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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?

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    As I read it, they're basically saying that a diagonal matrix is diagonal, or something. Strange problem. – 2017-02-11

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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.

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    @ Test123, did you mean $A = P^{-1}DP$? – 2017-02-11
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    @123456 $A=PDP^{-1}$ implies $P^{-1}AP=D=A$ – 2017-02-11
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    @123456 No doubt: Test123 has written the digonalizing identity in the right way. – 2017-02-11
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    It's impolite to ask, but just in case y'all happen to have time, can y'all please take a look at my other question? I am having a similar problem. Thanks. – 2017-02-11
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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.