Suppose $V$ is a finite dimensional vector space of dimension $n$ and $T$ is a linear operator on $V$ such that the characteristic polynomial of $T$ splits. Let $\lambda_1,\lambda_2,...,\lambda_k$ be the distinct eigenvalues of $T$. Further suppose that $T-\lambda_iI$ is idempotent for all $i\in\{1,2,...,k\}$. Then prove that $T$ is diagonalizable.

I would like a hint (only!) to start this problem. I am not aware of the result that $T=\lambda_1T_1+...+\lambda_kT_k$ or something like that. I mean, I am not allowed to use that since it has not been done in class. Hints excluding this will be appreciated.

  1. Show first that if a linear map $T$ is idempotent, then it is diagonalizable.

  2. Show then that if $T-\lambda I$ is idempotent, then $T$ is diagonalizable.

  • $\begingroup$ Kindly check my answer. Thanks!! $\endgroup$ – Landon Carter Feb 20 '15 at 3:52

So taking help from Mariano Suarez-Alvarez, I came up with the following solution:

We first shall show that if $T$ is idempotent, then $T$ is diagonalizable. We aim to show that $T$ has only $1$ and $0$ as eigenvalues.

Suppose $Rank(T)=r$ then, $Range(T)=L\{v_1,v_2,...,v_r\}$ where $v_1,v_2,...,v_r$ are the basis vectors for the range of $T$. Then, correspondingly, for some $u_1,u_2,...,u_r$ in $V$ we must have $T(u_i)=v_i$. But $T^2(u_i)=T(T(u_i))=T(v_i)$ and by idempotence, as $T(u_i)=T^2(u_i)$ it follows that $T(v_i)=v_i$ for all $i\in\{1,2,...,r\}$. Hence we have got the linearly independent eigenvectors corresponding to the eigenvalue $1$.

Now since $T$ is an operator, and because $Nullity(T)+Rank(T)=dim(V)$, it must happen that $V=Null(T)+Range(T)$ and the sum is a direct sum. We have got a basis for $Range(T)$ so all we now need is a basis for $Null(T)$. Since $Nullity(T)=dimV-Rank(T)$ such a basis is ensured. Let this basis be $\{v_{r+1},...v_n\}$. It is a set of linearly independent eigenvectors corresponding to the eigenvalue $0$.

The union of these two bases gives us a basis of eigenvectors for $T$. Thus, $T$ is diagonalizable.

Now we are given that for each $i$, $T-\lambda_i I$ is idempotent. So $T-\lambda_i I$ is diagonalizable, henceI can find ALL the linearly independent eigenvectors of $T-\lambda_iI$ for which the eigenvalue is $0$. Thus, for all such vectors $v$, I get $Tv=\lambda v$ and hence I get ALL linearly independent eigenvectors for the eigenvalue $\lambda_i$ for $T$. This holds for all $i$. Hence we get a basis of eigenvectors for $T$ by collecting all these linearly independent vectors corresponding to distinct eigenvalues. Therefore, $T$ is diagonalizable.

Please check whether the argument is correct.

EDIT: One thing that still bugs me is that, whether I can say that $0$ IS indeed an eigenvalue of each $T-\lambda_i I$. What if not? To be more precise, what if $1$ is the only eigenvalue of $T-\lambda_i I$?Then can we still get the basis of eigenvectors of $T$ by collecting the linearly independent eigenvectors of the other eigenvalues?

  • $\begingroup$ I think you are over-complicating the second part. If $T - \lambda I$ is diagonalizable for just one value of $\lambda$, then $T$ is diagonalizable. $\endgroup$ – Stephen Montgomery-Smith Feb 20 '15 at 5:02
  • $\begingroup$ Could you kindly elaborate a bit more? I am actually a beginner at these so it is a bit hard to understand these things. And thank you for checking it :) $\endgroup$ – Landon Carter Feb 20 '15 at 5:16
  • $\begingroup$ Suppose $T = PDP^{-1}$. Then $T-\lambda I = P(D-\lambda I)P^{-1}$. $\endgroup$ – Stephen Montgomery-Smith Feb 20 '15 at 13:18
  • $\begingroup$ Anyway I figured that out right after writing this comment. But I was travelling and hence could not delete it. The reasoning is: Since $T-\lambda_iI$ is idempotent and hence diagonalizable, there exists a basis of eigenvectors. Look at the matrix of $T-\lambda_iI$ w.r.t this basis of eigenvectors. Write $T=(T-\lambda_iI)+\lambda_iI$. Thus $T$ is the sum of two diagonal matrices which makes T diagonal w.r.t. this basis. $\endgroup$ – Landon Carter Feb 20 '15 at 14:52

If the characteristic polynomial is $$ p(\lambda) = (\lambda -\lambda_{1})^{r_{1}}\cdots(\lambda-\lambda_{k})^{r_{k}}, $$ then the Cayley-Hamilton Theorem gives $$ (T-\lambda_{1}I)^{r_{1}}\cdots(T-\lambda_{k})^{r_{k}}=0. $$ However, beccause $(T-\lambda_{j})^{2}=(T-\lambda_{j})$, then $$ (T-\lambda_{1}I)\cdots(T-\lambda_{k}I) = 0. $$ That means that the minimal polynomial for $T$ has no repeated factors.

  • $\begingroup$ Sorry I do not know about the minimal polynomial: I cannot use this concept right now. Can you kindly check my answer to see if it is correct? Also, if you could give a different method it would be wonderful!! $\endgroup$ – Landon Carter Feb 20 '15 at 4:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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