Find symmetric matrix containing no 0's, given eigenvalues

Question

I'm preparing for a final by going through the sample exam, and have been stuck on this:

$$Produce\ symmetric\ matrix\ A ∈ R^{3×3},\ containing\ no\ zeros.\ \\ A\ has\ eigenvalues\ λ_1 = 1,\ λ_2 = 2,\ λ_3 = 3$$

I know $A = S^{-1}DS$, where A is similar to the diagonal matrix D, and S is orthogonal.
The diagonal entries of D are the eigenvalues of A.

I also know that A & D will have the same determinant, eigenvalues, characteristic polynomial, trace, rank, nullity, etc. I am not sure where to go from here though. How cna A be found with only the two pieces of information? It seems like too little information is given...

Answer 1

You are correct in observing that "too little" information is given in the sense that there are infinitely many such matrices. But you need to produce just one. So start with the diagonal matrix $D = \operatorname{diag}(1,2,3)$ and conjugate it by a simple (but not too simple) orthogonal matrix $S$. You don't want $S$ to be a block matrix because then $S^{-1} D S$ will also be a block matrix and so will have zeroes. For example, we can take

$$ S = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{pmatrix}, S^{-1} = S^T = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0\end{pmatrix} $$

and define

$$ A = S^{-1} D S = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0\end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0\end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{2}{\sqrt{6}} & \frac{2}{\sqrt{6}} & -\frac{4}{\sqrt{6}} \\ -\frac{3}{\sqrt{2}} & \frac{3}{\sqrt{2}} & 0 \end{pmatrix} \\ = \begin{pmatrix} \frac{1}{3} + \frac{2}{6} + \frac{3}{2} & \frac{1}{3} + \frac{2}{6} - \frac{3}{2} & \frac{1}{3} - \frac{4}{6} \\ \frac{1}{3} + \frac{2}{6} - \frac{3}{2} & \frac{1}{3} + \frac{2}{6} + \frac{3}{2} & \frac{1}{3} - \frac{4}{6} \\ \frac{1}{3} - \frac{4}{6} & \frac{1}{3} - \frac{4}{6} & \frac{1}{3} + \frac{8}{6} \end{pmatrix} = \begin{pmatrix} \frac{13}{6} & -\frac{5}{6} & -\frac{1}{3} \\ -\frac{5}{6} & \frac{13}{6} & -\frac{1}{3} \\ -\frac{1}{3} & -\frac{1}{3} & \frac{5}{3} \end{pmatrix}. $$

Then $A$ is symmetric and has eigenvalues $1,2,3$ (because it is similar to $D$).

Answer 2

Spectral theorem states that if $A\in\mathbb{R}^{3\times 3}$ and is symmetric then $$ A=QDQ^T$$ where $D=\text{diag}(\lambda_1,\dots,\lambda_n)$ and the columns of $Q$ are the corresponding (normalized) eigenvectors.

You're going backwards and what you need is $Q$. Given an orthonormal base of $\mathbb{R}^3$, i.e. $(\mathbf{u_1},\mathbf{u_2},\mathbf{u_3})$, then $$ Q=[\mathbf{u_1}\quad\mathbf{u_2}\quad\mathbf{u_3}] $$

Now you just need to find one such that $A$ contains no zeros.

Answer 3

Indeed too litle information is given and such a matrix is not unique. I tried two run of the mill Orthogonal matrix, and it gave acceptable results.

The test didn't required you to find the unique matrix A satisfying these conditions.

Have a nice day :-)