of V.(b) Find the coordinates of v(t) = −t2 + 3 t + 1 in the basis B.(c)... | {
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1 and solving yields c1 = 1 , c 2 = 1, and c3 = −1. Hence, [v]B = (1 , 1, −1) (c) The matrix representation A of T is A = [[T(v1)] B [T(v2)] B [T(v3)] B ] Now we compute directly that T(v1) = −2t + 6 , T(v2) = −4t + 2 , T(v3) = −2t2 + 8 t − 2And then one computes that [T(v1)] B = −18 /54/50 , [T(v2)] B = −6/5 ... | {
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PB)[ x]B . If we multiply the equation [x]E = ( E PB)[ x]B on the left by the inverse of E PB we obtain (E PB)−1[x]E = [ x]B Hence, the matrix ( E PB)−1 maps standard coordinates to B-coordinates, see Figure 19.1 . It is natural then to introduce the notation > B PE = ( E PB)−1Change of Basis > b > x > b > [x]B > V=Rn ... | {
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−680 + 3 −63 = −823 ︸ ︷︷ ︸ > x # 19.2 Change of Basis We saw in the previous section that the matrix > E PB takes as input the B-coordinates [ x]B of a vector x and returns the coordinates of x in the standard basis. We now consider the situation of dealing with two basis B and C where neither is assum... | {
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(E PC )−1v2 · · · (E PC)−1vn ] . Therefore, the ith column of ( E PC )−1(E PB), namely (E PC )−1vi, is the coordinate vector of vi in the basis C = {w1, w2, . . . , wn}. To compute C PB we augment E PC and E PB and row reduce fully: [E PC E PB ] ∼ [In C PB ] . Example 19.2. Let B = {[ 1 −3 ] , [−24 ]} , C = {[ −79 ] , ... | {
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[ x]C we can solve the linear system > E PC [x]C = x Alternatively, since we now know [ x]B and C PB has been computed, to find [ x]C we simply multiply C PB by [ x]B:[x]C = C PB[x]B = [ 2 −3/2 −3 5/2 ] [ 21 ] = [ 5/2 −7/2 ] Let’s verify that [ x]C = [ 5/2 −7/2 ] are indeed the C-coordinates of x = [ 0 −2 ] : > E PC [x... | {
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u • v = v • u (b) (u + v) • w = u • w + v • w (c) (αu) • v = α(u • v) = u • (αv) (d) u • u ≥ 0, and u • u = 0 if and only if u = 0 153 Inner Products and Orthogonality Example 20.3. Let u = (2 , −5, −1) and let v = (3 , 2, −3). Compute u • v, v • u, u • u, and v • v. Solution. By definition: u • v = (2)(3) + ( −5)(2) +... | {
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> 154 Lecture 20 Notice that α = 1 > ‖u‖ is just a scalar and thus v is a scalar multiple of u. Then by Theorem 20.5 we have that ‖v‖ = ‖αu‖ = |α| · ‖ u‖ = 1 ‖u‖ · ‖ u‖ = 1 and therefore v is a unit vector, see Figure 20.1 . The process of taking a non-zero vector u and creating the new vector v = 1 > ‖u‖ u is sometime... | {
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v‖ = √(3 − 7) 2 + ( −2 + 9) 2 = √65 . > 155 Inner Products and Orthogonality # 20.2 Orthogonality In the context of vectors in R2 and R3, orthogonality is synonymous with perpendicularity. Below is the general definition. Definition 20.9: Two vectors u and v in Rn are said to be orthogonal if u • v = 0. In R2 and R3, t... | {
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. , up} is said to be an orthogonal set if any pair of distinct vectors ui, uj are orthogonal, that is, ui • uj = 0 whenever i 6 = j.In the following theorem we prove that orthogonal sets are linearly independent. > 156 Lecture 20 Theorem 20.12: Let {u1, u2, . . . , up} be an orthogonal set of non-zero vectors in Rn.Th... | {
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Solution. Compute u1 • u2 = (1)(0) + ( −2)(1) + (1)(2) = 0 u1 • u3 = (1)( −5) + ( −2)( −2) + (1)(1) = 0 u2 • u3 = (0)( −5) + (1)( −2) + (2)(1) = 0 Therefore, {u1, u2, u3} is an orthogonal set. By Theorem 20.12 , the set {u1, u2, u3} is linearly independent. To verify linear independence, we computed that det( [u1 u2 u3... | {
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n). By definition, the coordinates c1, c 2, . . . , c n satisfy the equation x = c1u1 + c2u2 + · · · + cnun. Taking the inner product of u1 with both sides of the above equation and using the fact that u1 • u2 = 0, u1 • u3 = 0, and u1 • un = 0, we obtain u1 • x = c1(u1 • u1) = c1(1) = c1 where we also used the fact tha... | {
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suspect then that U−1 = UT . This is indeed the case. To see this, let B = {u1, u2, . . . , un} be an orthonormal basis for Rn and put U = [ u1 u2 · · · un]. Consider the matrix product UT U, and recalling that ui • uj = uTi uj , we obtain UT U = uT > 1 uT > 2 ... uTn [u1 u2 · · · un ] = uT > 1 u1 ... | {
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v3} consists of three vectors then {v1, v2, v3} is basis for R3.(b) We compute that ‖v1‖ = √2, ‖v2‖ = √18, and ‖v3‖ = 3. Then let u1 = 1/√201/√2 , u2 = −1/√18 4/√18 1/√18 , u3 = 2/31/3 −2/3 Then B = {u1, u2, u3} is now an orthonormal set and thus since B consists of three vectors then B is an orthonor... | {
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w ∈ W it holds that (u1 + u2) • w = u1 • w + u2 • w = 0 + 0 = 0 . Therefore, u1 + u2 is also orthogonal to w and since w is an arbitrary vector in W then (u1 + u2) ∈ W⊥. Lastly, let α be any scalar and let u ∈ W⊥. Then for any vector w in W we have that (αu) • w = α(u • w) = α · 0 = 0 . Therefore, αu is orthogonal to w... | {
"page_id": null,
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any arbitrary vector x ∈ Rn and outputs a new vector Ax . In some cases, the new output vector Ax is simply a scalar multiple of the input vector x, that is, there exists a scalar λ such that Ax = λx.This case is so important that we make the following definition. Definition 21.1: Let A be a n × n matrix and let v be a... | {
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"title": "from dpo"
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find the eigenvalue of A associated to v: A = 2 −1 −1 −1 2 −1 −4 2 2 , v = 111 . Solution. We compute Av = 000 = 0. Hence, if λ = 0 then λv = 0 and thus Av = λv. Therefore, v is an eigenvector of A with corresponding eigenvalue λ = 0. How does one find the eigenvectors/eigenvalues of a matrix A? The g... | {
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Example 21.4. It is known that λ = 4 is an eigenvalue of A = −4 6 31 7 98 −6 1 . Find a basis for the eigenspace of A corresponding to λ = 4. Solution. First compute A − 4I = −4 6 31 7 98 −6 1 − 4 0 00 4 00 0 4 = −8 6 31 3 98 −6 −3 Find a basis for the null space of A − 4I: −8 6 31 3 98 −6 −3... | {
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−5 − 3 0 00 3 00 0 3 = 8 −4 −84 −2 −48 −4 −8 Now find the null space of A − 3I: 8 −4 −84 −2 −48 −4 −8 R1lR2 −−−→ 4 −2 −48 −4 −88 −4 −8 4 −2 −48 −4 −88 −4 −8 > −2R1+R2 > −2R1+R3 −−−−−−→ 4 −2 −40 0 00 0 0 Hence, any vector in the null space of A − 3I = 4 −2 −40 0 00 0 0 can be ... | {
"page_id": null,
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"title": "from dpo"
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. . , λ k} are distinct. Then, one of the eigenvectors vp+1 that is a linear combination of v1, . . . , vp,and {v1, . . . , vp} is linearly independent: vp+1 = c1v1 + c2v2 + · · · + cpvp. (21.1) Applying A to both sides we obtain Av p+1 = c1Av 1 + c2Av 2 + · · · + cpAv p and since Av i = λivi we can simplify this to λp... | {
"page_id": null,
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that (A − λ1I) = span −3 −43 > 167 Eigenvalues and Eigenvectors Hence, v1 = ( −3, −4, 3) is an eigenvector of A with eigenvalue λ1 = 1, and {v1} forms a basis for the corresponding eigenspace. Next, compute A − λ2I = −4 6 31 7 98 −6 1 + 1 0 00 1 00 0 1 = −3 6 31 8 98 −6 2 and one finds tha... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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condition is set to an eigenvector of the system matrix > 168 Lecture 22 # Lecture 22 The Characteristic Polynomial # 22.1 The Characteristic Polynomial of a Matrix Recall that a number λ is an eigenvalue of A ∈ Rn×n if there exists a non-zero vector v such that Av = λv or equivalently if v ∈ Null( A − λI). In other wo... | {
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Theorem 22.2: The characteristic polynomial p(λ) = det( A − λI) of a n × n matrix A is an nth degree polynomial. Solution. Recall that for the case n = 2 we computed that det( A − λI) = λ2 − (a11 + a22 )λ + a11 a22 − a12 a22 . Therefore, the claim holds for n = 2. By induction, suppose that the claims hold for n ≥ 2. I... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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− λ 0 00 λ 00 0 λ = −4 − λ −6 −73 5 − λ 30 0 3 − λ Then det( A − λI) = ( −4 − λ) ∣∣∣∣5 − λ 3 −λ 3 − λ ∣∣∣∣ − 3 ∣∣∣∣−6 −7 −λ 3 − λ ∣∣∣∣ = ( −4 − λ)[(3 − λ)(5 − λ) + 3 λ] − 3[ −6(3 − λ) − 7λ]= λ3 − 4λ2 + λ + 6 Factor the characteristic polynomial: p(λ) = λ3 − 4λ2 + λ + 6 = ( λ − 2)( λ − 3)( λ + 1) Therefor... | {
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is spanned by v3 = −210 and therefore v3 is an eigenvector of A with eigenvalue λ3. Notice that in this case, the 3 × 3matrix A has three distinct eigenvalues and the eigenvectors {v1, v2, v3} = −110 , −101 , −210 correspond to the distinct eigenvalues λ1, λ 2, λ 3, respectively. Therefore... | {
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"title": "from dpo"
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2 2 −2 0 1 Does R3 have a basis of eigenvectors of A? Solution. The characteristic polynomial of A is p(λ) = det( A − λI) = λ3 − 5λ2 + 8 λ − 4 = ( λ − 1)( λ − 2) 2 and therefore the eigenvalues are λ1 = 1 and λ2 = 2. Notice that although p(λ) is a polynomial of degree n = 3, it has only two distinct roots and hence ... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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polynomial. In the previous Example 22.7 , the characteristic polynomial was factored as p(λ) = ( λ − 1)( λ − 2) 2 and we found a basis for R3 of eigenvectors despite the presence of a repeated eigenvalue. In general, if p(λ) is an nth degree polynomial that can be completely factored into linear terms, then p(λ) can b... | {
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"source": 6828,
"title": "from dpo"
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1, respectively. The eigenvalue λ1 = 0 is repeated, while λ2 = 6 and λ3 = −2 are simple eigenvalues. In Example 22.7 , we had p(λ) = ( λ−1)( λ−2) 2 and thus λ1 = 1 has algebraic multiplicity k1 = 1 and λ2 = 2 has algebraic multiplicity k2 = 2. For λ1 = 1, we found one linearly independent eigenvector, and therefore λ1 ... | {
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Therefore, since rank( A − λ2I) = 2, the geometric multiplicity of λ2 = −2 is g2 = 1, which is less than the algebraic multiplicity k2 = 2. An eigenvector corresponding to λ2 = −2 is v2 = −110 Therefore, for the repeated eigenvalue λ2 = −2, we are able to find only one linearly inde-pendent eigenvector. Therefore... | {
"page_id": null,
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"title": "from dpo"
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we would like to begin classifying matrices. How should we decide if matrices A and B are of the same type or, in other words, are similar? Below is how we will decide. Definition 22.12: Let A and B be n × n matrices. We will say that A is similar to B if there exists an invertible matrix P such that A = PBP −1. If A i... | {
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a diagonal matrix. After this lecture you should know the following: • what the characteristic polynomial is and how to compute it • how to compute the eigenvalues of a matrix • that when a matrix A has distinct eigenvalues, we are guaranteed a basis of Rn con-sisting of the eigenvectors of A • that when a matrix A has... | {
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7 0 0 −1 0 0 −4 08 −2 3 0 7 .Diagonalization (a) Find the characteristic polynomial and the eigenvalues of A.(b) Find the geometric and algebraic multiplicity of each eigenvalue of A.We now introduce a very special type of a triangular matrix, namely, a diagonal matrix. Definition 23.3: A matrix D whose off-diag... | {
"page_id": null,
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called diagonalizable if it is similar to a diagonal matrix D. In other words, if there exists an invertible P such that A = PDP −1. How do we determine when a given matrix A is diagonalizable? Let us first determine what conditions need to be met for a matrix A to be diagonalizable. Suppose then that A is diag-onaliza... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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· · vn ] . Then P is invertible because {v1, v2, . . . , vn} are linearly independent. Let D = λ1 0 . . . 00 λ2 . . . 0... ... . . . ...0 0 . . . λn . > 181 Diagonalization Now, since Av i = λivi we have that AP = A [v1 v2 · · · vn ] = [Av 1 Av 2 · · · Av n ] = [λ1v1 λ2v2 · · · λnvn ] . Therefore, AP = [λ1v... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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. . . , λ n. Then A is diagonalizable. Proof. Each eigenvalue λi produces an eigenvector vi. The set of eigenvectors {v1, v2, . . . , vn} are linearly independent because they correspond to distinct eigenvalues (Theorem 21.6 ). Therefore, {v1, v2, . . . , vn} is a basis of Rn consisting of eigenvectors of A and then by... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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find a matrix P that diagonalizes A. A = −4 −6 −73 5 30 0 3 Solution. The characteristic polynomial of A is p(λ) = det( A − λI) = ( λ − 2)( λ − 3)( λ + 1) and therefore λ1 = 2, λ2 = 3, and λ3 = −1 are the eigenvalues of A. Since A has n =3 distinct eigenvalues, then by Theorem 23.6 A is diagonalizable. Eigenvecto... | {
"page_id": null,
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λ1 0 00 λ2 00 0 λ3 P−1 = A Example 23.10. Determine if A is diagonalizable. If yes, find a matrix P that diagonalizes A. A = 2 4 3 −4 −6 −33 3 1 Solution. The characteristic polynomial of A is p(λ) = det( A − λI) = −λ3 − 3λ2 + 4 = −(λ − 1)( λ + 2) 2 and therefore the eigenvalues of A are λ1 = 1 and λ2 = −2.... | {
"page_id": null,
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example, here is a 3 × 3 symmetric matrix: A = 1 −3 7 −3 2 87 8 4 . Symmetric matrices are ubiquitous in mathematics. For example, let f (x1, x 2, . . . , x n) be a function having continuous second order partial derivatives. Then Clairaut’s Theorem from multivariable calculus says that ∂f ∂x i∂x j = ∂f ∂x j ∂x i... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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the roots of which are clearly λ1 = √−1 = i and λ2 = −√−1 = −i. Thus, in general, a matrix whose entries are all real numbers may have complex eigenvalues. However, for symmetric matrices we have the following. Theorem 24.1: If A is a symmetric matrix then all of its eigenvalues are real numbers. The proof is easy but ... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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satisfy this condition and thus not > 188 Lecture 24 all matrices are diagonalizable. As it turns out, any symmetric A is diagonalizable and moreover (and perhaps more importantly) there exists an orthogonal eigenvector matrix P that diagonalizes A. The full statement is below. Theorem 24.3: If A is a symmetric matrix ... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
} |
diagonalizes A we must normalize the eigenvectors u1, u2, u3 to obtain an orthonormal basis {v1, v2, v3}. To that end, first compute uT > 1 u1 = 3, uT > 2 u2 = 2, and uT > 3 u3 = 6. Then let v1 = 1√3 u1, let v2 = 1√2 u2, and let v3 = 1√6 u3. Therefore, an orthogonal matrix that diagonalizes A is P = [v1 v2 v3 ] = ... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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The ranking is based on the hyperlinked or networked structure of the web, and the ranking is based on a popularity contest; if many pages link to page Pi then Pi must be an important page and should therefore have a high popularity score. In January 1998, John Kleinberg from IBM (now a CS professor at Cornell) present... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
} |
the inlink (vote) is important. The vote of each page should be divided by the total number of recommendations made by the page. The PageRank of page i, denoted xi,is the sum of all the weighted PageRanks of all the pages pointing to i: xi = ∑ > j→i xj |Nj | where (1) Nj is the number of outlinks from page j (2) j → i ... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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Hkx0 cycle between (0 , 0, 0, 0.28 , 0.40) and (0 , 0, 0, 0.40 , 0.28). Therefore, the sequence x0, x1, x2, . . . does not converge. The reason for this is that nodes 4 and 5 form a cycle . ## 1 ## 34 52 H = 0 1 > 3 0 0 0 0 0 1 > 2 0 0 0 1 > 3 0 0 0 0 1 > 31 > 2 0 1 0 0 0 1 0 Figure 25.2: Cycl... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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surfer reaches a dangling node, the surfer will “teleport” to any page in the web with equal probability. The new updated hyperlink matrix H∗ may still not have the desired properties. To deal with cycles, a surfer may abandon the hyperlink structure of the web by ocassionally moving to a random page by typing its addr... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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will prove a special case 4. Assume for simplicity that G is positive (this is the case of the Google Matrix). If x = Gx , and x has mixed signs, then |xi| = ∣∣∣∣∣ > n ∑ > j=1 Gij xj ∣∣∣∣∣ n ∑ > j=1 Gij |xj |. Then n∑ > i=1 |xi| n ∑ > i=1 > n ∑ > j=1 Gij |xj | = ∑ > j=1 |xj | which is a contradiction. Therefore, all ... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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= α(H + 1 > n 11 T ) + (1 − α) 1 > n 11 T = αH + ( αa + (1 − α)1) 1 > n 1T and H is very sparse and requires minimal storage. A vector-matrix multiplication generally requires O(n2) computation ( n ≈ 8, 000 , 000 , 000 in 2006). Estimates show that the average webpage has about 10 outlinks, so H has about 10 n non-zero... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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x0 is fixed, the remaining state vectors x1, x2, . . . , can be found by iterating the equation xk+1 = Ax k. # 26.2 Population Model Consider the dynamic system consisting of the population movement between a city and its suburbs. Let x ∈ R2 be the state population vector whose first component is the population of the ... | {
"page_id": null,
"source": 6828,
"title": "from dpo"
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= [0.95 0.03 0.05 0.97 ] [ 0.70 0.30 ] = [0.674 0.326 ] . Then, x2 = Ax 1 = [0.95 0.03 0.05 0.97 ] [ 0.674 0.326 ] = [0.650 0.349 ] . In a similar fashion, one can compute that up to 3 decimal places: x500 = [0.375 0.625 ] , x1000 = [0.375 0.625 ] . It seems as though the population distribution converges to a steady s... | {
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"source": 6828,
"title": "from dpo"
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long term behavior of the discrete dynamical system? We will know the answer once we perform an eigenvalue analysis on A (Lecture 22). As a preview, we will use the fact that xk = Akx0 and then write x0 in an appropriate basis that reveals how A acts on x0. To see how the last equation was obtained, notice that x1 = Ax... | {
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"source": 6828,
"title": "from dpo"
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· · + cnvn)= c1Akv1 + · · · + cnAkvn = c1λk > 1 v1 + · · · + cnλknvn. Since |λi| k→∞ λki = 0. Therefore, lim > k→∞ xk = lim > k→∞ (c1λk > 1 v1 + · · · + cnλknvn)= c1 ( lim > k→∞ λk > 1 ) v1 + · · · + cn ( lim > k→∞ λkn ) vn = 0 v1 + · · · + 0 vn = 0. This completes the proof. > 200 Lecture 26 As an example of an asymp... | {
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"source": 6828,
"title": "from dpo"
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Title: URL Source: Markdown Content: # CSE 311 Quiz Section: December 6, 2012 (Solutions) 1 Determining Countability Determine whether each of these sets is finite, countably infinite, or uncountable. For those that are countably infinite, exhibit a one-to-one correspondence between the set of positive integers and th... | {
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B be uncountable? Answer: Assume A − B is countable. Then, since A = ( A − B) ∪ (A ∩ B), and A ∩ B is countably infinite because B is countable, the elements of A can be listed in a sequence by alternating elements of A − B and the elements of A ∩ B (because they are both countably infinite, then a listing exists for e... | {
"page_id": null,
"source": 6829,
"title": "from dpo"
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continuing, you agree to ourUser Agreement * [Amazing]( * [Animals & Pets]( * [Cringe & Facepalm]( * [Funny]( * [Interesting]( * [Memes]( * [Oddly Satisfying]( * [Reddit Meta]( * [Wholesome & Heartwarming]( * Games * [Action Games]( * [Adventure Games]( * [Esports]( * [Gaming Consoles & Gear]( * [Gaming News & Discussi... | {
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RESOURCES * [About Reddit]( * [Advertise]( * [Reddit Pro BETA]( * [Help]( * [Blog]( * [Careers]( * [Press]( * * * * [Communities]( * [Best of Reddit]( * [Topics]( ![Image 4]( | {
"page_id": null,
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Title: Are there more numbers between 1 and 5 than between 1 and 2? If yes, how? Aren't both infinity? : r/askscience URL Source: Markdown Content: Are there more numbers between 1 and 5 than between 1 and 2? If yes, how? Aren't both infinity? : r/askscience =============== [Skip to main content]( there more numbers b... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
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* New * Controversial * Old * Q&A []( freemath/4). There are indeed an infinite amount of numbers between 1 and 5, but not every infinite set has the same amount of elements: there are, for example, more real numbers between 1 and 2 than there are integers, even though there are an infinite amount of both. The proof of... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
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for the sake of the post...) to 'usual' sets (intervals, boxes and so on), you can also calculate the measure of VERY bizarre sets, like the Cantor set (a kind of infinitely fine patchwork of points between 0 and 1) So to summarize, while (1,5) and (2,5) do share the same cardinality as has been explained, they do not... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
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same size if there exists a bijection between them. In this case, f(x)=4x-3 is a bijection (one-to-one and onto) between [1,2] and [1,5], i.e. any number a in [1,2] corresponds with a number b in [1,5] and vice versa through f, i.e f(a)=b and you cannot replace just a or b in the equation with any other number. Another... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
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