Page 361 - 35Linear Algebra

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(b) Yes, the matrix M is symmetric, so it will have a basis of eigenvectors
and is similar to a diagonal matrix of real eigenvalues.

a − λ b
To find the eigenvalues notice that det = (a − λ)(d −
b d − λ
2

2
λ) − b = λ − a+d 2 − b − a−d 2 . So the eigenvalues are
2 2
r r
a + d a − d 2 a + d a − d 2
2
2
λ = + b + and µ = − b + .
2 2 2 2
(c) The trick is to write
T
T
T
T
T
T
X MX +C X +X C = (X +C M −1 )M(X +M −1 C)−C M −1 C ,
so that
T
T
T
(X + C M −1 )M(X + M −1 C) = C MC − f .
T
Hence Y = X + M −1 C and g = C MC − f.
(d) The cosine of the angle between vectors V and W is given by
T
V W V W
√ = √ .
T
V V W W V V W W
T
T
So replacing V → PV and W → PW will always give a factor P P
T
inside all the products, but P P = I for orthogonal matrices. Hence
none of the dot products in the above formula changes, so neither does
the angle between V and W.
(e) If we take the eigenvectors of M, normalize them (i.e. divide them
by their lengths), and put them in a matrix P (as columns) then P
will be an orthogonal matrix. (If it happens that λ = µ, then we
also need to make sure the eigenvectors spanning the two dimensional
eigenspace corresponding to λ are orthogonal.) Then, since M times
the eigenvectors yields just the eigenvectors back again multiplied by
their eigenvalues, it follows that MP = PD where D is the diagonal
matrix made from eigenvalues.
T
T
T
T
T
T
(f) If Y = PZ, then Y MY = Z P MPZ = Z P PDZ = Z DZ

λ 0
where D = .
0 µ
(g) Using part (f) and (c) we have
2
2
λz + µw = g .
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