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

[aashish.v]

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

v^TAv=v^TA^Tv


So far I wasn't able to think of anyway for proving this... any help will be appreciated.

 

[aashish.v]

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,,

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

But I need rigorous proof for the same...

Thank you. :)

 

[ehild]

What is the dimension of v^TAv and v^TA^Tv?
What is (v^TAv)^T? How do you transpose a matrix product?



ehild

 

[Dick]

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,,

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

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.

 

[aashish.v]

So this is what I understand...

since the equation $$v'Av$$ is a number... we can write...

$$v'Av=(v'Av)'=v'(v'A)'=v'A'v$$

Am I correct?

 

[ehild]

Yes, that was I wanted you to do.

ehild

 

[Mark44]

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.