Page 336 - 35Linear Algebra

336 Sample Second Midterm

(d) Swapping two rows also leaves the permanent unchanged. The argu-
ment is almost the same as in the previous part, except that we need
j i
only reshuffle two matrix elements M and M (in the case where
σ(i) σ(j)
rows i and j were swapped). Then we use the fact that summing over
all permutations σ or over all permutations eσ obtained by swapping a
pair in σ are equivalent operations.

5. Firstly, lets call (1) = 1 (the 1 × 1 identity matrix). Then we calculate

T
T
T T
T T
T
T
T T
H = (I −2XX ) = I −2(XX ) = I −2(X ) X = I −2XX = H ,
which demonstrates the first equality. Now we compute
2
T
T
T
T
H = (I − 2XX )(I − 2XX ) = I − 4XX + 4XX XX T
T
T
T
T
T
= I − 4XX + 4X(X X)X = I − 4XX + 4X.1.X = I .
So, since HH = I, we have H −1 = H.
6. We know Mv = λv. Hence
2
2
M v = MMv = Mλv = λMv = λ v ,
and similarly
k
k
M v = λM k−1 v = . . . = λ v .
k
k
So v is an eigenvector of M with eigenvalue λ .
Now let us assume v is an eigenvector of the nilpotent matrix N with eigen-
value λ. Then from above
k
k
N v = λ v
but by nilpotence, we also have
k
N v = 0.
k
k
Hence λ v = 0 and v (being an eigenvector) cannot vanish. Thus λ = 0
and in turn λ = 0.
7. Let us think about the eigenvalue problem Mv = λv. This has solutions
when

3 − λ −5 2
0 = det = λ − 4 ⇒ λ = ±2 .
1 −3 − λ
336