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 $$ \textbf{A}=\textbf{P}^{-1}\textbf{D}\textbf{P} $$ ?


1 Answer 1


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.

  • $\begingroup$ Thanks for the answer. I will just set $\textbf{Q}=\textbf{P}^{-1}$. I am just over complicating things $\endgroup$
    – kowalski
    Jul 21 at 20:58

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