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I need to prove: matrix A is diagonalizable iff $\exp(A)$ is diagonalizble. exp means exponent function.

I know to prove that if $A$ is diagonalizable so $\exp(A)$ is diagonalizable, but have a problem with the other side.

can I write $P^{-1}.\exp(A).P=D$ (since $\exp(A)$ is diagonalizable ) and operate log on both sides of the equation? if yes I'm done, or any other hints?

Educ
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CnR
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2 Answers2

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Since you already know that "$A$ is diagonalisable" implies that "$\exp(A)$ is diagonalisable", we show that if $A$ is not diagonalisable, then neither is $\exp(A)$.

$A$ is not diagonalisable $\Rightarrow$ $\exp(A)$ is not diagonalisable.

Let $A$ be non-diagonalisable and let $A=XJX^{-1}$ be its Jordan form, where $J$ is a direct sum of elementary Jordan blocks: $J=\oplus_{i=1}^l J_i$. The function $\exp(A)$ can then be written in the form $\exp(A)=X\exp(J)X^{-1}$, where $\exp(J)$ is again a direct sum of exponentials of elementary Jordan blocks: $\exp(J)=\oplus_{i=1}^l \exp(J_i)$. Since $A$ is non-diagonalisable and so is $J$, $J$ contains at least one non-trivial Jordan block $$ \tilde{J}=\lambda I + N\in\mathbb{C}^{k\times k}, \quad k\geq 2, $$ where $\lambda\in\mathbb{C}$ is an eigenvalue of $A$ and $N$ is the nilpotent matrix $$ N=\begin{bmatrix} 0 & 1 & & & \\ & 0 & 1 & & \\ & & \ddots & \ddots & \\ & & & 0 & 1 \\ & & & & 0 \end{bmatrix}. $$ If we show that $\exp(\tilde{J})$ of such a non-trivial block is non-diagonalisable, we are done. Looking again here how $\exp(J)$ looks like, we see that $$ \exp(\tilde{J})=\exp(\lambda)\exp(N) =\exp(\lambda)\sum_{i=0}^{k-1}\frac{1}{i!}N^i. $$ Note that $1$ is the only eigenvalue of the upper triangular matrix $\exp(N)$. So $\exp(N)$ is diagonalisable iff the kernel of the $k\times k$ matrix $\exp(N)-I$ is $\mathbb{C}^{k}$. This is, however, not the case as $\exp(N)-I$ is not a zero matrix. Therefore, $\exp(N)$ and consequently $\exp(\tilde{J})$ and consequently $\exp(A)$ is not diagonalisable.

Algebraic Pavel
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    This is working over $\Bbb C$. But consider this example: $$\exp\left(\begin{bmatrix} 0&-\pi\\pi&0\end{bmatrix}\right)=\begin{bmatrix}-1&0\0&-1\end{bmatrix}.$$ I just noticed loup blanc gave the same example earlier. – Ted Shifrin May 09 '14 at 23:47
  • @TedShifrin Yes, $\mathbb{C}$ is what I had in mind. – Algebraic Pavel May 09 '14 at 23:51
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    @PavelJiranek I don't understand a couple of things: I learned that jordan form can be written as one diagonal matrix plus one nilpotent matrix which is consisted of nilpotent jordan blocks. it that was you meant? furthermore, why exp(N) is diagonalisable iff the kernel of exp(N)-I is C^k? I can't see why.. and why if exp(N) is not diagonalisble we get that exp(J wave) is not diagonalisble? – CnR May 10 '14 at 21:32
  • @CnR Well, Jordan form consists of a block diagonal matrix where each block is an "elementary" Jordan matrix (multiple of identity plus nilpotent matrix). For the other question, the eigenvalues $1$ is the only eigenvalue of $\exp(N)$ of the algebraic multiplicity $k$. For $\exp(N)$ to be diagonalisable, it is required that its geometric multiplicity is $k$ as well. Equivalently, $\dim \mathrm{ker}(exp(N)-I)=k$. It is not hard to see that if a $k\times k$ matrix has a $k$ dimensional kernel if and only if it is equal to the zero matrix. – Algebraic Pavel May 11 '14 at 00:29
  • @PavelJiranek and you're using here that if exp(N) is not diagonalisible so exp (J wave) is not diagonalisble , why it that? – CnR May 11 '14 at 07:02
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    @CnR hum, because one is a nonzero scalar multiple of the other? – Algebraic Pavel May 11 '14 at 09:53
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Try the Taylor series expansion for the exponent. For small $t$, $e^{tA} \approx I + tA$

PA6OTA
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  • Why has this answer received so many downvotes? What is wrong with it? – math.n00b May 09 '14 at 23:57
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    Given $e^A$, how does one compute $e^{tA}$? – robjohn May 10 '14 at 15:21
  • @ PA6OTA, I just read your post. You want to use that $A=\lim_{t\rightarrow 0}\dfrac{e^{tA}-I}{t}$. You have two problems: firstly, if $e^A$ is diagonalizable, then how do you show that $e^{tA}$ is diagonalizable ? Secondly, the limit of a sequence of diagonalizable matrices is not necessarily diagonalizable. Last remark: I'm more and more annoyed by those who downvote without explanations. –  May 10 '14 at 16:55
  • I understand this is not a "clean" answer. For $e^{tA}$ instead of $e^A$: you can see that if $B$ is diagonalizable then $B^{1/n}$ is, too (considering $B = P^{-1} \Lambda P$ then $B = P^{-1} \Lambda^{1/n} P$. So, setting $t = 1/n$ seems to work. But this works cleanly only for $\lambda_i$ positive, or maybe over $\mathbb{C}$ as noted in the other posts. The limit transitions are trickier. Downvotes: my theory is that the same trolls who posted bad comments in this thread earlier, were at work here. But you guys can upvote ;0 – PA6OTA May 10 '14 at 21:17