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This is a homework problem, but I feel like I'm struggling with not knowing facts from linear algebra. Apparently this is supposed to be an easy question but I hit a brick wall at the following point. Can anyone give me a tip or tell me if I'm going down the wrong road?

My approach:

Let $A$ be a symmetric $n\times n$ matrix, and then suppose $x$ is a real-valued column vector of dimension $n$ with not all entries equal to $0$. I need to show that $x^Te^{A}x> 0$.

(based on looking up the definition of positive definite on wikipedia)

Then

$\begin{eqnarray*} x^{T}(e^{A})x &=& x^{T}(\sum_{n=0}^{\infty}\frac{A^{n}}{n!})x\\ &=& \sum_{n=0}^{\infty}\frac{x^{T}A^{n}x}{n!}\\ &=& \sum_{n=0}^{\infty}\frac{x^{T}A^{n}x}{n!})\\ \end{eqnarray*}$

As noted below, from here on is incorrect:

$\begin{eqnarray*} &=& \sum_{n=0}^{\infty}\frac{(x^{T}Ax)^{n}}{n!\|x\|^{2(n-1)}})\\ &=& \sum_{n=0}^{\infty}\frac{(x^{T}A^{T}x)^{n}}{n!\|x\|^{2(n-1)}}\\ &=& \sum_{n=0}^{\infty}\frac{((Ax)^{T}x)^{n}}{n!\|x\|^{2(n-1)}}\\ &=& \sum_{n=0}^{\infty}\frac{((Ax)^{T}(x^{T})^{T})^{n}}{n!\|x\|^{2(n-1)}}\\ &=& \sum_{n=0}^{\infty}\frac{((x^{T}Ax)^{T})^{n}}{n!\|x\|^{2(n-1)}}\\ \end{eqnarray*}$

As you can see this brings me no closer to getting $x^Tx$ somewhere, which I may assume is greater than $0$.

So my conclusion is that using only the fact that $A$ is symmetric is not enough. Is there some result about symmetric matrices that I should use?

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1  
Can you detail the step from line 3 to line 4? Since $A$ is symmetric, then so is $e^A$. Then write $x^Te^Ax=x^Te^{A/2}e^{A/2}x=x^T(e^{A/2})^Te^{A/2}x=\lVert e^{A/2}x\rVert\geq 0$, with equality if and only if $x=0$. –  Davide Giraudo Oct 16 '11 at 19:27
    
Well sliding $x^T$ and $x$ into the power of $n$ would add $n-1$ extra scalar terms of the form $x\cdot x^{T}$ between all the $A$'s. since this is $\|x\|^{2}$. So to compensate for this I just divided on the numerator as well. (about to read your suggestion now) –  Kyle Schlitt Oct 16 '11 at 19:31
    
Oh wait. That wouldn't be a scalar so that's not true at all. $x^Tx$ is a scalar but $xx^T$ is $n\times n$. My mistake. –  Kyle Schlitt Oct 16 '11 at 19:32
    
Thanks Davide, I see how it works. BTW: wouldn't you get $\|e^{A/2}x\|^2$ ? Not that this would cause any problems since that's still $\geq 0$. –  Kyle Schlitt Oct 16 '11 at 19:37
    
Another dumb question, in order to prove that $e^{A}$ is symmetric. I need to flip a power with a transpose, can I do that? (I couldn't find the result anywhere yet but it sounds true.) I mean, is it true that $(A^{T})^{n} = (A^{n})^{T}$? –  Kyle Schlitt Oct 16 '11 at 19:46

3 Answers 3

up vote 3 down vote accepted

Since $A$ is symmetric and the operator $\cdot^T\colon A\in \mathcal M_n(\mathbb R)\mapsto A^T\in\mathcal M_n(\mathbb R)$ is continuous (as a linear operator in a finite dimensional vector space), the matrix $e^{A/2}$ is symmetric. Therefore, we have for $x\in\mathbb R^n$: $$x^Te^Ax=x^Te^{A/2}e^{A/2}x=x^T(e^{A/2})^Te^{A/2}x =(e^{A/2}x)^Te^{A/2}x=\lVert e^{A/2}x\rVert^2\geq 0,$$ and since $e^{A/2}$ is invertible, we have the equality if and only if $x=0$, which shows that $e^A$ is positive definite.

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I didn't think of using continuity of $\cdot^{T}$, but instead I just expanded $e^{A}$ and then took the transpose of each term. Would this work as well? (See my question above as a comment) –  Kyle Schlitt Oct 16 '11 at 19:54
1  
Right, because finite sums can be switched with transpose, so you need continuity to switch it with the infinite sum. But that doesn't deal with the issue of switching transpose and finite powers? Is that a standard property? –  Kyle Schlitt Oct 16 '11 at 20:00
    
Sorry don't answer that. Of course it is. –  Kyle Schlitt Oct 16 '11 at 20:10
    
Question: We would have equality if and only if $e^{A/2}x = 0$, no? This may cause problems even if $x\neq 0$? –  Kyle Schlitt Oct 16 '11 at 20:22
    
Yes, but the exponential of a matrix is invertible. –  Davide Giraudo Oct 16 '11 at 20:23

If $A$ is symmetric, it is diagonalizable. So write $$A = PDP^{-1}$$ where $D = \text{diag}(\lambda_1, ..., \lambda_n)$ is a diagonal matrix whose entries are the eigenvalues of $A$. Thus $$e^A = Pe^DP^{-1}$$ and it is easy to see that $e^D = \text{diag}(e^{\lambda_1}, ..., e^{\lambda_n})$. But since the exponential is always positive, this means that all the eigenvalues of $e^A$ are positive. Hence $e^A$ is positive definite.

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Thank you. These results you pointed out are probably exactly the ones I would need to use. And once you state them they are fairly easy to proof (I think). –  Kyle Schlitt Oct 16 '11 at 19:40

I assume that $A$ is a real matrix here. Here is a way that requires some knowledge about linear algebra but no calculation. By spectral theorem $A=U \Lambda U^{-1}$ where $\Lambda$ is the diagonal matrix, the entries of which are the eigenvalues of $A$ (hence non negative), call these $\lambda_i\geq 0$. Take now $e^A$ which is clearly symmetric, its eigenvalues are $e^{\lambda_i} > 0$, and you are done.

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