Invertible Idempotent Matrix is the Identity Matrix

Idempotent Matrix Problems and Solutions in Linear Algebra

Problem 1

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

Show that a square invertible idempotent matrix is the identity matrix.

LoadingAdd to solve later

Sponsored Links


Let $A$ be an $n \times n$ invertible idempotent matrix.

Since $A$ is invertible, the inverse matrix $A^{-1}$ of $A$ exists and it satisfies $A^{-1} A=I_n$, where $I_n$ is the $n\times n$ identity matrix.

Since $A$ is idempotent, we have $A^2=A$.
Multiplying this equality by $A^{-1}$ from the left, we get $A^{-1}A^2=A^{-1}A$. Using the fact that $A^{-1} A=I_n$, we obtain $A=I_n$.

The proof is completed.

Related Question.

Give it a try with the following problems about idempotent matrices.

(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).

The proofs are given in the post ↴
Unit Vectors and Idempotent Matrices

Show that

(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.

LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.