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

Rodrigo de Azevedo
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smanoos
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  • Have you determined the eigenvectors and eigenvalues of T? – Adam Saltz Oct 19 '11 at 02:33
  • No. As a matter of fact I do not know the matrix for T. – smanoos Oct 19 '11 at 02:35
  • Ok. What is your definition of projection? – Adam Saltz Oct 19 '11 at 02:48
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    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 02:54
  • 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

3 Answers3

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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 $T$.

Michael Hardy
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  • 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 03:03
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    @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 04:36
  • @ArturoMagidin : I can see how the Kernel contains $ S:={Pv-v} $, but not sure how to show S equals the kernel. How would we show this. Sorry for the necropost. – MSIS Jan 31 '24 at 02:39
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    @MSIS So, you are asking me about a post and answer from 2011, that I didn't even write in the first place, but you also do it with reference to some $P$ that does not appear in either the original post nor this answer. Who is $P$? Is $P$ supposed to be the projection $T$? If $x$ is in the kernel, then $x= T(-x)-(-x)$. – Arturo Magidin Jan 31 '24 at 02:43
  • Sorry, @Arturo Magidin, I meant T as in this post; it's the Orthogonal Projection from a finite-dimensional Vector Space V onto a subspace . But I got it from your comment, it cleared my doubt, thanks. – MSIS Jan 31 '24 at 02:52
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@Michael Hardy's answer is nice and complete. I'd like to write down how I think about this question.

Let $P:{\mathbb R}^m\to{\mathbb R}^m$ be the projection transformation. By rank–nullity theorem,

$$\dim(\ker(P))+\dim(\operatorname{range}(P))=m $$ 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 $\operatorname{range}(P)$ is the eigenspace of $\lambda=1$. Therefore, we have an independent set of $m$ eigenvectors, which implies that $P$ is diagonalizable.

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If $P$ is a projection and $J$ is the Jordan matrix associated with $P$ (that is, for some nonsingular $Q$, $P=QJQ^{-1}$) then $$J^2=\left(Q^{-1}PQ\right)\left(Q^{-1}PQ\right)=Q^{-1}P^2Q=Q^{-1}PQ=J$$ This is possible only when $J$ is diagonal.

Maximilian Janisch
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user165536
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