# Construction of a Symmetric Matrix whose Inverse Matrix is Itself

## Problem 556

Let $\mathbf{v}$ be a nonzero vector in $\R^n$.
Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$.
Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by
$A=I-a\mathbf{v}\mathbf{v}^{\trans},$ where $I$ is the $n\times n$ identity matrix.

Prove that $A$ is a symmetric matrix and $AA=I$.
Conclude that the inverse matrix is $A^{-1}=A$.

## Proof.

### $A$ is symmetric

We first show that the matrix $A$ is symmetric.
We calculate using properties of transpose
\begin{align*}
A^{\trans}&=(I-a\mathbf{v}\mathbf{v}^{\trans})^{\trans} && \text{by definition of $A$}\\
&=I^{\trans}-(a\mathbf{v}\mathbf{v}^{\trans})^{\trans}\\
&=I-a(\mathbf{v}^{\trans})^{\trans}\mathbf{v}^{\trans}\\
&=I-a\mathbf{v}\mathbf{v}^{\trans}\\
&=A && \text{by definition of $A$}.
\end{align*}
Hence we have $A^{\trans}=A$, and thus $A$ is symmetric.

### $AA=I$ and $A^{-1}=A$

Next, we prove that $AA=I$.

We compute
\begin{align*}
AA&=(I-a\mathbf{v}\mathbf{v}^{\trans})(I-a\mathbf{v}\mathbf{v}^{\trans})\\
&=I(I-a\mathbf{v}\mathbf{v}^{\trans})-a\mathbf{v}\mathbf{v}^{\trans}(I-a\mathbf{v}\mathbf{v}^{\trans})\\
&=I-a\mathbf{v}\mathbf{v}^{\trans}-a\mathbf{v}\mathbf{v}^{\trans}+a^2\mathbf{v}\mathbf{v}^{\trans}\mathbf{v}\mathbf{v}^{\trans}\\
&=I-2a\mathbf{v}\mathbf{v}^{\trans}+a^2\mathbf{v}\mathbf{v}^{\trans}\mathbf{v}\mathbf{v}^{\trans}. \tag{*}
\end{align*}

Note that we have
\begin{align*}
\mathbf{v}\mathbf{v}^{\trans}\mathbf{v}\mathbf{v}^{\trans}&=\mathbf{v}(\mathbf{v}^{\trans}\mathbf{v})\mathbf{v}^{\trans}\\
&=\mathbf{v}\left(\, \frac{2}{a} \,\right)\mathbf{v}^{\trans} &&\text{by definition of $a\neq 0$}\\
&=\frac{2}{a}\mathbf{v}\mathbf{v}^{\trans}.
\end{align*}

Plugging this relation into (*), we obtain
\begin{align*}
AA&=I-2a\mathbf{v}\mathbf{v}^{\trans}+a^2\frac{2}{a}\mathbf{v}\mathbf{v}^{\trans}=I.
\end{align*}
Thus we get $AA=I$.
This implies that the inverse matrix of $A$ is $A$ itself: $A^{-1}=A$.

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