A certian Linear Algebra gimmick needed for a part of my project

aashish.v
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1. I need to prove that for any matrix A(n,n) and a vector v(n,1) the following is true...

vTAv=vTATv


So far I wasn't able to think of anyway for proving this... any help will be appreciated.
 
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I do understand intuitively how this can be true...
Because of the arrangement of the equation, both side of the equation gets the same value.
I am able to show that with small example,,

<br /> A=\left[\begin{array}{cc}<br /> p &amp; q\\<br /> r &amp; s<br /> \end{array}\right],v=\left[\begin{array}{cc}<br /> a &amp; b\end{array}\right]<br /> <br /> \Rightarrow v&#039;Av=v&#039;A&#039;v=a^{2}p+abr+abq+b^{2}s<br />

But I need rigorous proof for the same...

Thank you. :)
 
What is the dimension of vTAv and vTATv?
What is (vTAv)T? How do you transpose a matrix product?



ehild
 
aashish.v said:
I do understand intuitively how this can be true...
Because of the arrangement of the equation, both side of the equation gets the same value.
I am able to show that with small example,,

<br /> A=\left[\begin{array}{cc}<br /> p &amp; q\\<br /> r &amp; s<br /> \end{array}\right],v=\left[\begin{array}{cc}<br /> a &amp; b\end{array}\right]<br /> <br /> \Rightarrow v&#039;Av=v&#039;A&#039;v=a^{2}p+abr+abq+b^{2}s<br />

But I need rigorous proof for the same...

Thank you. :)

For two matrices A and B, ##\left(AB\right)^T=B^TA^T##. If you don't already know that prove it using sigma notation.
 
So this is what I understand...

since the equation v&#039;Av is a number... we can write...

v&#039;Av=(v&#039;Av)&#039;=v&#039;(v&#039;A)&#039;=v&#039;A&#039;v

Am I correct?
 
Yes, that was I wanted you to do.

ehild
 
aashish.v, when you post a problem, include your work in the inital post. Without the work you show in your 2nd post, your 1st post would normally earn you a warning for an unacceptable homework request.
 
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