Linear Algebra

Diagonalizable by an Orthogonal Matrix Implies a Symmetric Matrix

Diagonalization Problems and Solutions in Linear Algebra

Problem 210

Let $A$ be an $n\times n$ matrix with real number entries.

Show that if $A$ is diagonalizable by an orthogonal matrix, then $A$ is a symmetric matrix.

Proof.

Suppose that the matrix $A$ is diagonalizable by an orthogonal matrix $Q$.
The orthogonality of the matrix $Q$ means that we have
\[Q^{\trans}Q=QQ^{\trans}=I, \tag{*}\] where $Q^{\trans}$ is the transpose matrix of $Q$ and $I$ is the $n\times n$ identity matrix.

Since $Q$ diagonalizes the matrix $A$, we have
\[Q^{-1}AQ=D,\] where $D$ is a diagonal matrix.
Equivalently, we have
\[A=QDQ^{-1} \tag{**}.\] Taking transpose of both sides, we obtain
\begin{align*}
A^{\trans}&=(QDQ^{-1})^{\trans}\\
&=(Q^{-1})^{\trans}D^{\trans} Q^{-1}\\
&=(Q^{-1})^{\trans}D Q^{-1} \text{ since } D \text{ is diagonal.}\tag{***}
\end{align*}

By (*), we observe that the inverse matrix of $Q$ is the transpose $Q^{\trans}$, that is, $Q^{-1}=Q^{\trans}$.
It follows from this observation and (***) that we have
\[A^{\trans}=QDQ^{-1}.\] (Note that $(Q^{-1})^{\trans}=Q^{\trans \trans}=Q$.)

Comparing this with (**), we obtain
\[A^{\trans}=A,\] and hence $A$ is a symmetric matrix.

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