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I am struggling with understanding this:

For an inner product of $\mathbb{R}^3$ defined by $\langle x,y\rangle = 2x_1y_1 -x_1y_2 -x_2y_1 + 5x_2y_2$ the matrix relative to the standard basis is:-

$$\begin{pmatrix}2&-1\\-1& 5\end{pmatrix}$$

if the substitutions $$x_1 = (2/3)x_1' + (1/3)x_2'$$ $$x_2 = (1/3)x_1' - (1/3)x_2'$$

and $$y_1 = (2/3)y_1' + (1/3)y_2'$$ $$y_2 = (1/3)y_2' - (1/3)y_2'$$ are made,then the inner product takes the simple form $\langle x,y\rangle = x_1'y_1' + x_2'y_2' = x'^ty'$ . I understand why it works and I understand the use of eigenvectors to form an orthonormal vectors. Have tried this but the eigenvalues are messy.

How do I arrive at the above substitutions?

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I have edited it to incorporate LaTeX.But I am not sure about the last equation.Please verify if that's what you wanted to say. –  Vishesh Sep 28 '13 at 10:46
    
Yes. When the subs are made and the inner product is multiplied out according to the definition, then the resulting matrix of the inner product is the identity matrix. So there are no x1y2 and x2y1 terms. So the inner product is in a simpler form. My question is how do I arrive at the correct substitutions in a systematic way. –  Paul Williams Sep 28 '13 at 10:56

1 Answer 1

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Define $\langle x,y\rangle_A:=x^tAy$ for a positive definit symmetric matrix $A$. Find a square root $N$ of $A$, i.e., $N^2=A$. (You know how to do this?). Then it is easy to derive that $$\langle x,y\rangle_A=\langle Nx,Ny\rangle_I,$$ where $I$ is the identity matrix.

In your example: $A=\begin{pmatrix}2&-1\\-1&5\end{pmatrix}$, $N=\begin{pmatrix}1&1\\ 1&-2\end{pmatrix}$ and btw. $N^{-1}=\frac 13 \begin{pmatrix}2&1\\ 1&-1\end{pmatrix}$.

Michael

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I see how all this works. Although there has been a change in the RHS <Nx,Ny> - originally it was N inverse in this. However using N inverse it does all work out. But where is the intuition here? Is there a more strightforward way of deriving the required substitution? In the text I am studying it was just introduced as if it were trivial. –  Paul Williams Sep 30 '13 at 10:03
    
If $N^2=A$, then $x^tAy=x^tN^2y=(x^tN)(Ny)=(N^tx)^t(Ny)=(Nx)^t(Ny)$ since $N$ is symmetric. So define $x'=Nx$ and $y'=Ny$ to obtain $x^tAy=x'y'$. So my solution stays correct. –  Michael Hoppe Sep 30 '13 at 11:19
    
I was not querying the validity of your solution. I see how it works. I was asking if there was a simpler way. It may help if I refer you to the text math.ust.hk/~mabfchen/Math111/Week13-14.pdf. The relevant section is on page 3. It seems to introduce the substitutions as if it were trivial. –  Paul Williams Sep 30 '13 at 14:35
    
In the mentioned text the transition matrix pops up like a Deus ex machina. Bad style. I showed you a way to derive that matrix -- which in the example is hard to find, you'll do calculations with ugly roots. –  Michael Hoppe Sep 30 '13 at 14:56
    
Excellent. Thank you. I thought I was missing something obvious. It has been driving me to distraction. –  Paul Williams Sep 30 '13 at 15:20

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