Find 3x3 matrix by determinant and 2 eigenvalues/-vectors

Question

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.

Answer 1

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 $$ A=\begin{pmatrix} 2 & -1 & 1 \cr -1 & 5/2 & -1/2 \cr 1 & -1/2 & 5/2 \end{pmatrix} $$

Answer 2

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?

Answer 3

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 $A=uu'+2vv'+4ww'.$