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I want to find a symmetric matrix $A$, whose eigenvalues are $4$ and $-1$. One of the eigenvectors corresponding to the eigenvalue $4$ is $(2,3)$. I want to find an eigenvector corresponding to the eigenvalue $-1$ and then find the matrix $A$.

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    $\begingroup$ What do you know about the eigenspaces of a real symmetric matrix? $\endgroup$ – amd Mar 17 at 5:13
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Consider general form

$$\begin{bmatrix} x & d \\ d & y \end{bmatrix}= \begin{bmatrix} 2/\sqrt{13} & a \\ 3/\sqrt{13} & b \end{bmatrix} \begin{bmatrix} 4 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 2/\sqrt{13} & a \\ 3/\sqrt{13} & b \end{bmatrix}^T $$

Additionally vector $v=\begin{bmatrix} a \\ b \end{bmatrix} $ can be normalized to unit length (as $[2 \ \ 3]^T$ was normalized) and it is orthogonal to $w=[2 \ \ 3]^T $ i.e. $w^Tv=0$.

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HINT

Pick your favorite eigenvector , say $(1, 0)^T$ and remember that the diagonalized form of $A$ looks like $A= V^{-1}DV$ for special $D$ and $V$. Can you reconstruct $A$?

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    $\begingroup$ 'symmetric matrix'. $\endgroup$ – user647486 Mar 17 at 4:33
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Recall that the eigenspaces of a symmetric matrix are mutually orthogonal. Thus, all eigenvectors with eigenvalue $-1$ are orthogonal to all eigenvectors with eigenvalue $4$. Can you come up with a nonzero vector that’s orthogonal to $(2,3)$? Once you’ve done that, you have a basis of $\mathbb R^2$ that consists of eigenvectors of $A$, therefore $A$ is diagonalizable. Can you construct $A$ from its diagonalization?

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