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 diagonalizable if there are $n$ linearly independent eigenvectors. So if eigenvectors of $B$ span $\R^n$, then $B$ is diagonalizable.
We prove that $\R^n$ is spanned by eigenspaces. Every vector $\mathbf{v}\in \R^n$ can be expresses as
\[\mathbf{v}=(\mathbf{v}-A\mathbf{v})+A\mathbf{v}=\mathbf{v}_0+\mathbf{v}_1,\]
where we put $\mathbf{v}_0=\mathbf{v}-A\mathbf{v}$ and $\mathbf{v}_1=A\mathbf{v}$.
We claim that $\mathbf{v}_0$ and $\mathbf{v}_1$ are elements in the eigenspaces corresponding to (possible) eigenvalues $0$ and $1$, respectively.
To see this, we compute
\begin{align*}
A\mathbf{v}_0&=A(\mathbf{v}-A\mathbf{v})\\
&=A\mathbf{v}-A^2\mathbf{v}\\
&=A\mathbf{v}-A\mathbf{v} && \text{since $A$ is idempotent}\\
&=O=0\mathbf{v}_0.
\end{align*}
Thus, we have $A\mathbf{v}_0=0\mathbf{v}_0$, and this means that $\mathbf{v}_0$ is a vector in the eigenspace corresponding to the eigenvalue $0$.
(If $0$ is not an eigenvalue of $A$, then $\mathbf{v}_0=\mathbf{0}$.)
We also have
\begin{align*}
A\mathbf{v}_1=A(A\mathbf{v})=A^2\mathbf{v}=A\mathbf{v}=\mathbf{v}_1,
\end{align*}
where the third equality holds as $A$ is idempotent.
This implies that $\mathbf{v}_1$ is a vector in the eigenspace corresponding to eigenvalue $1$. (If $1$ is not an eigenvalue of $A$, then $\mathbf{v}_1=\mathbf{0}$.)
It follows that every vector $\mathbf{v}\in \R^n$ is a sum of eigenvectors (or the zero vector).
That is, $\R^n$ is spanned by eigenvectors.
By the general fact mentioned at the beginning of the proof, we conclude that the idempotent matrix $A$ is diagonalizable.
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 […]
Unit Vectors and Idempotent Matrices
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 […]
Maximize the Dimension of the Null Space of $A-aI$
Let
\[ A=\begin{bmatrix}
5 & 2 & -1 \\
2 &2 &2 \\
-1 & 2 & 5
\end{bmatrix}.\]
Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix.
Your score of this problem is equal to that […]
Idempotent Matrix and its Eigenvalues
Let $A$ be an $n \times n$ matrix. We say that $A$ is idempotent if $A^2=A$.
(a) Find a nonzero, nonidentity idempotent matrix.
(b) Show that eigenvalues of an idempotent matrix $A$ is either $0$ or $1$.
(The Ohio State University, Linear Algebra Final Exam […]
Given All Eigenvalues and Eigenspaces, Compute a Matrix Product
Let $C$ be a $4 \times 4$ matrix with all eigenvalues $\lambda=2, -1$ and eigensapces
\[E_2=\Span\left \{\quad \begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix} \quad\right \} \text{ and } E_{-1}=\Span\left \{ \quad\begin{bmatrix}
1 \\
2 \\
1 \\
1
[…]
Quiz 13 (Part 1) Diagonalize a Matrix
Let
\[A=\begin{bmatrix}
2 & -1 & -1 \\
-1 &2 &-1 \\
-1 & -1 & 2
\end{bmatrix}.\]
Determine whether the matrix $A$ is diagonalizable. If it is diagonalizable, then diagonalize $A$.
That is, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that […]
How to Diagonalize a Matrix. Step by Step Explanation.
In this post, we explain how to diagonalize a matrix if it is diagonalizable.
As an example, we solve the following problem.
Diagonalize the matrix
\[A=\begin{bmatrix}
4 & -3 & -3 \\
3 &-2 &-3 \\
-1 & 1 & 2
\end{bmatrix}\]
by finding a nonsingular […]