Show $T$ is diagonalizable if $T-\lambda I$ is idempotent

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.

• Kindly check my answer. Thanks!! Feb 20, 2015 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?

• 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. Feb 20, 2015 at 5:02
• 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 :) Feb 20, 2015 at 5:16
• Suppose $T = PDP^{-1}$. Then $T-\lambda I = P(D-\lambda I)P^{-1}$. Feb 20, 2015 at 13:18
• 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. Feb 20, 2015 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.

• 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!! Feb 20, 2015 at 4:38