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I have the following expression:

$$ \begin{bmatrix} 1 & 2 & -1 \\ 2 & -2 & -0.5 \\ 2 & 1 & 1 \end{bmatrix} \begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix} = \begin{bmatrix} 6 \\ -3 \\ 0 \end{bmatrix} $$

For which I have found $p_1 = 0$, $p_2 = 2$, $p_3 = -2$. The columns of the above matrix are the eigenvectors $e_1$, $e_2$, and $e_3$ of this matrix $A$: $$ A = \begin{bmatrix} 4 & -2 & 0 \\ -2 & 3 & -2 \\ 0 & -2 & 2 \end{bmatrix} $$

By expressing $y = p_1e_1 + p_2e_2 + p_3e_3 $ and $x = q_1e_1 + q_2e_2 + q_3e_3 $, I want to solve the equation $Ax = y$.

I have tried a brute force approach by simply constructing $x$ as a vector formed from the 3 eigenvectors, since I know the values. But this result in a very complicated algebra which involves 3 variables at each term in the matrix on the left hand side, which cannot be discernably solved using a cleaner linear algebra method. How should I proceed?

oldselflearner1959
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  • Have you tried using coordinate vectors with respect to the two basis and the diagonal representation of $A $ (wrt to the eigenvector basis). – AnyAD May 14 '18 at 22:30
  • Are coordinate vectors the same thing as eigenvectors? – oldselflearner1959 May 14 '18 at 22:32
  • No, coordinate vector of $x$ wrt to basis ${b_1,b_2,b_3}$ would be $(u_1,u_2,u_3)$ where $u'i's $ satisfy $x=u_1b_1+u_2b_2+u_3b_3$ – AnyAD May 14 '18 at 22:41

2 Answers2

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Let $[A]=D$ be the representation of $A $ wrt to the basis of eigenvectors. This is diagonal. The coordinate vectors of $x,y$, with respect to the same basis, are $(q_1,q_2,q_3)$ and $(p_1,p_2,p_3)$ respectively. So you can use these: [D(q_1,q_2,q_3)=(p_1,p_2,p_3).]

AnyAD
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Here is what AnyAD is alluding to:

If $ X $ is the matrix of eigenvectors of $ A $, then $ X^{-1} A X = D $ where $ D $ is the diagonal matrix with eigenvalues on the diagonal.

Now, if you want to solve $ A x = y $ you can instead solve $ X^{-1} A X X^{-1} x = X^{-1} y $ or, equivalently, $ D X^{-1} x = X^{-1} y $.

Now, you are saying that $ x = X q $ and $ y = X p $. So, $ q = X^{-1} x $ and $ p = X^{-1} y $. So, we find that $ D q = p $.

Anyway, you may be able to now piece it all together.

Robert van de Geijn
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  • Thanks for clarifying the answer. This is such a surprising find. I tried solving it and I got $q_1 = 6$ $q_2 = -3$ $q_3 = 0$. Why does it happen to be equal to $y$? Is this a coincidence or is there a deeper relationship? – oldselflearner1959 May 15 '18 at 00:44
  • I am not a details person, so I haven't worked through the problem to that degree. – Robert van de Geijn May 15 '18 at 01:25