Page 327 - 35Linear Algebra

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T
T
Since M M −1 6= I, it follows M 6= M so M is not symmetric. Finally

2 1 2 1
2
T
trf(M) = trf(M) = tr(M − I) = tr − trI
3 −1 3 −1
= (2 · 2 + 1 · 3) + (3 · 1 + (−1) · (−1)) − 2 = 9 .
5. First

cos θ sin θ x
T

X (MX) = X MX = x y
− sin θ cos θ y

x cos θ + y sin θ
2 2
= x y = (x + y ) cos θ .
−x sin θ + y cos θ
√ p
T
2
2
Now ||X|| = X X = x + y and (MX) (MX) = XM MX. But

cos θ − sin θ cos θ sin θ
T
M M =
sin θ cos θ − sin θ cos θ
2 2
cos θ + sin θ 0
= 2 2 = I .
0 cos θ + sin θ
p
2
2
Hence ||MX|| = ||X|| = x + y . Thus the cosine of the angle between X
and MX is given by
2
2
X (MX) (x + y ) cos θ
= cos θ .
= p p
||X|| ||MX|| x + y 2 x + y 2
2
2
In other words, the angle is θ OR −θ. You should draw two pictures, one
where the angle between X and MX is θ, the other where it is −θ.
|X (MX)|
For Cauchy–Schwartz, = | cos θ| = 1 when θ = 0, π. For the
||X|| ||MX||
triangle equality MX = X achieves ||X + MX|| = ||X|| + ||MX||, which
requires θ = 0.
6. This is a block matrix problem. Notice the that matrix M is really just

I I
M = , where I and 0 are the 3×3 identity zero matrices, respectively.
0 I
But

I I I I I 2I
2
M = =
0 I 0 I 0 I
and

I I I 2I I 3I
3
M = =
0 I 0 I 0 I
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