Unit Vectors and Idempotent Matrices

Problem 527

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

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(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.

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(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.

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(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).

 
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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.

See the post ↴
Idempotent Matrix and its Eigenvalues
for solutions of this problem.

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