Symmetric matrix properties If $A$ is a symmetric matrix then prove that the elements in the main diagonal of $AA^T$ are not negative.
Can someone help me with this one? I know properties of symmetric matrices but I don't know how to start proving this. Any help would be appreciated. Thank you!
 A: $$e_i^TAA^Te_i = \left\| A^Te_i\right\|^2 \ge 0$$
Note that the left hand side is the $i$-th diagonal entries of the matrix $AA^T$.
A: If your matrix is$$\begin{pmatrix}a_{11}&a_{12}&\ldots&a_{1n}\\a_{21}&a_{22}&\ldots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\a_{n1}&a_{n2}&\ldots&a_{nn}\\\end{pmatrix}\text,$$then the upper left entry of $A.A^t=A^2$ is ${a_{11}}^2+a_{12}a_{21}+a_{13}a_{31}+\cdots+a_{1n}a_{n1}$. But since $A$ is symmetric, this is equal to ${a_{11}}^2+{a_{22}}^2+\cdots+{a_{nn}}^2\geqslant0$. A similar argument applies to the other entries of the main diagonal.
A: Hint: Since $A$ is symmetric, that means $A = A^T$. Then $A A^T = AA$.
Can you write what the diagonal entries of $AA$ look like? That is, can you find the question marks below?
\begin{align*}
AA= 
\begin{bmatrix}
a_{11} & ... & a_{1n} \\
\vdots& \ddots & \vdots \\
a_{n1} & ... & a_{nn}
\end{bmatrix}
\begin{bmatrix}
a_{11} & ... & a_{1n} \\
\vdots& \ddots & \vdots \\
a_{n1} & ... & a_{nn}
\end{bmatrix}
= 
\begin{bmatrix}
? & ... &* \\
\vdots & \ddots & \vdots \\
* & ... & ?
\end{bmatrix}
\end{align*}
