Let $V$ be a finite dimensional vector space and let $T:V\rightarrow V$ be a linear map. Suppose that $T$ is diagonalizable. Show that $Ch_T(T)=0$ without using the Cayley-Hamilton theorem. We are told as a hint to use: $f(T)\vec{v}=f(\lambda)\vec{v}$ where $\lambda\text{ and }\vec{v}$ are an eigenvalue and corresponding eigenvector of T. This was proved in the prior question. I would like to know if this is a valid proof.


Since $T$ is diagonalizable, $\exists ~~S\in\mathbb{R}^{n\times n}\text{ such that }D=S^{-1}TS$ Where $D$ is a diagonal matrix. Thus Matrices $D$ and $T$ are similar and they have the same characteristic equation and the same eigenvalues i.e. $Ch_T=Ch_D\text{ and } \lambda_T=\lambda_D$.

We have that $f(T)\vec{v}=f(\lambda)\vec{v}$, now let $f = Ch_T $.


$$Ch_T(T)\vec{v}=Ch_D(\lambda_D)\vec{v}$$ Since $D$ is diagonal $Ch_D(\lambda_D)=0$ because there will be at least one multiple of zero in its factors. Thus:$$Ch_T(T)\vec{v}=0$$And since $\vec{v}$ is non-zero (its an eigenvector) this implies that $Ch_T(T)=0._{~~\square}$

Attempt 2:

Since $T$ is diagonalizable, there exists a basis $B$ consisting of eigenvectors of $T$ such that $[T]_B$ is diagonal. Let $B=\{b_1,b_2,\dots,b_n\}$. In the basis $B$ the characteristic equation of $T$ has the following factorisation: $Ch_T(x)=(x-\lambda_1)(x-\lambda_2)\dots(x-\lambda_n)$. Need to show that $Ch_T(T)\vec{v}=0$. Also note that $v=\sum_{k=1}^n a_kb_k$ where $a_k$ is an element of the field.

$$Ch_T(T)\vec{v}=\prod_{i=1}^n[T-\lambda_iI]\sum_{k=1}^n[ a_kb_k]$$ $$~~~~~~~~~~~~~~~~~~=\sum_{k=1}^n\prod_{i=1}^n[Ta_kb_k-\lambda_iIa_kb_k]$$ $$~~~~~~~~~~~~~~~~~~=\sum_{k=1}^n\prod_{i=1}^n[a_k\lambda_kb_k-a_k\lambda_ib_k]$$ $$~~~~~~~~~~~~=\sum_{k=1}^n\prod_{i=1}^n(\lambda_k-\lambda_i)a_kb_k$$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=0_{~~\square}$

That last step is because in every term in the sum, there will be a factor of zero in the product when k=i.

  • $\begingroup$ I thought diagonalizing a linear map and finding a basis in which it is a diagonal matrix were the same thing :S $\endgroup$ Nov 19, 2015 at 13:06

2 Answers 2


Your approach will lead to the right answer (but you may need to add the facts about an eigenbasis mentioned by @MorganRodgers)

An alternate approach is to observe that $A^n=SD^nS^{-1}$ where $A$ is the matrix of the linear transformation $T$ (after choosing a basis). Then, plugging in $A$ into the characteristic polynomial, you have $$ Ch_T(A)=S\begin{bmatrix}Ch_T(\lambda_1)\\&\ddots\\&&Ch_T(\lambda_n)\end{bmatrix}S^{-1} $$ But, since the roots of the characteristic polynomial are the eigenvalues, the diagonal matrix is actually the zero matrix. Therefore, $Ch_T(A)=0$.

  • $\begingroup$ Why do we need that $A^n = SD^nS^{-1}$? $\endgroup$ May 28 at 0:14
  • $\begingroup$ @user129393192 It's easiest to understand this via an example, but the main idea is that you can factor $S$ and $S^{-1}$ out of the polynomial. $\endgroup$ May 29 at 14:02
  • $\begingroup$ I believe I understand what you’re getting at. My proof was along the lines of $(\lambda x^0 - x)\dots$ being the characteristic polynomial, which led to $(\lambda [I] - [A])\dots$ and then $(\lambda [S][S]^{-1} - [S][D][S]^{-1})\dots$ and finally factoring it out, so I didn’t need to use that $\endgroup$ May 29 at 17:07

Let $P(X)$ be the characteristic polynomial of an $n \times n$ matrix $A$. You want to show that $$P(A)v = 0 \mbox{ for all } v \in \mathbb{R}^{n}.$$ To do this, use the fact that $A$ is diagonalizable (that is, that there exists a basis for $V$ consisting entirely of eigenvectors of $A$) along with the factorization $$P(X) = (x-\lambda_{1})(x-\lambda_{2})\cdots(x-\lambda_{n})$$ where the $\lambda_{i}$ are the eigenvalues of $A$ (possibly with repeats).

  • $\begingroup$ Hi Morgan, I think this the approach I took in my attempt, could you check if it is valid? Thanks. $\endgroup$ Nov 19, 2015 at 13:01
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    $\begingroup$ I agree, I think that one of the main steps that is missing in the original attempt (and is present in this answer) is that there is an eigenbasis, so that every $v\in V$ can be written in terms of eigenvectors. $\endgroup$ Nov 19, 2015 at 13:08
  • $\begingroup$ Ok I will make another attempt, using the advice you both have given me. I will update the original post. If you would like a bit more background knowledge of what $f$ is and $v$ see this link: math.stackexchange.com/questions/1535883/… $\endgroup$ Nov 19, 2015 at 13:23

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