# Idempotent Linear Transformation and Direct Sum of Image and Kernel

## Problem 327

Let $A$ be the matrix for a linear transformation $T:\R^n \to \R^n$ with respect to the standard basis of $\R^n$.
We assume that $A$ is idempotent, that is, $A^2=A$.
Then prove that
$\R^n=\im(T) \oplus \ker(T).$

## Proof.

To prove the equality $\R^n=\im(T) \oplus \ker(T)$, we need to prove
(a) $\R^n=\im(T) + \ker(T)$, and
(b) $\im(T) \cap \ker(T)=\{0\}$.

By definition, the image $\im(T)$ and $\ker(T)$ are subspaces of $\R^n$, hence $\im(T) + \ker(T) \subset \R^n$.
To prove the reverse inclusion, for any $x\in \R^n$, we write
\begin{align*}
x=Ax+(x-Ax).
\end{align*}
Then the first term is in $\im(T)$ since
$Ax=T(x)\in \im(T).$ The second term $x-Ax$ is in $\ker(T)$ since
\begin{align*}
T(x-Ax)&=A(x-Ax)\\
&=Ax-A^2x\\
&=Ax-Ax && (\text{since $A$ is idempotent})\\
&=0.
\end{align*}

Thus we have
$x=\underbrace{Ax}_{\in \im(T)}+\underbrace{(x-Ax)}_{\in \ker(T)}\in \im(T) + \ker(T) .$ Since $x$ is arbitrary element in $\R^n$, we have
$\R^n\subset \im(T) + \ker(T),$ and putting the two inclusions together yields
$\R^n= \im(T) + \ker(T),$ and we proved (a).

To prove (b), let $x\in \im(T) \cap \ker(T)$. Thus $x\in \im(T)$ and $x\in \ker(T)$.
Since $x\in \im(T)$, there exists $x’\in \R^n$ such that $T(x’)=x$, or equivalently, $Ax’=x$.

Then, we have
\begin{align*}
&0=T(x)=Ax\\
&=A(Ax’)=A^2x’\\
&=Ax’ && (\text{since $A$ is idempotent})\\
&=x.
\end{align*}

Hence we have proved an arbitrary element $x$ in the intersection is $x=0$, and thus we have
$\im(T) \cap \ker(T)=\{0\}.$ So (b) is proved.

The facts (a), (b) implies that we have
$\R^n=\im(T) \oplus \ker(T),$ as required.

Let $T$ be the linear transformation from the $3$-dimensional vector space $\R^3$ to $\R^3$ itself satisfying the following relations. \begin{align*}...