# Symmetric matrix properties [closed]

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!

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• This question confuses me a little bit because in fact this does not require the symmetry of $A$ itself. – Ian Jun 27 '17 at 20:46
• Do you mean $A^*A^T$ (i.e. $A^*$ is the conjugate transpose) or do you mean $AA^T$? The asterisk has a meaning in linear algebra, and if you are just multiplying, leave it out. – Doug M Jun 27 '17 at 20:47
• #Ian can you prove that ? – Maths Survivor Jun 27 '17 at 20:56
• @Linda Sionge Thye Goh already did. The only thing left to show is that in general $e_i^T B e_i = B_{ii}$, but that is just a simple calculation. – Ian Jun 28 '17 at 9:41

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.
$$e_i^TAA^Te_i = \left\| A^Te_i\right\|^2 \ge 0$$
Note that the left hand side is the $i$-th diagonal entries of the matrix $AA^T$.
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*}