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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.

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    $\begingroup$ Note that they didn't say that $(1, 0, 0, 1)$ spans the eigenspace with eigenvalue $2$. I think they are being sly, and omitting one eigenvector (and even trying to hide its existence as much as they can). $\endgroup$ – Arthur Feb 1 '16 at 11:57
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    $\begingroup$ Hint: consider $\langle (1,0,0,-1), (0, 1, 1, 0) \rangle^{\perp}$. $\endgroup$ – Andreas Caranti Feb 1 '16 at 12:00
  • $\begingroup$ @Arthur if so, how do I find that other eigenvector? $\endgroup$ – Aidan Leith Feb 1 '16 at 12:02
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    $\begingroup$ See my comment. You should know that eigenvectors relative to distinct eigenvalues are orthogonal. $\endgroup$ – Andreas Caranti Feb 1 '16 at 12:03
  • $\begingroup$ @AndreasCaranti Sorry Andreas, that notation seems beyond me, this is only first year university mathematics, so it may not have been covered. $\endgroup$ – Aidan Leith Feb 1 '16 at 12:03
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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 $a, b$.

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  • $\begingroup$ I think I understand, so therefore the remaining eigenvector relative to eigenvalue 2, must be (0,1,-1,0), correct? $\endgroup$ – Aidan Leith Feb 1 '16 at 12:22
  • $\begingroup$ Your vector is perfectly fine, and it is probably the answer your instructors expect. But any vector of the form $a (1, 0, 0, 1) + b (0, 1, -1, 0)$, with $b \ne 0$ would do. (And thanks.) $\endgroup$ – Andreas Caranti Feb 1 '16 at 12:59

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