Eigenvalues of a $A^T A$

Given the matrix of order $$1\times{n}$$, $$A=(a_1, a_2, ..., a_n)$$ , where $$a_i$$ are real; The question is to find all eigenvalues of $$A^T A$$.

I have proved that it is a non-invertible matrix, therefore $$0$$ is one of the values. And the product matrix is an $$nxn$$ matrix with the diagonal elements being $$a_1^2, a_2^2,...,a_n^2$$. I am struggling with finding the other eigenvalues, tried by calculating the det of $$A - aI$$, but didn't go anywhere.

• Your might check out math.stackexchange.com/questions/3137885/…. I left an answer there which I believe addresses your question. – Robert Lewis Mar 12 '19 at 16:52
• Also, by "A(transpose)*" do you mean $(A^T)^* = A^\dagger$? Cheers! – Robert Lewis Mar 12 '19 at 16:54
• @RobertLewis I think the * is probably supposed to indicate multiplication. – saulspatz Mar 12 '19 at 17:00
• @RobertLewis By that I mean multiplying A transpose with A. – prism Mar 12 '19 at 17:01

$$A^tA = \begin{pmatrix} a_1a_1 & a_1a_2 & a_1a_3 & \dots & a_1a_n\\ a_2a_1 & a_2 a_2 & a_2a_3 & \dots & a_2a_n \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ a_na_1 & a_na_2 & a_na_3 & \dots & a_na_n \end{pmatrix}$$

All the columns can be obtained by mutiplying by a scalar the first column, so $$rank(A^tA)=1$$. So $$0$$ is an eigen value with multiplicity $$n-1$$. The sum of eigen values is equal to the trace of the matrix, thus you can easily find the last eigenvalue.

• Thank you very much; this has been the most helpful! – prism Mar 12 '19 at 18:07

Hint: Suppose that $$v$$ is an $$n \times 1$$ vector orthogonal to $$A$$ (or more precisely, $$w$$ is a $$1 \times n$$ vector orthogonal to $$A$$, and $$v = w^t$$). What do you get when you compute $$A^{*}Av$$?

• Thanks for your time! I managed to solve the problem :) – prism Mar 12 '19 at 18:08

Consider the definition of an eigenvalue / eigenvector pair, $$A^T A v = \lambda v.$$ Now extend $$A$$ to an orthogonal basis: so let $$A_1, A_2, \ldots, A_n$$ be an orthogonal basis for $$\mathbb{R}^n$$, where $$A_1 = A^T$$. Write $$v$$ in this basis: $$v = a_1 A^T + a_2 A_2 + a_3 A_3 + \cdots + a_n A_n.$$

Then, $$A^T Av = a_1 A^T A A^T + a_2 A^T A A_2 + a_3 A^T A A_3 + \cdots + a_n A^T A A_n.$$

Because the basis is orthogonal, $$A A_k = 0$$ for any $$k \ge 2$$. Also $$A A^T$$ = $$\|A\|^2$$. So this just reduces to $$A^T A v = a_1 A^T \|A\|^2.$$ The only way this can be equal to $$\lambda v$$ is if $$\lambda = 0$$ and the component $$a_1 = 0$$, or if $$\lambda = \|A\|^2$$ and the components $$a_2, a_3, \ldots, a_n$$ are all $$0$$.

So the only eigenvalues are $$0$$ and $$\|A\|^2$$, and we get the corresponding eigenvectors as well.