# Proving $P$ exists and is invertible for $e^{A} = P\Lambda P^{-1}$

Given $$A \in M_{n}(\mathbb{R})$$ a diagonalizable matrix with real and distinct eigenvalues $$\lambda_1 , \dots, \lambda_{n}$$. How can I prove that there exists an invertible matrix $$P \in M_{n}(\mathbb{R})$$ such that $$e^{A} = P \Lambda P^{-1},$$ where $$\begin{equation*} \Lambda \equiv \begin{bmatrix} e^{\lambda_{1}} & \dots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \dots & e^{\lambda_{n}} \end{bmatrix}. \end{equation*}$$

My attempt :

We know $$e^{A} = \sum_{k=0}^{\infty} \frac{t^{k}}{k!} \begin{bmatrix} \lambda_{1} & & \\ & \ddots & \\ & & \lambda_{n} \end{bmatrix} =\begin{bmatrix} \sum_{k=0}^{\infty} \frac{t^{k}}{k!}\lambda_{1} & & \\ & \ddots & \\ & & \sum_{k=0}^{\infty} \frac{t^{k}}{k!}\lambda_{n} \end{bmatrix}$$ $$= \begin{bmatrix} e^{\lambda_{1}} & & \\ & \ddots & \\ & & e^{\lambda_{n}} \end{bmatrix}$$ This $$\implies \{e^{\lambda_{i}}\}$$ are the corresponding eigenvalues of $$e^{A}$$ since $$\Lambda$$ is a diagonal matrix. For the initial equality to be satisfied $$e^{A} = P \Lambda P^{-1}$$ , I conclude that both $$P$$ and $$P^{-1}$$ must be the Identity matrices. Moreover, $$P$$ exists since $$\Lambda$$ is diagonal .

Is my reasoning correct , if not, could someone perhaps guide me ?

• “We know $e^A=\dots$” No, we don’t. The matrix $A$ is diagonalizable, but not necessarily diagonal.
– amd
Feb 16, 2020 at 7:26

Diagonalize $$A$$ via $$A=PDP^{-1}.$$

$$P$$ is a matrix that diagonalizes $$A$$. Take $$P$$ to have columns that are the eigenvectors of $$A$$.

$$e^A=e^{PDP^{-1}}=\sum_{k=0}^\infty \frac{(PDP^{-1})^k}{k!} = I + PDP^{-1} + \frac{1}{2!} (PDP^{-1})(PDP^{-1}) + \frac{1}{3!} (PDP^{-1})(PDP^{-1})(PDP^{-1})+ \cdots$$

$$e^A= I + PDP^{-1} + \frac{1}{2!} (PD^2P^{-1}) + \frac{1}{3!} (PD^3P^{-1})+ \cdots$$

$$e^A = P \sum_{k=0}^\infty \frac{D^k}{k!} P^{-1} = Pe^{D}P^{-1}= P\Lambda P^{-1}.$$

I suggest you try to prove the following lemmas and then it will be trivial to prove your question.

Lemma 1. If $$\lambda_1 \neq \lambda_2$$, then corresponding eigenvectors are linearly independent.

Corollary. If all eigenvalues of $$A$$ are distinct, then it has $$n$$ linearly independent eigenvectors.

Lemma 2. If $$A$$ has $$n$$ linearly independent eigenvectors $$x_i$$, then $$\Lambda = P^{-1} A P = \operatorname{diag} \{ \lambda_i \}$$ where $$P=\begin{bmatrix}x_1 & \dots & x_n\end{bmatrix}$$.

Lemma 3. If $$A=P \Lambda P^{-1}$$, then $$A^n = P \Lambda^n P^{-1}$$.

Corollary. If $$A=P \Lambda P^{-1}$$ and $$f(x)$$ is an analytic function, then $$f(A) = P f(\Lambda) P^{-1}$$.