# Another Inner Product Question.

• May 2nd 2008, 07:01 PM
pearlyc
Another Inner Product Question.
Let A be a real invertible $\displaystyle n$ x $\displaystyle n$ matrix. Show that

<x,y> ≡ $\displaystyle y^T$$\displaystyle A^T$$\displaystyle Ax$ = $\displaystyle (Ay)^T$$\displaystyle (Ax)$

defines an inner product in ℝ^n, where x and y are column vectors in ℝ^n. What happens when A is not invertible? (Note : $\displaystyle M^T$ is the transpose of matrix M, obtained by interchaging the rows and colums of M).
• May 3rd 2008, 12:06 AM
pearlyc
Anyone at all :\?
• May 3rd 2008, 01:02 AM
flyingsquirrel
Hi

What has to be shown : ($\displaystyle x,\,y,\,\in\mathbb{R}^n$, $\displaystyle \alpha\in \mathbb{R}$)
• $\displaystyle <x,y>\,=\,<y,x>$ (use $\displaystyle ^tA\cdot B=^t( ^tB\cdot A)$, and remember that two $\displaystyle 1 \cdot 1$ matrices (=two real numbers) are equal iff their transposes are equal)
• $\displaystyle <x,y+z>\,=\,<x,y>+<x,z>$
• $\displaystyle <\alpha x,y>\,=\,<x,\alpha y>\,=\,\alpha<x,y>$
• $\displaystyle <x,x>\,\geq \, 0$ and $\displaystyle <x,x>\,=\,0 \Leftrightarrow x=0$. To show both, use $\displaystyle <x,x>\,=\,^t(Ax)Ax$ and the definition of the product of matrices which involves a sum : $\displaystyle (Ax)_{i}=\sum_{k=1}^n a_{i,k}x_k$
• May 3rd 2008, 05:45 AM
pearlyc
Hmm, I still have no idea how to approach this :\ Can you demonstrate how to do the first property and the last property for me to see? Perhaps I could understand from there. Thanks!
• May 3rd 2008, 06:04 AM
flyingsquirrel
Let $\displaystyle x,\,y\in\mathbb{R}^n$.

First one :
• $\displaystyle <x,y>\,=\,^ty ^tAAx\,=\, ^t(^t(Ax)Ay)$
• $\displaystyle <y,x>\,=\,^tx ^tAAy\,=\, ^t(Ax)Ay$
But $\displaystyle <x,y>$ is a real numbers (i.e. $\displaystyle 1\cdot 1$ matrix) hence $\displaystyle ^t<x,y>\,=\,<x,y>$ and we get what we want.

Last one :
Denote $\displaystyle Ax=(Ax_i)=\left(\begin{array}{c} Ax_1\\ \vdots \\Ax_n\end{array}\right)$, $\displaystyle x=\left(\begin{array}{c} x_1\\ \vdots \\x_n\end{array}\right)$ and $\displaystyle A=(a_{i,j})$.
• $\displaystyle <x,x>\,=\,^t(Ax)Ax=\sum_{i=1}^n(Ax)_i^2\geq 0$
• If $\displaystyle x=0$, we check that $\displaystyle <x,x>\,=\,0$ because $\displaystyle (Ax)_i=\sum_{k=1}^na_{i,k}x_k=0$
• If $\displaystyle <x,x>\,=\,0\,=\,\sum_{i=1}^n(Ax)_i^2,$ hence, for all $\displaystyle i$ such that $\displaystyle 1\leq i \leq n$, $\displaystyle (Ax)_i=0$. This exactly means that $\displaystyle x$ is in the nullity of $\displaystyle A$. As $\displaystyle A$ is invertible, $\displaystyle x=0$.
• May 3rd 2008, 05:58 PM
pearlyc
Hmm, sorry but I still don't quite understand :(

Do I have to put like,

x = [ x1 x2 and y = [ y1 y2
x3 x4] y3 y4]

and insert them into whatever you provided above and work from there?

This is such a tough question :(
• May 3rd 2008, 11:06 PM
flyingsquirrel
Quote:

Originally Posted by pearlyc
Do I have to put like,

x = [ x1 x2 and y = [ y1 y2
x3 x4] y3 y4]

and insert them into whatever you provided above and work from there?

No, there is no need to add this. (at least, I don't know why it would be interesting to introduce this)
Quote:

This is such a tough question :(
Tell me what you don't understand and I'll try to explain it or to make the post #5 clearer.
• May 4th 2008, 12:07 AM
pearlyc
• May 4th 2008, 12:38 AM
flyingsquirrel
Quote:

Originally Posted by pearlyc

There may have a problem of notation : I write $\displaystyle ^tA$ the transpose of $\displaystyle A$. Anyway :

$\displaystyle <x,y>\,=\,^ty ^tAAx$ (that's the definition)

$\displaystyle <x,y>\,=\,^t(Ay)Ax$ because $\displaystyle ^t(BC)=^tC\cdot ^tB$ for any two matrices $\displaystyle B$ and $\displaystyle C$.

hence $\displaystyle <x,y>\,=\,^t(^t(Ax)Ay)$ because $\displaystyle ^t( ^t(Ax)Ay)= ^t(Ay) \cdot^t( ^t(Ax))\,=\, ^t(Ay)Ax$ (previous rule with $\displaystyle B=^t(Ax)$ and $\displaystyle C=Ay$)

Quote:

Can you show me how to do the other two as well ? Properties 2 & 3.
No. These ones are really basic questions that you should be able to answer by yourself. The only things you need to know is that $\displaystyle A\cdot (B+C)= A\cdot B+A\cdot C$ and that $\displaystyle (\lambda B)\cdot C=B\cdot(\lambda C)=\lambda (B\cdot C)$ with $\displaystyle A,\,B,\,C\in\mathcal{M}_n(\mathbb{R})$ and $\displaystyle \lambda\in\mathbb{R}$. Try to do it and, if you want, post your answer, I'll tell you if it's correct or not. :)
• May 4th 2008, 05:57 AM
pearlyc
Okay, I have a slightly clearer view of it now considering the confusion was about the notation. Hm, to prove Property 1 only require that two steps ?

I don't know how to explain how did it go from <x,y> to <y,x>.

I am quite doomed right now, this is my assignment question which dues tomorrow and I am pretty much still quite blur about it (Crying)
• May 4th 2008, 06:28 AM
flyingsquirrel
Quote:

Originally Posted by pearlyc
I don't know how to explain how did it go from <x,y> to <y,x>.

We got $\displaystyle <x,y>\,=\, ^t(^t(Ax)Ay)$ and we also know that $\displaystyle <x,y>$ is a real number so it can be seen as the one-one matrix $\displaystyle \begin{pmatrix}<x,y>\end{pmatrix}$. This gives us $\displaystyle ^t<x,y>\,= \,^t\begin{pmatrix}<x,y>\end{pmatrix} =\begin{pmatrix}<x,y>\end{pmatrix} =\,<x,y>$.
Hence $\displaystyle <x,y>\,=\,^t<x,y>\,=\, ^t\left( ^t( ^t(Ax)Ay)\right)= ^t(Ax)Ay=<y,x>$.

Quote:

I am quite doomed right now, this is my assignment question which dues tomorrow and I am pretty much still quite blur about it
Does your assignment only depend on one question ? :confused:
• May 4th 2008, 06:57 AM
pearlyc
Unfortunately, it does depend on this question.

So stressed out! :( I still don't know how to do.(Worried)
• May 4th 2008, 07:06 AM
flyingsquirrel
Quote:

Originally Posted by pearlyc
Unfortunately, it does depend on this question.

I know, you said it. :D I wanted to know it it *only* depends on it. (as you said "*my* assignment question")

Quote:

So stressed out! :( I still don't know how to do.(Worried)
What don't you know how to do ? The two remaining properties ? Show me what you've tried and I'll help you.
• May 4th 2008, 07:10 AM
pearlyc
Sigh. It is quite bad shape now.

What for ? $\displaystyle x$ is a vector that's to say a $\displaystyle n\cdot 1$ matrix. What would it change to say "x=a matrix" ?