MAT224 ASSIGNMENT 3 SOLUTIONS DUE BY FRIDAY JULY 24, 2020, 11:59 PM
Remark: 5 of the following questions will be marked, but you must do all of them. Each question that is marked is out of 5 for a total of 25 marks.
Question 1. Let U = span{(1,0,1,0),(1,1,1,0),(1,1,0,0)} and v = (2,0,−1,3). Find the vector in U closest to v. Solution. Ifa(1,0,1,0)+b(1,1,1,0)+c(1,1,0,0)=(0,0,0,0),then
a+b+c=0, b+c=0, a+c=0⇒a=b=c=0.
Hence, {(1, 0, 1, 0), (1, 1, 1, 0), (1, 1, 0, 0)} is independent, and thus a basis for U . We use the Gram-Schmidt method to
get an orthogonal basis for U; let f1 = (1,0,1,0),
f2 = (1,1,1,0)−(1,1,1,0)·(1,0,1,0)(1,0,1,0)=(0,1,0,0), ∥(1, 0, 1, 0)∥2
f3 = (1,1,0,0)−(1,1,0,0)·(1,0,1,0)(1,0,1,0)−(1,1,0,0)·(0,1,0,0)(0,1,0,0)=1(1,0,−1,0). ∥(1, 0, 1, 0)∥2 ∥(0, 1, 0, 0)∥2 2
So {f1,f2,f3} is an orthogonal basis for U. By Theorem 8.1.3(2),
projU(2,0,−1,3)
= (2, 0, −1, 0) is the vector in U closest to (2, 0, −1, 3).
= (2,0,−1,3)·(1,0,1,0)(1,0,1,0)+ (2,0,−1,3)·(0,1,0,0)(0,1,0,0) ∥(1, 0, 1, 0)∥2 ∥(0, 1, 0, 0)∥2
(2, 0, −1, 3) · 1 (1, 0, −1, 0) 1
+ 2 (1,0,−1,0)
∥1(1,0,−1,0)∥2 2 2
Question 2. If U is a subspace of Rn, show that U⊥⊥ = U.
Solution. Notice that dimU + dimU⊥ = n = dimU⊥⊥ + dimU⊥. So dimU = dimU⊥⊥. Moreover, if x ∈ U, then x·y = 0 for all y ∈ U⊥. This implies that x ∈ U⊥⊥. Hence, U ⊆ U⊥⊥, and since their dimensions are equal, U = U⊥⊥.
Question 3. Let P be an orthogonal matrix.
(a) Prove that detP = 1 or detP = −1.
Solution. Notice that 1 = detI = det(PP−1) = det(PPT) = detPdetPT = (detP)2. Hence, detP = 1 or det P = −1.
(b)IfdetP =−1,showthatI+P hasnoinverse. Hint: PT(I+P)=(I+P)T. Solution.NoticethatPT(I+P)=PTI+PTP=PT +I=PT +IT =(P+I)T =(I+P)T.IfdetP=−1,then
−det(I+P)=detPdet(I+P)=detPT det(I+P)=det(PT(I+P))=det(I+P) ⇒ det(I+P)=0
⇒ I + P is not invertible.
Question 4. Let A be a positive definite matrix. If a ∈ R, prove that aA is positive definite if and only if a > 0. Solution. For any real number x, we have (xA)T = xAT = xA. In particular, xA is symmetric for any real number x. Suppose a > 0, then vT (aA)v = a(vT Av) > 0 for any 0 ̸= v ∈ Rn. Hence, aA is positive definite.
Conversely, suppose aA is positive definite. Pick any non-zero vector v ∈ Rn. Since a(vT Av) = vT (aA)v > 0 and vT Av > 0, we get a > 0.
a b Question 5. (a) Show that a real 2×2 matrix is normal if and only if it is either symmetric or has the from −b a .
ab HT
Solution. Suppose N = c d is normal, where a,b,c,d ∈ R. Notice that N = N (since N is a real matrix).
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2 MAT224 ASSIGNMENT 3 SOLUTIONS DUE BY FRIDAY JULY 24, 2020, 11:59 PM
Moreover,
a ba c T T a ca b a2 +b2 ac+bd
a2 +c2 ab+cd = ab+cd b2+d2 .
c d
In particular,
b d =NN =N N= b d c d ⇒ ac+bd c2+d2
a2+b2 = a2+c2, ac+bd = ab+cd, c2+d2 = b2+d2.
(0.1)
(0.2)
(0.3)
Equations (0.1) and (0.3) both imply that b2 = c2. So b = ±c.
a b a b (a)Supposeb=c.ThenN= c d = b d .Hence,Nissymmetric.
a b a 0 (b)Supposeb=−c.Ifb=0,thenc=0andN= c d = 0 d issymmetric.Ifb̸=0,thenc̸=0,and
a b subbing b = −c into equation (0.2) gives −2ba = −2bd. Dividing by −2b gives a = d. Hence, N = −b a .
a b
Hence, N is either symmetric or has the form −b a . Conversely, if N is a real 2 × 2 matrix and it is symmetric
ab HTTH orhastheform −b a ,aquickcalculationshowsthatineithercaseN N=N N=NN =NN ,thatis,N
is normal.
(b) Let U be a unitary matrix. Prove that |λ| = 1 for every eigenvalue λ of U (where |λ| is the norm of the complex number λ).
Solution. Suppose λ is an eigenvalue of U and x is an eigenvector corresponding to λ. Then Ux = λx. Since U is unitary we have ∥x∥ = ∥Ux∥. In particular, ∥x∥ = ∥Ux∥ = ∥λx∥ = |λ| ∥x∥. Since ∥x∥ ̸= 0 (x is an eigenvector, hence non-zero) we have |λ| = 1.