I have the $n \times n$ matrix $A = (a_{ij})$, where $a_{i,j}= \sqrt{a_{ii}a_{jj}} $ and $a_{i,i} \geq 0$ for $i=1,2,3 \ldots n$. for example $A= \begin{pmatrix} 1 &3 &2\\ 3 &9 &6\\ 2 &6 &4 \end{pmatrix} $. We are also given that the matrix is diagonalizable
I want to prove the following:
- The matrix has only Eigenvalues $0$ and $Trace(A)$.
- The dimension of the Eigenspace for $0$ is $n-1$, or the dimension of Eigenspace for $Trace(A)$ is$ 1$. And hence, there exists a basis where A can be diagonalized and written as $ \begin{pmatrix} 0 &0 &0& \ldots &0\\ 0 &0 &0 &\ldots &0\\ . &. &. &\ldots &.\\ 0 &0 &0 &\ldots &Trace(A) \end{pmatrix} $
I have obtained 1 using minimal polynomials as follows: Observe that $A(A- Trace(A) \cdot I)=0$. This was obtained using the definition of matrix multiplication for $C=A \cdot B$ where $c_{i,j}= \sum_{k=1}^n a_{ik}b_{kj} $.
However for the dimension of the Eigenenspace for $Trace(A)$, using some examples, I have the following observation that the eigenvector is $\begin{pmatrix} \sqrt{a_{11}}\\ \sqrt{a_{22}}\\.\\.\\.\\\sqrt{a_{nn}} \end{pmatrix}$.
Can someone help me with this?