Let $\mathbb{K}$ be a field, $1\leq n\in \mathbb{N}$ and let $V$ be a $\mathbb{K}$-vector space with $\dim_{\mathbb{R}}V=n$.

Let $\phi :V\rightarrow V$ be a linear map.

I want to show that $\phi$ is diagonalizable if and only if there is a basis $(b_1, \ldots , b_n)$ of $V$ and $\lambda_1, \ldots, \lambda_n\in \mathbb{K}$ such that $\phi (b_i)=\lambda_i b_i$.

Could you give me a hint how we could show this equivalence?

  • $\begingroup$ A linear map is defined to be diagonalizable if its matrix representation is diagonalizable. Here is a paper in which Theorem 1.1 proves the result for matrices: math.jhu.edu/~bernstein/math201/EIGEN.pdf $\endgroup$
    – wormram
    Jun 20, 2020 at 7:51
  • $\begingroup$ If that is a little terse, you can find this result all over the internet with a good Google search. You're really looking to prove, a linear map is diagonalizable if and only if there is an eigenbasis consisting only of eigenvectors of that linear map. $\endgroup$
    – wormram
    Jun 20, 2020 at 7:53
  • $\begingroup$ The condition given is exactly the same as saying that the matrix of $\phi$ relative to the given basis is diagonal. In my answer here, I explain why this is the case in detail for a $2 \times 2$ matrix. $\endgroup$ Jun 20, 2020 at 12:13

1 Answer 1


We say that a linear operator $T:V\to V$ is diagonalizable if and only if there exits an ordered basis $\mathcal{B} = \{b_{1},b_{2},\ldots,b_{n}\}$ of $V$ such that $[T]_{\mathcal{B}}$ is a diagonal matrix.

Based on such definition, let us prove the implication $(\Rightarrow)$ first.

If $T$ is diagonalizable, then there is a basis $\mathcal{B} = \{b_{1},b_{2},\ldots,b_{n}\}$ such that \begin{align*} [T]_{\mathcal{B}} = \begin{bmatrix} \lambda_{1} & 0 & \cdots & 0\\ 0 & \lambda_{2} & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_{n} \end{bmatrix} \end{align*}

But we also know that $[T]_{\mathcal{B}} = [[T(b_{1})]_{\mathcal{B}},[T(b_{2})]_{\mathcal{B}},\ldots,[T(b_{n})]_{\mathcal{B}}]$.

Consequently, $T(b_{j}) = \lambda_{j}b_{j}$ and we are done with the first part.

Let us prove the converse implication $(\Leftarrow)$ now.

Let us suppose there exists a basis $\mathcal{B} = \{b_{1},b_{2},\ldots,b_{n}\}$ and scalars $\lambda_{j}\in\textbf{F}$ s.t. $T(b_{j}) = \lambda_{j}b_{j}$.

Thus it results that $$[T]_{\mathcal{B}} = [[T(b_{1})]_{\mathcal{B}},[T(b_{2})]_{\mathcal{B}},\ldots,[T(b_{n})]_{\mathcal{B}}] = [[\lambda_{1} b_{1}]_{\mathcal{B}},[\lambda_{2}b_{2}]_{\mathcal{B}},\ldots,[\lambda_{n}b_{n}]_{\mathcal{B}}] = \text{diag}(\lambda_{1},\lambda_{2},\ldots,\lambda_{n})$$

whence we conclude that $T$ is diagonalizable, and we are done.

Hopefully this helps.


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