$T^2=I$ implies $T$ is diagonalizable

Question

Suppose $T:V\rightarrow V$ is linear and $T^2=I$. Prove that $T$ is diagonalizable.

First, I know that $T$ has only eigenvalues 1 or -1. Also I observed that $(T-I)(T+I)=0$, does this fact help to show that $T$ is diagonalizable?

@Omnomnomnom: Not really. I will try john's method and see if it works. — user112358, Jun 08 '15 at 20:11
From Jordan form it follows immediately $ PJP^{-1} PJP^{-1}=I $,Squarred Jordan block is identity only if it is diagonal. — Widawensen, Dec 13 '18 at 09:10

score 8 · Accepted Answer · edited Dec 13 '18 at 01:10

8

In this case you can split $V = V_1 \oplus V_{-1}$ where $T$ acts as either $1$ or $-1$. Then take the direct sum over each subspace. $T = T_1 \oplus T_{-1}$

Since $T^2 v = v$ for all $v \in \mathbb{V}$ (an involution), we can decompose every vector as:

$$ v = \underbrace{\tfrac{1}{2}(v + Tv)}_{E_1} + \underbrace{\tfrac{1}{2}(v - Tv)}_{E_{-1}},$$

where $E_\lambda$ is an eigenspace with eigenvalue $\lambda$, so this eigenspace decomposition exists for all vector.

edited Dec 13 '18 at 01:10

Zau

3,909
12
29

answered Jun 08 '15 at 20:04

cactus314

24,438

how do I know that $V$ is the sum of the eigenspaces? Thanks. – user112358 Jun 08 '15 at 20:16

score 2 · Answer 2 · answered Jun 08 '15 at 20:02

2

Since the minimum polynomial factors over the field into distinct linear factors this implies that $T$ is diagonalisable.

answered Jun 08 '15 at 20:02

Rogelio Molina

3,380

score 1 · Answer 3 · edited Apr 13 '17 at 12:20

1

According to this answer, the Jordan Canonical Form of $A$ is a diagonal matrix with only $+1$ or $-1$ along the diagonal. That is, $A=SDS^{-1}$ where $D$ a diagonal matrix where each element on the diagonal is $+1$ or $-1$. Thus, $A$ is diagonalizable.

edited Apr 13 '17 at 12:20

Community

1

answered Jun 09 '15 at 21:15

robjohn

345,667

mathreadler · Answer 4 · 2015-06-08T21:23:03.077

0

Your proposition is in general false unless you specify which elements the matrices are allowed to take.

The matrix $T = \left(\begin{array}{cc} 0&1\\1&0\end{array}\right)$ with elements in non-negative reals or integers has $T^2 = I$ without being diagonalizable.

This particular T gives $aT$ for positive real a is a representation for negative reals if $aI$ represent positive a. Furthermore if we define addition as the subspace spanned by all matrices except $\left(\begin{array}{cc} 1&1\\1&1\end{array}\right)$ we can perform both addition and multiplication - we have constructed new numbers!

Example, add 2 and -1: Just ordinary matrix addition: $\left(\begin{array}{cc} 2&1\\1&2\end{array}\right)$, but after stripping the projection on $\left(\begin{array}{cc} 1&1\\1&1\end{array}\right)$, we are left with $\left(\begin{array}{cc} 1&0\\0&1\end{array}\right)$ which represents 1.

edited Jun 08 '15 at 21:23

answered Jun 08 '15 at 20:54

mathreadler

25,824

Aren't ${(1,1), (1,-1)}$ eigenvectors? – cactus314 Jun 08 '15 at 21:00
Not if -1 is not allowed as an element. – mathreadler Jun 08 '15 at 21:02
Which field? Non-negative reals $\mathbb{R}_{\geq 0}$ are not a field, since $1$ does not have an additive inverse. $\mathbb{Z}$ is not a field since $\frac{1}{2} \notin \mathbb{Z}$. – cactus314 Jun 08 '15 at 21:03
Yes you are right, it's not a field. Let's just say the set of elements is limited to positive reals. – mathreadler Jun 08 '15 at 21:05
If we use $\mathbb{Z}$ instead of $\mathbb{R}$ we get a $\mathbb{Z}$-module. In the other case, $\mathbb{R}_{\geq 0}$ is not even a group, so non-negative matrices are a world of their own. – cactus314 Jun 08 '15 at 21:09
Is there even a theory of diagonalisation for matrices over semirings? – Jun 08 '15 at 21:46

$T^2=I$ implies $T$ is diagonalizable

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