# Finding other eigenvector and matrix $A$ given eigenvalues

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 $$A$$.

• What do you know about the eigenspaces of a real symmetric matrix? – amd Mar 17 at 5:13

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. $$w^Tv=0$$.

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 $$A$$?

• 'symmetric matrix'. – user647486 Mar 17 at 4:33

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?