MATH 54 QUIZ 7 SOLUTIONS
MAR 19, 2014
GSI: MINSEON SHIN
(Last edited March 20, 2014 at 3:45pm.)
a −b
3
Problem 1. Find an invertible matrix P and a matrix C of the form
such that A =
b a
2
−1
A = P CP .
−5
has the form
5
Solution. The characteristic polynomial of A is
3 − t −5
det(A − tI) = det
= (3 − t)(5 − t) − (2)(−5) = (t − 4)2 + 32 = (t − (4 + 3i))(t − (4 − 3i))
2
5−t
so the eigenvalues of A are 4 ± 3i. Let’s choose λ = 4 − 3i, so that a = 4 and b = 3. The eigenspace corresponding to the
eigenvalue 4 − 3i is
3i − 1
−5
3i + 1
E4−3i = Nul(A − (4 − 3i)I) = Nul
= Span
2
3i + 1
−2
3i + 1
1
3
1 3
4 −3
so let’s choose v =
, in which case Rev =
and Imv =
. Thus we can set P =
and C =
.
−2
−2
0
−2 0
3 4
3i + 1
−3 + i
3i + 1
−3 1
(If we chose v = i
=
instead of
, then we would get P =
.)
−2
−2i
−2
0 −2
Problem 2. Give an example of a 2 × 2 matrix (with real coefficients) that has no eigenvectors in R2 .
0 1
Solution. Let A =
; since the characteristic polynomial of A is t2 + 1, it has no real eigenvalues, hence no real
−1 0
eigenvectors.
Problem 3. Suppose x is an eigenvector of A corresponding to an eigenvalue λ.
(i) Show that x is an eigenvector of 5I − A. What is the corresponding eigenvalue?
(ii) Show that x is an eigenvector of 5I − 3A + A2 . What is the corresponding eigenvalue?
Solution. Since x is an eigenvector of A corresponding to eigenvalue λ, we have Ax = λx.
(i) We have
(5I − A)x = 5x − Ax = 5x − λx = (5 − λ)x .
Hence x is an eigenvalue of 5I − A corresponding to eigenvalue 5 − λ.
(ii) We have
(5I − 3A + A2 )x = 5x − 3Ax + A2 x = 5x − 3λx + λ2 x = (5 − 3λ + λ2 )x .
Hence x is an eigenvalue of 5I − 3A + A2 corresponding to eigenvalue 5 − 3λ + λ2 .
Problem 4. Give an example of a vector in R4 orthogonal to v = [1, 4, 2, 6]T .
Solution. Possible examples: [−4, 1, 0, 0]T , [0, 1, −2, 0]T , [0, 0, 3, −1]T , or anything else in Nul[1 4 2 6].
Problem 5. Give an example of three vectors u1 , u2 , u3 in R2 such that u1 · u2 = 0 and u2 · u3 = 0 but u1 · u3 6= 0.
     
1
0 
 1
1
0
1
Solution. One possible answer:
,
,
. (For R3 , a valid answer is 1 , −1 , 1 .)
0
0
1


0
1
1
−3
1
Problem 6. Let y =
and u =
. Compute the distance from y to the line through u and the origin.
9
2
Solution. Let ` be the line passing through u and the origin. The projection of y onto `
−3
1
·
9
2
y·u
15
u= u=
u = 3u .
u·u
5
1
1
·
2
2
√
√
Hence the distance from y to ` is the length of the vector y − 3u = [−6, 3]T , which is 62 + 32 = 3 5.
(A common mistake was to compute the distance between (the endpoint of) y and (the endpoint of) u, which is not the
same thing; consider (0, 1) and (1, 0); the line through (1,√0) and the origin is the x-axis so the distance from (0, 1) to this
line is 1, whereas the distance between (0, 1) and (1, 0) is 2.)


a1 a2 a3
Problem 7. Does there exist an orthonormal matrix a4 a5 a6  such that |a1 | + |a2 | + · · · + |a9 | > 3? If yes, give an
a7 a8 a9
example; if no, give a proof explaining why.
√


 √

cos θ − sin θ 0
1/ √2 1/√2 0
cos θ 0 where θ is not of the form n π2 for
Solution. One example is −1/ 2 1/ 2 0, or more generally  sin θ
0
0
1
0
0
1
any integer n. This is because if θ 6= n π2 for any integer n, then both cos θ and sin θ are nonzero, so (| cos θ| + | sin θ|)2 =
(cos θ)2 + (sin θ)2 + 2| cos θ|| sin θ| = 1 + 2| cos θ|| sin θ| > 1, so that | cos θ| + | sin θ| > 1.