# Diagonalization of a projection

If I have a projection, T on a finite dimensional vector space V, how do I show that T is diagonalizable?

-
Have you determined the eigenvectors and eigenvalues of T? –  Adam Saltz Oct 19 '11 at 2:33
No. As a matter of fact I do not know the matrix for T. –  smanoos Oct 19 '11 at 2:35
Ok. What is your definition of projection? –  Adam Saltz Oct 19 '11 at 2:48
If $V$ is the direct sum of $M$ and $N$, so that every $z$ in $V$ may be written uniquely in the form $z=x+y$ with $x$ in $M$ and $y$ in $N$, the projection on $M$ along $N$ is the transformation $T$ defined by $Tz=x$. That is the definition of projection. –  smanoos Oct 19 '11 at 2:54
Sometimes the definition is taken to be that $T$ is a projection precisely if $T^2=T$, i.e. $T$ is idempotent. That's actually exactly equivalent. –  Michael Hardy Oct 19 '11 at 21:26

If $T$ is a projection, that means there's a subspace $W$ onto which it projects. It maps every vector in $W$ to itself. Therefore every vector in $W$ is an eigenvector with eigenvalue $1$. Every vector not in $W$ is mapped to a vector in $W$. Take any vector $v$ and write $$v = Tv + (v-Tv),$$ so the first term $Tv$ is in $W$. It is easy to see that the second term, $v-Tv$, is in the kernel of $T$: the first term is mapped to $Tv$, and the second is mapped to $Tv-T^2v$. But since $Tv$ is in $W$, it must be fixed by $T$, so $T^2v=Tv$; thus $T(v-TV)=0$. In this way, every vector $v$ is written as the sum of a vector in $W$, which is an eigenvector with eigenvalue $1$, and a vector in the kernel of $T$, which is an eigenvector with eigenvalue $0$. So form a basis of the whole space by taking the union of a basis of $W$ and a basis of the kernel of $T$, and the matrix of $T$ with respect to that basis is $$\begin{bmatrix} 1 \\ & 1 \\ & & 1 \\ & & & \ddots \\ & & & & 1 \\ & & & & & 0 \\ & & & & & & \ddots \\ & & & & & & & 0 \end{bmatrix}$$ (and all off-diagonal entries are $0$) where the number of $1$s is the dimension of $W$ and the number of $0$s is the dimension of the kernel of $T$.

-
You are perfectly right. But can you throw more light on how to set up this matrix. That is where I have a problem. –  smanoos Oct 19 '11 at 3:03
@smanoos: Using your notation: take a basis $m_1,\ldots,m_k$ for $M$, a basis $n_{k+1},\ldots,n_r$ for $N$. Show that $m_1,\ldots,m_k,n_{k+1},\ldots,n_r$ is a basis for $V$, and that the matrix of $T$ relative to that basis is the one that Michael Hardy writes above. –  Arturo Magidin Oct 19 '11 at 4:36
Thanks for your help. –  smanoos Oct 19 '11 at 9:38
Let $P:{\mathbb R}^m\to{\mathbb R}^m$ be the projection transformation. By rank–nullity theorem,
$$\dim(\ker(P))+\dim(range(P))=m$$ On the other hand, by definition, $P^2=P$, which implies that the eigenvalues of $P$ are $\lambda=0$ or $1$. It's not hard to show that $\ker(P)$ is the eigenspace of $\lambda=0$ and $range(P)$ is the eigenspace of $\lambda=1$. Therefore, we an independent set of $m$ eigenvectors, which implies that $P$ is diagonalizable.