Simple eigendecomposition of matrix

A real symmetric matrix $$\textbf{A}$$ can be decomposed such that $$\textbf{A}=\textbf{P}\textbf{D}\textbf{P}^{-1}$$ where $$\textbf{P}$$ is the orthonormal matrix ($$\textbf{P}^{-1}=\textbf{P}^{\text{T}}$$) consisting of the columns of the eigenvectors of $$\textbf{A}$$ and $$\textbf{D}$$ is the diagonal matrix with the entries as eigenvalues of $$\textbf{A}$$.

How can I show that the decomposition of $$\textbf{A}$$ can also be written as $$\textbf{A}=\textbf{P}^{-1}\textbf{D}\textbf{P}$$ ?

In general, you can't. We know the eigenvectors of $$\bf P D P^{-1}$$ are the columns $$\bf v_i$$ of $$\bf P$$, $$i = 1, \ldots, n$$. Let's call the corresponding eigenvalues $$\lambda_i$$.
Now, observe that, if $$\bf \eta_i$$ is the $$i$$-th column of $$\bf P^{-1}$$, we have $$\bf{P^{-1}DP \eta_i = P^{-1}D e_i} = P^{-1}(\lambda_i e_i) = \lambda_i\eta_i,$$ where $$\bf e_i$$ is the $$i$$-th member of the canonical basis, i.e., $$\bf e_i = [0 \ \cdots \ 0 \ 1 \ 0 \ \cdots 0]^T$$, where the $$1$$ is in the $$i$$-th place.
Therefore, if $$\bf A$$ is also equal to $$\bf P^{-1}DP$$, then eigenvectors corresponding to the same eigenvalues must match, that is, we must have $$\bf v_i = \eta_i$$ for all $$i$$, which implies $$\bf P = P^{-1}$$. This is a very restrictive condition which does not happen in general.
However, if all you want is to be able to write $$\bf A = Q^{-1} D Q$$, where $$\bf Q$$ need not be equal to $$\bf P$$, than just set $$\bf Q = P^{-1}$$ and you are done.
• Thanks for the answer. I will just set $\textbf{Q}=\textbf{P}^{-1}$. I am just over complicating things Jul 21 at 20:58