# 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).

## 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. Sponsored Links ### More from my site • Inner Product, Norm, and Orthogonal Vectors Let \mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 are vectors in \R^n. Suppose that vectors \mathbf{u}_1, \mathbf{u}_2 are orthogonal and the norm of \mathbf{u}_2 is 4 and \mathbf{u}_2^{\trans}\mathbf{u}_3=7. Find the value of the real number a in […] • Idempotent Matrices are Diagonalizable Let A be an n\times n idempotent matrix, that is, A^2=A. Then prove that A is diagonalizable. We give three proofs of this problem. The first one proves that \R^n is a direct sum of eigenspaces of A, hence A is diagonalizable. The second proof proves […] • Orthogonality of Eigenvectors of a Symmetric Matrix Corresponding to Distinct Eigenvalues Suppose that a real symmetric matrix A has two distinct eigenvalues \alpha and \beta. Show that any eigenvector corresponding to \alpha is orthogonal to any eigenvector corresponding to \beta. (Nagoya University, Linear Algebra Final Exam Problem) Hint. Two […] • Idempotent (Projective) Matrices are Diagonalizable Let A be an n\times n idempotent complex matrix. Then prove that A is diagonalizable. Definition. An n\times n matrix A is said to be idempotent if A^2=A. It is also called projective matrix. Proof. In general, an n \times n matrix B is […] • Determine Eigenvalues, Eigenvectors, Diagonalizable From a Partial Information of a Matrix Suppose the following information is known about a 3\times 3 matrix A. \[A\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}=6\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \quad A\begin{bmatrix} 1 \\ -1 \\ 1 […] • Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given Let \mathbf{a} and \mathbf{b} be vectors in \R^n such that their length are \[\|\mathbf{a}\|=\|\mathbf{b}\|=1 and the inner product $\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.$ Then determine the length $\|\mathbf{a}-\mathbf{b}\|$. (Note […]
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