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