Question :
If A is a real m × n matrix, show that $AA^T$ is orthogonally diagonalizable.
My answer:
$AA^T$ is m×m real symmetric matrix.
Thus $AA^T$ is diagonalizable.
$AA^T =PDP^-1$
where P =[a b c] and a b c are eigen vectors of $AA^T$. (Also D is a diagonal matrix)
Now I wish to prove that $P^T = P^-1$ to show that $AA^T$ is orthogonally diagonalizable. So I took the product $P^T•P$ . For A to be orthogonally diagonalizable
$P^T•P$ = I where I is the identity matrix.
Now the issue that I am facing is that this will be true only if eigen vectors a b c are orthonormal.
How can I prove that they are orthonormal?
The short answer to your last question is " you can't prove they are orthonormal, you have to make them orthornormal". Disclaimer: What you need is the spectral theorem. The spectral theorem not only states that a symmetric matrix $A$ is diagonalizable but it states also that $A$ admits an orthonormal base of eigenvectors. Therefore, in my answer I will basically go through a part of the proof of the spectral theorem and I will adapt it to your case. Instead of $AA^{T}$ I will consider a symmetric matrix $A$ of size $3\times 3.$
Firstly, there is a well known fact in linear algebra that two eigenvectors $v_1$ and $v_2$ of a symmetric matrix $A$ which are relative to two different eigenvalues $\lambda_{1}$ and $\lambda_2$ must be orthogonal. Let $v_{1}$ and $v_{2}$ be two eigenvectors relative to two eigenvalues $\lambda_{1}$ and $\lambda_{2}$ with $\lambda_{1}\neq\lambda_{2}$: $$ \begin{gather*} \langle \lambda_{1}v_{1},\,v_{2}\rangle=\langle A(v_{1}),\,v_{2} \rangle = \langle v_{1},\,A^{T}(v_{2}) \rangle \\ =\langle v_{1},\,A(v_{2})\rangle = \langle v_{1},\,\lambda_{2}\,v_{2}\rangle \\ \iff \lambda_{1}\cdot \langle v_{1}\,v_{2}\rangle =\lambda_{2}\cdot \langle v_{1}\,v_{2}\rangle \iff \langle v_{1},\,v_{2}\rangle=0. \tag{1} \end{gather*} $$ Now, we have to consider three cases.
Firstly, the case in which the eigenvectors $a,\,b,\,c$ are all relative to different eigenvalues. By $(1)$, we obtain that they are already orthogonal, if you renormalize them you can even make them orthonormal. Secondly, we consider the case in which the eigenvectors $a,\,b,\,c$ are all relative to the same eigenvalue, then you have to use the orthogonalization process of Gram-Schimdt to obtain an orthonormal base of eigenvectors $a',\,b',\,c'$. Finally, there is the case in which one of the eigenvectors, say $a$, is relative to an eigenvalue $\lambda_{1}$ and $b$ and $c$ are relative to an eigenvalue $\lambda_{2}\neq \lambda_1.$ What you need do is to divide $a$ by its norm to obtain $a'$ and you have to apply the Gram-Schimdt algorithm to $b,\,c$ to obtain an orthonormal set of eigenvectors $b',\,c'$ relative to $\lambda_2.$ By $(1)$, $a'$ is orthogonal both to $b'$ and $c',$ so the set $\{a',b',c'\}$ is made of orthonormal vectors of $\mathbb{R}^{3},$ so is a base of $\mathbb{R}^{3}.$ In conclusion, in each of the tree cases you have obtained an orthonormal base of eigenvectors of $A.$ So in your example you should pick $P=[a',\,b',\,c'].$