# Help with critical points

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• Dec 6th 2011, 10:58 PM
tubetess123
Help with critical points
Assuming that A is a real symmetric n x n matrix, and http://latex.codecogs.com/png.latex?... e^{||x||^{2}}

prove that f has a critical point at x if and only if Ax = ?.

I need to find out what Ax must equal for a critical point, and I just can't come up with anything.

I've tried taking the gradient and such, but I always end up with

http://latex.codecogs.com/png.latex?Ax = -(Ax \cdot x)x

I feel like that's not the right answer, though.
• Dec 7th 2011, 12:03 AM
CaptainBlack
Re: Help with critical points
Quote:

Originally Posted by tubetess123
Assuming that A is a real symmetric n x n matrix, and http://latex.codecogs.com/png.latex?... e^{||x||^{2}}

prove that f has a critical point at x if and only if Ax = ?.

I need to find out what Ax must equal for a critical point, and I just can't come up with anything.

I've tried taking the gradient and such, but I always end up with

http://latex.codecogs.com/png.latex?Ax = -(Ax \cdot x)x

I feel like that's not the right answer, though.

Write $x=\sum_i x_i \hat{e}_i$, now write $f(x)$ in tems of the components, ...

When you have found the gradient set it to zero, and simplify and then translate back in to vector/matrix form

CB
• Dec 7th 2011, 06:27 AM
tubetess123
Re: Help with critical points
Can you please explain why multiplying by a unit vector and writing in summation notatioin works? I don't understand how this makes a difference.
• Dec 7th 2011, 11:36 AM
tubetess123
Re: Help with critical points
I don't understand how I get the gradient now. Doesn't setting x as that make it a scalar? Doesn't it need to be a vector? Can you please help me? I really have no clue.

Thanks!
• Dec 7th 2011, 03:48 PM
HallsofIvy
Re: Help with critical points
No, that doesn't make x a scalar. $\sum_i x_i\hat(e)_i$ is just the vector written in component form. In three dimensions, it would be $\vec{x}= x_1\vec{i}+ x_2\vec{j}+ x_3\vec{k}$.
• Dec 7th 2011, 04:49 PM
tubetess123
Re: Help with critical points
But how then, do I know what to do with A when I take the gradient? I learned that that gradient of $Ax \cdot x$ is 2Ax. But then I can't figure out what Ax must be in order for there to be a critical point. I don't know how Writing the vector in component form changes anything either. I'll keep trying though.
• Dec 7th 2011, 05:01 PM
tubetess123
Re: Help with critical points
I got down to $0 = Ax_i\hat{e}_i + x_i\hat{e}_i (Ax \cdot x)$. So I don't know how to determine what value of Ax will make that equation equal to 0. Also, to translate it back to regular form, is
$0 = Ax + x(Ax \cdot x)$ correct?

Thanks!
• Dec 7th 2011, 09:02 PM
CaptainBlack
Re: Help with critical points
Quote:

Originally Posted by tubetess123
Can you please explain why multiplying by a unit vector and writing in summation notatioin works? I don't understand how this makes a difference.

Because it turns your function into:

$f({\bf{x}})=\left[ \sum_i \sum_j A_{i,j}x_i x_j\right]e^{\sum_kx_k^2}$

CB
• Dec 7th 2011, 09:22 PM
tubetess123
Re: Help with critical points
I don't get what $A_{ij}$ is. Can you please explain where that comes from? I also don't get what I'd do to take the gradient of that equation. Can you please explain what I need to do to take the gradient? I guess part of my problem is I don't understand what the A component of the sum is.

Thanks! Sorry for all the lack of understanding. I'm doing my best.
• Dec 8th 2011, 12:37 AM
CaptainBlack
Re: Help with critical points
Quote:

Originally Posted by tubetess123
I don't get what $A_{ij}$ is. Can you please explain where that comes from? I also don't get what I'd do to take the gradient of that equation. Can you please explain what I need to do to take the gradient? I guess part of my problem is I don't understand what the A component of the sum is.

Thanks! Sorry for all the lack of understanding. I'm doing my best.

$A_{i,j}$ is the value in the $i$-th row and $j$-th column of $A$.

Note: because of the given condition $A_{i,j}$ is real and $A_{i,j}=A{j,i}$

CB
• Dec 8th 2011, 12:40 AM
CaptainBlack
Re: Help with critical points
Quote:

Originally Posted by tubetess123
I also don't get what I'd do to take the gradient of that equation. Can you please explain what I need to do to take the gradient? I guess part of my problem is I don't understand what the A component of the sum is.

Thanks! Sorry for all the lack of understanding. I'm doing my best.

What definition of Gradient are you working with?

CB
• Dec 8th 2011, 12:59 AM
tubetess123
Re: Help with critical points
I'm don't know what you mean by definition of gradient. I just do it as if I'm differentiating, but I call it the gradient.
• Dec 8th 2011, 01:14 AM
tubetess123
Re: Help with critical points
At this point I got it down to

$0 = \sum_j{A_{ij}x_{j}}+x_{i}\sum_{m,j}{A_{mj}x_{m}x_{ j}$.

But I don't know how to simplifiy it past that point. This is where I had gotten it to when not in component form I think. But now I just don't know where to go.

I'm so confused. Thank you for all your help so far. It's so greatly appreciated.
• Dec 8th 2011, 04:13 AM
CaptainBlack
Re: Help with critical points
Quote:

Originally Posted by tubetess123
At this point I got it down to

$0 = \sum_j{A_{ij}x_{j}}+x_{i}\sum_{m,j}{A_{mj}x_{m}x_{ j}$.

But I don't know how to simplifiy it past that point. This is where I had gotten it to when not in component form I think. But now I just don't know where to go.

I'm so confused. Thank you for all your help so far. It's so greatly appreciated.

Well if that is right, what you have is:

${\bf{Ax}}+({\bf{x^tAx }}){\bf{x}}=\bf{0}$

$[{\bf{A}}+({\bf{x^tAx}}){\bf{I}}]{\bf{x}}=\bf{0}$
• Dec 8th 2011, 06:30 AM
tubetess123
Re: Help with critical points
Is that right? I have no idea. And if it is, I still have no idea what Ax has to be for that equation to be 0.
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