5
$\begingroup$

This question already has an answer here:

Let $T$ be a linear transformation $T$ such that $T\colon V \to V$. Also, let $T = T^2$. What are the possible eigenvalues of $T$?

I am not sure if the answer is only $1$, or $0$ and $1$.

It holds that $T = T^2$, thus $T(T(x)) = T(x)$. Let's call $T(x) = v$, so $T(v) = v$. which means that $\lambda=1$. But I am not sure about this, while I have seen a solution that says that $0$ is possible as well.

Thanks in advance !

$\endgroup$

marked as duplicate by Najib Idrissi, Martin R, Claude Leibovici, Community May 20 '15 at 8:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

8
$\begingroup$

Let $v\neq 0$ be an eigenvector of $T$ with eigenvalue $\lambda$, so $Tv=\lambda v$. Using $T=T^2$ we have $$ Tv = T^2 v = T(Tv) = T(\lambda v) = \lambda(Tv) = \lambda^2 v. $$ Hence, $\lambda v = \lambda^2 v$. Since $v\neq 0$ we conclude $\lambda = \lambda^2$. The only solutions to this equation are $0$ and $1$.

$\endgroup$
6
$\begingroup$

Think of this as follows:

$$T^2=T\implies T(T-I)=0$$

Thus, $\;T\;$ is a root of $\;x(x-1)\;$ and thus the characteristic polynomial of $\;T\;$ can only have $\;0\;$ or $\;1\;$ as its roots, and thus these precisely are the only possible eigenvalues of $\;T\;$ .

So you were half right...:)

$\endgroup$
2
$\begingroup$

Another side remark: You say that you are not sure if 1, or both 0 and 1 can be eigenvalues.

In some cases, it is worthwhile to think of specific examples and see what they can tell us. So what are some examples of matrices $T$ that satisfy $T^2 = T$?

Well, the identity is certainly one, and its eigenvalues are all 1.

However, another such matrix is the zero matrix! It also trivially satisfies $\mathbf{0}^2 = \mathbf{0}$. Its eigenvalues are all zero, so zero can certainly be an eigenvalue as well.

Anyhow, this of course just tells you that both 0 and 1 are possible eigenvalues of such a matrix, but not that they are the only possible eigenvalues. For that, the other answers provide a full solution.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.