The Spectral Theorem

MAT 224 Linear Algebra | Tutorial 7 | 04.11.13 page 1

Nikita Nikolaev

Problem 7.1Let: A ∈ Matn×n(C), and let λ1, . . . , λn be its eigenvalues.

Then:

det(A) = λ1 · . . . · λn and trace(A) = λ1 + . . .+ λn.

Proof.The eigenvalues λ1, . . . , λn are roots of the characteristic polynomial of A, which is pA(λ) = det(A−λI).This polynomial splits over C as follows:

(λ1 − λ)(λ2 − λ) · . . . · (λn − λ) = 0.

Thus, at λ = 0, its value is pA(0) = det(A), but also pA(0) = λ1 · . . . · λn.

I’ll only give the idea of how to establish the second formula. Expanding the characteristic polynomialpA(λ), one finds that the coefficient of λn−1 is precisely λ1 + . . .+λn. Similarly, expanding det(A−λI),one also finds that the coefficient of λn−1 is the trace of A. Equating the coefficients, we have the desiredformula. Q.E.D.

Problem 7.2Let: A ∈ Matn×n(C) be a Hermitian matrix.The function 〈�,�〉 : Cn × Cn → C, defined by

〈x, y〉 := x∗Ay,

is an inner product on Cn if and only if all eigenvalues of A are positive.

Proof.The linearity of 〈�,�〉 is straightforward (in fact, without any assumption on A). The function 〈�,�〉Ais conjugate-symmetric if and only if A is Hermitian (i.e., A∗ = A): indeed, for all x, y ∈ Cn,

〈y, x〉 = y∗Ax = y∗A∗x = (x∗Ay)∗ = 〈x, y〉∗ = 〈x, y〉.

To check that 〈�,�〉 is positive-definite, for any x = (xi)ni=1 ∈ Cn,

〈x, x〉 = x∗Ax = λ1|x1|2 + . . . λn|xn|2. (?)

Thus, 〈x, x〉 > 0 for all x if and only if λi > 0 for all i. Q.E.D.

MAT 224 Linear Algebra | Tutorial 7 | 04.11.13 page 2

Note: have another look at (?). If all eigenvalues of A are 1, then 〈�,�〉 defined this way is thestandard inner product on Cn. Thus, defining an inner product using a Hermitian matrix A is likedefining a weighted standard inner product with weighting given by the eigenvalues of A.

RemarkLet A be a square matrix, and let A∗ be its adjoint.

The matrix A + A∗ is Hermitian, because (A + A∗)∗ = A∗ + A∗∗ = A∗ + A. Similarly, the matrixA−A∗ satisfies the identity (A−A∗)∗ = −(A−A∗); a matrix S with the property that S∗ = −S∗

is called skew-Hermitian.

Every square matrix A can be decomposed into the sum of a Hermitian and a skew-Hermitianmatrices as follows:

A =1

2(A+ A∗) +

1

2(A− A∗).

Problem 7.3Let: (V, 〈�,�〉) be a finite-dimensional inner product space, and let T, S : V → V be two linearmaps.

Suppose: for all x, y ∈ V, 〈T (x), y〉 = S(x), y.

Then: T = S.

Proof.Let {v1, . . . , vn} be a basis for V. Fix i; since for all j, 〈T (vi), vj〉 = 〈S(vi), vj〉, it follows that T (vi) =

S(vi). Since this is true for all i, the maps T and S agree on a basis of V, hence T = S. Q.E.D.