# Linear map is diagonalizable

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?

• 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 Jun 20, 2020 at 7:51
• 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. Jun 20, 2020 at 7:53
• 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. Jun 20, 2020 at 12:13

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.