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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 ?

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

2 Answers 2

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

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

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

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

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

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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 $f(A) = P f(\Lambda) P^{-1}$.

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