Mathematics Some Questions and Answers

1. Suppose u and v are linearly independent vectors in R2 . Let A be the matrix with u and v as columns, i.e. A = (u v). Prove that det A 6= 0. Solution: By definition of linear independence, u and v can only satisfy the equation αu + βv = 0, (1) if and only if α = β = 0. Rewriting equation (1), we have:       u1 v 0 α +β 1 = u2 v2 0      u1 v1 α 0 =⇒ = . u2 v2 β 0  The only way to get a unique solution of α = β = 0 is if the coefficient matrix A = That is, the vectors u and v are linearly independent provided det A 6= 0. 2. Let A be a 2 × 2 matrix and λ be an eigenvalue of A. The λ-eigenspace of A is  Eλ (A) := v ∈ R2 : Av = λv ∪ {0} . Prove that Eλ (A) is a vector subspace of R2 . That is, show that (i) 0 is an element of Eλ (A) (ii) if u, v ∈ Eλ (A), then u + v ∈ Eλ (A) (iii) if u ∈ Eλ (A) and k ∈ R, then ku ∈ Eλ (A) Solution: (i) By definition, 0 ∈ Eλ (A) (ii) Let u, v ∈ Eλ (A). Then A(u + v) = Au + Av = λu + λv = λ(u + v). Thus, u + v ∈ Eλ (A). (iii) Let u ∈ Eλ (A) and k ∈ R. Then A(ku) = kAu = kλu = λ(ku). Thus, ku ∈ Eλ (A) u1 v1 u2 v2  is invertible. 3. Consider the standard rotation matrix   cos θ − sin θ R := , sin θ cos θ for 0 < θ < π. Show that R has no real eigenvalues. Can you explain this geometrically? Solution: The eigenvalues of R are roots of the characteristic polynomial p(λ) = det(R − λI). p(λ) = det(R − λI) = (cos θ − λ)2 + sin2 θ. So the eigenvalues are λ = cos θ ± i sin θ. In particular, if 0 < θ < π, then sin θ 6= 0 and the eigenvalues are complex. Geometrically, this means that R with 0 < θ < π applied to any vector always changes the direction of the vector. Note: we get real eigenvalues of ±1 when θ = 0, π. That is, R does nothing to a vector when θ = 0 and R reverses the direction of a vector when θ = π. 4. Find the eigenvalues and eigenspaces of the matrix   −2 1 M := . 3 2 Solution: Eigenvalues of M are roots of the characteristic polynomial p(λ) = det(M − λI) = (−2 − λ)(2 − λ) − 3 = λ2 − 7 √ so the eigenvalues are λ = ± 7. √ Eigenvector for λ = 7:  (M − λI)v = 0 =⇒ −2 − 3 √ 7 1√ 0 2− 7 0  . By definition of eigenvalue-eigenvector pairs, the matrix M − λI is not invertible. So the rows of M − λI are linearly dependent (i.e. scalar multiples of each other). This means row reduction of M − λI will lead to a row of zeros. So we can simply choose to look at one of the rows and safely ignore the other. For matrices of size 3 × 3 or larger, you cannot simply ignore the other rows. The first row tells us √ √ (−2 − 7)v1 + v2 = 0 =⇒ v2 = (2 + 7)v1 . Thus, the λ-eigenspace is  Eλ (M ) =  v1 1√ 2+ 7   : v1 ∈ R . √ Eigenvector for λ = − 7:  (M − λI)v = 0 =⇒ −2 + 3 2 √ 7 1√ 0 2+ 7 0  . Again, by definition of eigenvalue-eigenvector pairs, M − λI is not invertible so that the rows of M − λI are linearly dependent and we can safely ignore one of the rows (because it will row reduce to a row of zeros). The first row gives √ √ (−2 + 7)v1 + v2 = 0 =⇒ v2 = (2 − 7)v1 . Thus, the λ-eigenspace is   Eλ (M ) = v1   1√ : v1 ∈ R 2− 7 The eigenspaces are shown in Figure 1 below. 5 -2 1 -1 2 -5 Figure 1: Blue: λ-eigenspace for λ = √ √ 7. Red: λ-eigenspace for λ = − 7. 5. Let A be a 2 × 2 symmetric matrix. That is, A is a matrix of the form   a b . A := b d Let u and v be eigenvectors associated with eigenvalues λ and µ, respectively. Prove that if λ 6= µ, then u and v are orthogonal. That is, prove that u · v = 0, where · denotes the dot product. Solution: The characteristic equation of A is given by (λ − a)(λ − d) − b2 = 0. That is, the eigenvalues λ and µ satisfy (λ − a)(λ − d) − b2 = 0, 2 (µ − a)(µ − d) − b = 0.  p Note that λ, µ = 21 a + d ± (a + d)2 − 4(ad − b2 ) . Eigenvector for λ:   a−λ b 0 (A − λI)u = 0 =⇒ . b d−λ 0 (2) (3)  By definition of eigenvalue-eigenvector pairs, A − λI is not invertible so the rows of A − λI are scalar multiples of each other. Thus, we can compute an eigenvector by just looking at row 1: (a − λ)u1 + bu2 = 0 =⇒ bu2 = (λ − a)u1 . 3   b . Similarly, an eigenvector of A for the λ−a Thus, an eigenvector of A for the eigenvalue λ is u =   b eigenvalue µ is v = . µ−a Now, we compute the dot product:     b b u·v = · = b2 + (λ − a)(µ − a). λ−a µ−a Since µ satisfies the characteristc equation, we can use equation (3) (or equation (2) – it doesn’t matter) to get u · v = (µ − a)(µ − d) + (λ − a)(µ − a) = (µ − a) (µ − d + λ − a) , = (µ − a)(λ + µ − (a + d)) =0 since λ + µ = a + d. 6. Let A be a 2 × 2 matrix and let λ1 , λ2 be the eigenvalues of A, with λ1 6= λ2 . Let v 1 and v 2 be the eigenvectors corresponding to the eigenvalues λ1 and λ2 , respectively. Show that v 1 and v 2 are linearly independent. Solution: Suppose the eigenvectors v 1 and v 2 satisfy αv 1 + βv 2 = 0. (4) We must show that α = β = 0. First, note that since v 1 and v 2 are eigenvectors, they must be non-zero. Moreover, we have (A − λ1 I)v 1 = 0, (A − λ2 I)v 2 = 0. Multiplying both sides of equation (4) by A − λ1 I gives: α(A − λ1 I)v 1 + β(A − λ1 I)v 2 = 0 α0 + β(Av 2 − λ1 v 2 ) = 0 β(λ2 − λ1 )v 2 = 0. Since λ1 6= λ2 and v 2 6= 0, we must have β = 0. Similarly, multiplying equation (4) by A − λ2 I leads to α = 0. Thus, v 1 and v 2 are linearly independent. 7. Suppose that v is an eigenvector for the n × n matrix A with eigenvalue λ. (i) Show that v is also an eigenvector for A2 and determine the corresponding eigenvalue. (ii) Suppose A is invertible. Show that v is an eigenvector for A−1 and determine the corresponding eigenvalue. Solution: v is an eigenvector for A for the eigenvalue λ implies Av = λv. (i) A2 v = A(Av) = A(λv) = λAv = λ2 v, which shows that v is an eigenvector for A2 corresponding to an eigenvalue of λ2 . 4 (ii) Av = λv =⇒ v = A−1 λv =⇒ A−1 v = λ−1 v Thus, v is an eigenvector for A−1 corresponding to the eigenvalue λ1 . 8. (The Cayley-Hamilton Theorem)  Show that the characteristic equation of the square matrix A =  a b is c d λ2 − (a + d)λ + (ad − bc) = 0. Show (by direct substitution) that A satisfies the matrix equation A2 − (a + d)A + (ad − bc)I2 = 02 , where I2 and 02 are the 2 × 2 identity and zero matrices, respectively. Hence, use (5) to show that   1 d −b −1 A = . det A −c a Solution: The characteristic polynomial is p(λ) = det(A − λI) = (a − λ)(d − λ) − bc = λ2 − (a + d)λ + ad − bc So the characteristic equation of A is p(λ) = 0 =⇒ λ2 − (a + d)λ + ad − bc = 0.   a b into the characteristic polynomial gives Substituting A = c d p(A) = A2 − (a + d)A + (ad − bc)I2    2   1 0 a b a b + (ad − bc) − (a + d) = 0 1 c d c d  2   2    a + bc ab + bd a + ad ab + bd ad − bc 0 = − + ac + dc bc + d2 ac + cd ad + d2 0 ad − bc   0 0 = 0 0 Thus, we have A2 − (a + d)A + (ad − bc)I2 = 02 A2 − (a + d)A = −(ad − bc)I2 A (A − (a + d)I2 ) = −(ad − bc)I2 Note that det A = ad − bc:  A − 1 det A A (A − (a + d)I2 ) = −(det A)I2  (A − (a + d)I2 ) = I2 Thus, it follows that A −1 1 −1 =− (A − (a + d)I2 ) = det A det A       1 a b a+d 0 d −b − = c d 0 a+d det A −c a 5 (5) 9. (i) Let X and Y be 2 × 2 matrices. Show that det(XY ) = (det X)(det Y ) (ii) Suppose now that X is invertible. Using part (i), show that det(X −1 ) = 1 det X (iii) Now, let A and P be 2 × 2 matrices, with P invertible. Show that A and P AP −1 have the same characteristic equation (Hint: the identity matrix I can be rewritten as I = P P −1 ). Implication: You have just proved that A and P AP −1 have the same eigenvalues. Solution:  (i) Let X = x11 x12 x21 x22    y11 y12 and Y = . The product is y21 y22   x11 y11 + x12 y21 x11 y12 + x12 y22 XY = x21 y11 + x22 y21 x21 y12 + x22 y22 Thus, the determinant is det(XY ) = (x11 y11 + x12 y21 ) (x21 y12 + x22 y22 ) − (x11 y12 + x12 y22 ) (x21 y11 + x22 y21 ) = x11 x21 y11 y12 + x11 x22 y11 y22 + x12 x21 y21 y12 + x12 x22 y21 y22 − x11 x21 y11 y12 − x11 x22 y12 y21 − x12 x21 y11 y22 − x12 x22 y21 y22 = x11 x22 (y11 y22 − y12 y21 ) + x12 x21 (y12 y21 − y11 y22 ) = (x11 x22 − x12 x21 )(y11 y22 − y12 y21 ) = (det X)(det Y ) (ii) If X is invertible, then we have XX −1 = I2 . Taking the determinant of both sides, we have det(XX −1 ) = det I2 = 1. By part (i) above, we have det(XX −1 ) = (det X)(det X −1 ) = 1 =⇒ det(X −1 ) = 1 . det X (iii) Characteristic equation of P AP −1 is

det P AP −1 − λI = det P AP −1 − λP P −1
= det P (A − λI) P −1 = (det P ) (det(A − λI)) (det(P −1 )) 1 = (det P ) (det(A − λI)) det P Recall that the determinant of a matrix is just a scalar (so we can reorder the determinant by commutativity of scalar multipilcation). Thus, we have
1 (det(A − λI)) det P AP −1 − λI = (det P ) det P = det (A − λI) That is, the characteristic polynomials of P AP −1 and A are identical. Consequently, P AP −1 and A have the same eigenvalues. 6