# Need help solving for this matrix

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Oct 28th 2010, 12:55 AM
Need help solving for this matrix
FIRST LET ME SAY: I would like to be able to do this problem for myself, so if you want to actually do the problem, please don't post the answer right away so I can try it myself first.

The problem is: suppose the vector x =
[1
0
2]

and A is a (3x3) matrix such that (Atranspose)(A)(x) =
[2
2
0]

what is || Ax || ^2

So far I'm pretty much completely lost. I was thinking I could set (Atranspose)(A) equal to C and then solve for Cx=[2,2,0] but to be honest I wasn't sure even how to do that. I know that if x had been a square matrix I could have multiplied both sides by the inverse of x but obviously x isn't a square matrix.

• Oct 28th 2010, 02:30 AM
Ackbeet
Think about the way matrices act inside inner products. You know that

$\|Ax\|^{2}=\langle Ax,Ax\rangle.$

Does this give you any ideas?
• Oct 28th 2010, 02:40 AM
No sorry :( Correct me if I'm wrong, but isn't there a way to solve for A, or at least (Atranspose)A?

How would I go about solving Bx = C if I know what x and C are but x isn't square?
• Oct 28th 2010, 02:48 AM
Ackbeet
Even using the fact that $A^{T}A$ is symmetric won't give you enough equations to solve for the elements of $A^{T}A$.

Quote:

How would I go about solving Bx = C if I know what x and C are but x isn't square?
You can't even write the equation Bx = C, if both B and C are matrices. The dimensions don't agree. The LHS is a vector, the RHS is a matrix.

Look at this:

$\langle x,A^{T}Ax\rangle=x^{T}A^{T}Ax=(Ax)^{T}Ax=\|Ax\|^{2 }.$ What does that tell you?
• Oct 28th 2010, 10:21 AM
I'm a little confused with your use of
$\langle x,A^{T}Ax\rangle$

What exactly does that represent? I thought at first you were writing (x, $a^{T}Ax$) but that doesn't make sense. Perhaps something I haven't learned yet?

Looking at what else you have written though, if $|| A x || ^2 = (Ax)^{T} Ax$ then I could multiply both sides of my original equation by $x^{T}$ and get an answer of: [2]
Am I on the right train of thought here? Also, I haven't ever seen a theorem that would help me prove that $||Ax||^2 = (Ax)^{T} Ax$ Is there an easy way to prove this?
• Oct 28th 2010, 10:25 AM
Ackbeet
It's the inner product. In this case, inner product = dot product. I like the angled brackets, because it enables me to distinguish from ordered pairs, or intervals. Parentheses are used for so many darn things!

So, I'm using the notation $\langle \mathbf{x},\mathbf{y}\rangle\equiv \mathbf{x}\cdot \mathbf{y}=\mathbf{x}^{T}\mathbf{y},$ where the last multiplication is matrix multiplication.

Make sense?
• Oct 28th 2010, 10:37 AM
Ahhh yes that makes much more sense. Ok I guess the only part I'm confused about now is :
$\|Ax\|^{2}=\langle Ax,Ax\rangle.$

I'm assuming this comes from some theorem I missed. Is there a way to prove it using steps that I would already know?
• Oct 28th 2010, 10:40 AM
Ackbeet
Well, it's the definition of Euclidean length, really. You have, for a vector $\mathbf{x}=(x_{1},x_{2},\dots,x_{n}),$ that

$\|\mathbf{x}\|=\sqrt{x_{1}^{2}+x_{2}^{2}+\dots+x_{ n}^{2}}=\sqrt{\mathbf{x}\cdot\mathbf{x}}.$
• Oct 28th 2010, 10:50 AM
oooooh, Ok I think I got it now.

So for this problem all I would have to do is show that $||Ax|| ^2 = (Ax)^{T} Ax$ and with that, I would get that $||Ax||^2 = [2,2,0] * x^{T}$. Correct?

edit: oh and that [2,2,0] matrix should be the (3x1) matrix I had before. I didn't know how to make it a 3x1 and not a 1x3
• Oct 28th 2010, 11:15 AM
Ackbeet
Almost, but not quite. You need to do

$(Ax)^{T}Ax=(x^{T}A^{T})Ax=x^{T}(A^{T}Ax).$

Now do you see what to do?
• Oct 28th 2010, 11:26 AM
I belive so. I think that's what I meant though. If we multiply the original equation by $x^{T}$ then we can simplify down to (Ax)^T (Ax).
• Oct 28th 2010, 11:57 AM
Ackbeet