Linear Algebra II (pdf, 500 kB)

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10.8. Theorem. Let V be a finite-dimensioanl real inner product space, and let
f : V → V be linear. Then f has an orthonormal basis of eigenvectors if and only
if f is self-adjoint.
Proof. If f has an ONB of eigenvectors, then its matrix with respect to this basis
is diagonal and so symmetric, hence f is self-adjoint.
On the other hand, if f is self-adjoint, we know that it is diagonalizable over C
with all eigenvalues real, so it is diagonalizable (over R). Lemma 10.2, (3) then
shows that the eigenspaces are orthogonal in pairs.
In terms of matrices, this reads as follows.
10.9. Theorem. Let A be a square matrix with real entries. Then A is orthogonally
similar to a diagonal matrix (i.e., there is an orthogonal matrix P : P P ⊤ = I,
such that P −1 AP is a diagonal matrix) if and only if A is symmetric. In this case,
we can choose P to be orientation-preserving, i.e., to have det P = 1 (and not −1).
Proof. The first statement follows from the previous theorem. To see that we can
take P with det P = 1, assume that we already have an orthogonal matrix Q such
that Q −1 AQ = D is diagonal, but with det Q = −1. The diagonal matrix T with
diagional entries (−1, 1, . . . , 1) is orthogonal and det T = −1, so P = QT is also
orthogonal, and det P = 1. Furthermore,
P −1 AP = T −1 Q −1 AQT = T DT = D ,
so P has the required properties.
This statement has a geometric interpretation. If A is a symmetric 2 × 2-matrix,
then the equation
(1) x ⊤ Ax = 1
defines a conic section in the plane. Our theorem implies that there is a rotation
P such that P −1 AP is diagonal. This means that in a suitably rotated coordinate
system, our conic section has an equation of the form
a x 2 + b y 2 = 1 ,
where a and b are the eigenvalues of A. We can use their signs to classify the
geometric shape of the conic section (ellipse, hyperbola, empty, degenerate).
The directions given by the eigenvectors of A are called the principal axes of the
conic section (or of A), and the coordinate change given by P is called the principal
axes transformation. Similar statements are true for higher-dimensional quadrics
given by equation (1) when A is a larger symmetric matrix.
10.10. Example. Let us consider the conic section given by the equation
The matrix is
5 x 2 + 4 xy + 2 y 2 = 1 .
A =
5 2
.
2 2
We have to find its eigenvalues and eigenvectors. The characteristic polynomial
is (X − 5)(X − 2) − 4 = X 2 − 7 X + 6 = (X − 1)(X − 6), so we have the two