Ways to diagonalize a diagonal matrix

Question

Matrix $\mathbf{D}$ is a full rank diagonal matrix. $\mathbf{ADA}^T=\mathbf{D}$.

Can we conclude that $\mathbf{A}^2=\mathbf{I}_n$? (edited from $\mathbf{A}=\pm\mathbf{I}$)

It's almost sure that some one has asked this question before, but I cannot find proper keywords to search for that post.

Answer 1

No. Consider $$ A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\\ D = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} $$

Often looking at the $2 \times 2$ case for problems like this will help you get a good idea of what's going on in general.

NB: I'm assuming that this wasn't a HW problem, because I'd hate to think that somewhere out there is a prof who writes things this badly.

NB 2: This answer the question whether you meant $A^t$ or $A^{-1}$, as they are the same in this case.

Answer 2

I don't know if this answer addresses your question, but if we consider the matrices $A = D= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, then it holds: $$A \cdot D \cdot A^T = D,$$ but $A^2 = A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}.$

Answer 3

For the edited version, you can let $A=\begin{bmatrix}1&2\\0&3\end{bmatrix}$ and $D=\begin{bmatrix}1&0\\0&0\end{bmatrix}$;

then $ADA^{T}=\begin{bmatrix}1&0\\0&0\end{bmatrix}=D$, but $A^2\ne I$.

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For the case where D has full rank, we can let $A=\begin{bmatrix}\frac{1}{2}&-\frac{\sqrt{6}}{4}\\\frac{\sqrt{6}}{2}&\frac{1}{2}\end{bmatrix}$ and $D=\begin{bmatrix}1&0\\0&2\end{bmatrix}$;

then $ADA^{T}=\begin{bmatrix}1&0\\0&2\end{bmatrix}=D$, but $A^2\ne I$.