Prove or give a counterexample: if $V$ is a complex vector space and $\dim V = n$ and $T \in \mathcal{L}(V)$, then $T^n$ is diagonalizable.

I have a sketch of a solution, but I'm not convinced that it's sufficiently rigorous.

Consider a counterexample in $V = \mathbb{R}^2$: \begin{align*} T & = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \\ T^2 & = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}. \end{align*} The characteristic polynomial of $T^2$ is: \begin{align*} p(\lambda) = \det(\lambda I - T^2) = (\lambda - 1)^2. \end{align*} Since $p(\lambda)$ has a repeated root, $T^2$ lacks an eigenbasis, so it has only one linearly independent eigenvector, and is therefore not diagonalizable.

Here is my problem with this solution. If an operator is diagonalizable, then there has to exist a diagonal matrix with respect to some basis, but I just chose the standard basis. How do I establish from this that there is no basis wherein I can find two eigenvectors?

  • $\begingroup$ Having a repeated root, by itself, does not preclude diagonalizability. The identity matrix is definitely diagonalizable, and its characteristic polynomial is $(x-1)^2$, with a repeated root. You are correct that your matrix is not diagonalizable, but your reasoning is flawed. $\endgroup$ Nov 26, 2020 at 21:34
  • $\begingroup$ Thank you for pointing this out. Could you explain how I should go about showing that it isn't diagonalizable? Should I try to compute the eigenvectors directly? $\endgroup$
    – Ted Baker
    Nov 26, 2020 at 21:35
  • $\begingroup$ The question itself isn't well-written. You can guess the answer, from which you know the way to think. The conclusion (diag...) is an adjective that qualifies a certain class of linear operators. The assumption on $T$ in too weak. I wonder if there's a condition like ${\rm rank}(T^n) = \dim V = n$ missing. $\endgroup$ Nov 26, 2020 at 21:43
  • $\begingroup$ See this post. The two questions are exactly the same. $\endgroup$
    – Sky subO
    Mar 17, 2021 at 16:52

1 Answer 1


Your reasoning is incorrect. While it is true that if the characteristic polynomial has no repeated roots then the matrix is diagonalizable, the converse is false. A matrix can have a characteristic polynomial with repeated roots and still be diagonalizable. The simplest example is the $n\times n$ identity matrix, which is itself diagonal, but has characteristic polynomial $(-1)^n(x-1)^n$. (Note: I prefer the definition of characteristic polynomial as $\det(A-\lambda I)$ because it requires fewer sign changes when computing the determinant by hand). Of course, you can construct diagonal matrices with whatever characteristic polynomial you want, including one with several multiple roots.

The argument at the point you are in is pretty easy. Note that the matrix for $T^2-I$ is $$\left(\begin{array}{cc} 0&2\\ 0&0 \end{array}\right)$$ which has rank $1$; thus, $N(T^2-I)$ has rank $1$, which means the eigenspace of $1$ has dimension $1$. Thus, there aren’t enough linearly independent eigenvectors for a basis.


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