matrices - Finding a matrix given eigenvalues and eigenvectors.

I am asked to construct a $4 \times 4$ symmetric matrix, with given eigenvalues and eigenvectors. I understand how to actually get $A$ as a product of $P^T, D$ and $P$, when $D$ is the diagonal matrix, and $P$ is a matrix with the eigenvectors as columns.

The problem is that there is only three given eigenvectors, along with three eigenvalues (one is repeated), so my question is, how do you construct a $4 \times 4$ matrix with three eigenvectors?

For more information here is the actual question:

Let $A$ be a symmetric $4 \times 4$ matrix with real entries whose eigenvalues
are $−1$ and $2$. If $(1, 0, 0, −1)$, $(0, 1, 1, 0)$ is a basis for the eigenspace of eigenvalue $-1$ and $(1, 0, 0, 1)$ is an eigenvector of $A$ with eigenvalue $2$, find the matrix $A$.

Thank you.

Answer

Consider the usual dot product
$$(x_{1}, x_{2}, x_{3}, x_{4}) \cdot (y_{1}, y_{2}, y_{3}, y_{4}) = x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} + x_{4} y_{4}.$$
Then it is easy to see that $A$ is symmetric iff for all $x, y$
$$x A \cdot y = x \cdot y A.$$
(just take the $x = e_{i}, y = e_{j}$, where $e_{i}$ is the vector which is all zero except for a $1$ in the $i$-th position).

Now prove that if $x$ is an eigenvector with respect to $\lambda$, $y$ is an eigenvector with respect to $\mu$, and $\lambda \ne \mu$, then $x, y$ are orthogonal:
$$\lambda (x \cdot y) = (\lambda x) \cdot y = x A \cdot y = x \cdot y A = x \cdot (\mu y) = \mu (x \cdot y)$$
and $\lambda \ne \mu$ implies $x \cdot y = 0$.

Therefore in your case the missing eigenvector $v$ relative to the eigenvalue $2$ must be orthogonal to both $(1, 0, 0, −1)$ and $(0, 1, 1, 0)$. It is easy to check that this means
$$v = (a, b, -b, a) = a (1, 0, 0, 1) + b (0, 1, -1, 0),$$
for some $a, b$.