# Unit Vectors and Idempotent Matrices

## Problem 527

A square matrix $A$ is called **idempotent** if $A^2=A$.

**(a)** Let $\mathbf{u}$ be a vector in $\R^n$ with length $1$.

Define the matrix $P$ to be $P=\mathbf{u}\mathbf{u}^{\trans}$.

Prove that $P$ is an idempotent matrix.

**(b)** Suppose that $\mathbf{u}$ and $\mathbf{v}$ be unit vectors in $\R^n$ such that $\mathbf{u}$ and $\mathbf{v}$ are orthogonal.

Let $Q=\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans}$.

Prove that $Q$ is an idempotent matrix.

**(c)** Prove that each nonzero vector of the form $a\mathbf{u}+b\mathbf{v}$ for some $a, b\in \R$ is an eigenvector corresponding to the eigenvalue $1$ for the matrix $Q$ in part (b).

Contents

## Proof.

### (a) Prove that $P=\mathbf{u}\mathbf{u}^{\trans}$ is an idempotent matrix.

The length of the vector $\mathbf{u}$ is given by

\[\|\mathbf{u}\|=\sqrt{\mathbf{u}^{\trans}\mathbf{u}}.\]
Since $\|\mathbf{u}\|=1$ by assumption, it yields that

\[\mathbf{u}^{\trans}\mathbf{u}=1.\]

Let us compute $P^2$ using the associative properties of matrix multiplication.

We have

\begin{align*}

P^2&=(\mathbf{u}\mathbf{u}^{\trans})(\mathbf{u}\mathbf{u}^{\trans})\\

&=\mathbf{u}(\mathbf{u}^{\trans}\mathbf{u})\mathbf{u}^{\trans}\\

&=\mathbf{u}( 1 ) \mathbf{u}^{\trans}=\mathbf{u}\mathbf{u}^{\trans}=P.

\end{align*}

Thus, we have obtained $P^2=P$, and hence $P$ is an idempotent matrix.

### (b) Prove that $Q=\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans}$ is an idempotent matrix

Since $\mathbf{u}$ and $\mathbf{v}$ are unit vectors, we have as in part (a)

\[\mathbf{u}^{\trans}\mathbf{u}=1 \text{ and } \mathbf{v}^{\trans}\mathbf{v}=1.\]
Since $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, their dot (inner) product is $0$.

Thus we have

\[\mathbf{u}^{\trans}\mathbf{v}=\mathbf{v}^{\trans}\mathbf{u}=0.\]

Using these identities, we compute $Q^2$ as follows.

We have

\begin{align*}

Q^2&=(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})\\

&=\mathbf{u}\mathbf{u}^{\trans}(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})

+\mathbf{v}\mathbf{v}^{\trans}(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})\\

&=\mathbf{u}\underbrace{\mathbf{u}^{\trans}\mathbf{u}}_{1}\mathbf{u}^{\trans}+\mathbf{u}\underbrace{\mathbf{u}^{\trans}\mathbf{v}}_{0}\mathbf{v}^{\trans}

+\mathbf{v}\underbrace{\mathbf{v}^{\trans}\mathbf{u}}_{0}\mathbf{u}^{\trans}+\mathbf{v}\underbrace{\mathbf{v}^{\trans}\mathbf{v}}_{1}\mathbf{v}^{\trans}\\[6pt]
&=\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans}=Q.

\end{align*}

It follows that $Q$ is an idempotent matrix.

### (c) Prove that $a\mathbf{u}+b\mathbf{v}$ is an eigenvector

Let us first compute $Q\mathbf{u}$. We have

\begin{align*}

Q\mathbf{u}&=(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})\mathbf{u}\\

&=\mathbf{u}\underbrace{\mathbf{u}^{\trans}\mathbf{u}}_{1}+\mathbf{v}\underbrace{\mathbf{v}^{\trans}\mathbf{u}}_{0}=\mathbf{u}.

\end{align*}

Note that $\mathbf{u}$ is a nonzero vector because it is a unit vector.

Thus, the equality $Q\mathbf{u}=\mathbf{u}$ implies that $1$ is an eigenvalue of $Q$ and $\mathbf{v}$ is a corresponding eigenvector.

Similarly, we can check that $\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $1$.

Now let $a\mathbf{u}+b\mathbf{v}$ be a nonzero vector.

Then we have

\begin{align*}

Q(a\mathbf{u}+b\mathbf{v})&=aQ\mathbf{u}+bQ\mathbf{v}=a\mathbf{u}+b\mathbf{v}.

\end{align*}

It follows that $a\mathbf{u}+b\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $1$.

#### Another way to prove (c)

Another way to see this is as follows.

As we saw above, the vectors $\mathbf{u}$ and $\mathbf{v}$ are eigenvectors corresponding to the eigenvalue $1$.

Hence $\mathbf{u}, \mathbf{v} \in E_{1}$, where $E_1$ is an eigenspace of the eigenvalue $1$

Note that $E_1$ is a vector space, hence $a\mathbf{u}+b\mathbf{v}$ is a nonzero vector in $E_{1}$.

Thus, $a\mathbf{u}+b\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $1$ as well.

## Related Question.

**Problem**.

**(a)**Find a nonzero, nonidentity idempotent matrix.

**(b)** Show that eigenvalues of an idempotent matrix $A$ is either $0$ or $1$.

See the post ↴

Idempotent Matrix and its Eigenvalues

for solutions of this problem.

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