A question about a proof that has to do with diagonal matrices 
Show that if $A$ is a diagonal matrix then orthogonal diagonalising matrix $Q = \text {Identity}.$
Proof: Let $A$ be a diagonal matrix and if $Q = I,$ then $Q^{-1}AQ = I^{-1}AI = IAI = A.$ Therefore the identity matrix $I$ orthogonally diagonalizes $A$. Thus our orthogonal matrix $Q = I$.

I don't get the proof here. It looks like they have proved any matrix is orthogonally similar to itself. Can someone elaborate? Maybe I am missing something important?
 A: What is the reference here?  I do not believe that statement is true for the simple reason that it is incorrect to refer to the orthogonal diagonalizing matrix $Q$.  Take this simple example: $A = 
\begin{bmatrix} 
1 & 0 \\ 
0 & 2 
\end{bmatrix} $
So $A$ is diagonal, and of course conjugating $A$ by the identity leaves it diagonal, as demonstrated in the equations you've listed in the proof.
But here's another orthogonal diagonalizing matrix: $Q = 
\begin{bmatrix} 
0 & 1 \\ 
1 & 0 
\end{bmatrix}.$  Check for yourself that $Q$ is orthogonal and that $QAQ^{-1} = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$.  In a sense, $Q$ is just the linear transformations that re-indexes the basis vectors, calling $e_1' = e_2$ and $e_2' = e_1$.  So of course it leaves $A$ diagonal, but it's certainly not the identity.
So here's the point: when a matrix $A$ is diagonalizable, that diagonalization is not unique.  What is unique is what entries appear along the diagonal: these are the eigenvalues of $A$ and they are intrinsic to the linear operator $A$ represents.  No matter what, when you diagonalize $A$ those eigenvalues will appear on the diagonal.  But what order they appear in is completely arbitrary because you can always re-index your basis and make them appear in a different order (as I have done above).
So your reference is incorrect in referring to the orthogonal diagonalizing matrix $Q$.  When $A$ is diagonal, the identity matrix is a diagonalizing orthogonal matrix, but others exist as well.
A: A diagonalizable matrix $A$ is a matrix such that for invertible matrix $P$ and $D$ diagonal you have $A=PDP^{-1}$. Now for $A$ diagonal you can take $D=A$ and $P=I$ and this proves that $A$ is diagonalizable (as trivially expected).
Moreover orthogonally diagonalizable is when the matrix $P$ is orthogonal. For diagonal matrices $A$ the matrix of change of basis $P$ is the identity since $A$ is already in diagonal form. So it says that a diagonal matrix is orthogonally diagonalizable since $Q=I$ satisfies $A=QDQ^{-1}$ with $D=A$ diagonal and $Q$ orthogonal.
