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I have two eigenvectors: $(2, 1, -1)'$ with eigenvalue $1$, and $(0, 1, 1)'$ with eigenvalue $2$. The corresponding determinant is $8$. How can I calculate the $3\times3$ symmetric matrix $A$ and $AP$?

I cannot solve several variables of the matrix.

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    $\begingroup$ What do you mean by $AP$? $\endgroup$ – Dietrich Burde Feb 25 at 14:54
  • $\begingroup$ The product of the matrix and a standardized eigenvector. It would already be great to know A. The P vector is used for diagonalization. $\endgroup$ – B..H Feb 25 at 14:57
  • $\begingroup$ Is the matrix assumed to by symmetric? $\endgroup$ – Klaus Feb 25 at 14:58
  • $\begingroup$ Yes it is symmetric and positive definite. $\endgroup$ – B..H Feb 25 at 14:58
  • $\begingroup$ @DietrichBurde If $A$ is not symmetric, there are not enough constraints. $\endgroup$ – Klaus Feb 25 at 15:09
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This follows by an easy computation with linear equations. Denote the matrix coefficients of $A$ by $a_1,\ldots ,a_9$. Then the first eigenvalue equation gives $$ a_7=2a_1 + a_4 - 2,\; a_8=2a_2 + a_5 - 1,\; a_9=2a_3 + a_6 + 1. $$ The second equation gives $$ a_4=1-a_1,\; a_5= \frac{1}{2}(3-2a_2), \; a_6=\frac{1}{2}(1-2a_3). $$ Now we have $\det(A)=2(a_1-a_2+a_3)$. If you assume that $A$ is symmetric, we have $a_4=a_2$, $a_7=a_3$ and $a_8=a_6$. So we obtain $$ A=\begin{pmatrix} 2 & -1 & 1 \cr -1 & 5/2 & -1/2 \cr 1 & -1/2 & 5/2 \end{pmatrix} $$

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  • $\begingroup$ Could you please provide more detailed steps? I would really appreciate. $\endgroup$ – B..H Feb 25 at 15:06
  • $\begingroup$ Just write down $A\cdot (2,1,-1)^T=(2,1,-1)^T$ in the coefficients of $A$. This gives $a_7=2a_1+a_4-2$ etc. $\endgroup$ – Dietrich Burde Feb 25 at 15:06
  • $\begingroup$ You edited your answer a gazillion times, so maybe it comes from a previous version. Anyway, something is wrong with your statement $\det(A) = a_1 - a_2 + a_3$. It does not match with the final solution. $\endgroup$ – Klaus Feb 25 at 15:18
  • $\begingroup$ @Klaus Yes, I am sorry to edit my typos. Now it matches with the final solution, the matrix has determinant $8$. A typo can occur if one is too fast, although I find your "gazillion times" and the downvote too harsh for this. $\endgroup$ – Dietrich Burde Feb 25 at 15:20
  • $\begingroup$ Sorry, I didn't mean to say that editing a lot is a bad thing. In fact, it is good to correct typos and errors. $\endgroup$ – Klaus Feb 25 at 15:21
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The determinant is the product of the eigenvalues, hence you can compute the third eigenvalue. Moreover, you know two eigenvectors. As the matrix is assumed to be symmetric, you can complete the eigenvectors to an orthogonal basis. So you know the diagonalized form of $A$ and the transformation matrix. Can you take it from here?

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Note that these two eigen vectors are orthogonal, which should be. The determinant is just multiple of eigen values. So third eigen value is 4. As this is different from the other eigen values, the eigen vector corresponding to it should be orthogonal to the other eigen vectors. So the eigen vector corresponding to 4 is $(1,-1,1).$ Normalize these 3 vectors to get orthonormal vectors which are $u=\frac1{\sqrt{6}}(2,1,-1), v=\frac1{\sqrt{2}}(0,1,1), w= \frac1{\sqrt{3}}(1,-1,1).$ From spectral theorem $A=uu'+2vv'+4ww'.$

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