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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.

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  • $\begingroup$ Your might check out math.stackexchange.com/questions/3137885/…. I left an answer there which I believe addresses your question. $\endgroup$ – Robert Lewis Mar 12 '19 at 16:52
  • $\begingroup$ Also, by "A(transpose)*" do you mean $(A^T)^* = A^\dagger$? Cheers! $\endgroup$ – Robert Lewis Mar 12 '19 at 16:54
  • $\begingroup$ @RobertLewis I think the * is probably supposed to indicate multiplication. $\endgroup$ – saulspatz Mar 12 '19 at 17:00
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    $\begingroup$ @RobertLewis By that I mean multiplying A transpose with A. $\endgroup$ – prism Mar 12 '19 at 17:01
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$$A^tA = \begin{pmatrix} a_1a_1 & a_1a_2 & a_1a_3 & \dots & a_1a_n\\ a_2a_1 & a_2 a_2 & a_2a_3 & \dots & a_2a_n \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ a_na_1 & a_na_2 & a_na_3 & \dots & a_na_n \end{pmatrix}$$

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.

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  • $\begingroup$ Thank you very much; this has been the most helpful! $\endgroup$ – prism Mar 12 '19 at 18:07
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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 $A^{*}Av$?

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

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