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Given three eigenvectors and three eigenvalues, how would you go about finding BOTH non-symmetric matrix A and symmetric matrix B?

EDIT 7/27: Sorry for not being specific enough~ In the problem I am given three linearly independent 4x1 eigenvectors u1, u2, and u3 and their respective eigenvectors. I found out that every vector pairing is orthogonal EXCEPT between u1 and u2 because their dot product is non-zero.

Then, I am asked to find a non-symmetric matrix A and a symmetric matrix B. Matrices A and B share these eigenvectors and eigenvalues.

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    $\begingroup$ Could you be a little more precise ? What should be the size of the matrices ? $\endgroup$
    – M. Dus
    Commented Jul 26, 2017 at 20:42
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    $\begingroup$ Can we presume that $\{ {\bf A}, {\bf B} \} \in \mathbb{M}^{3 \times 3}$? $\endgroup$
    – David G. Stork
    Commented Jul 26, 2017 at 20:43
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    $\begingroup$ Also, are there any assumptions on your vectors? Are they allowed to be equal? $\endgroup$
    – M. Dus
    Commented Jul 26, 2017 at 20:48

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The question is really unclear. In the following, I assume that the matrices you're looking for are 3x3 matrice, otherwise the answer is trivial.

If your given eigenvectors are linearly independent, then your matrix is completely determined by those three vectors. It is symmetric if and only if they are orthogonal.

Now, if they span a two dimensional vector space, then there are two cases.

If they generate two distinct eigenspaces (each of dimension 1), you can find a symmetric matrix with those eigenvectors if and only if the eigenspaces are orthogonal. You can easily check if this is the case, because in this situation, one the three vectors is proportional to another one. You can also find a non-symmetric matrix by declaring that a third eigenspace is not orthogonal to the first ones, or declaring that the matrix is not diagonalizable.

In the second case, they generate only one eigenspace. You can then find a symmetric matrix declaring that a second eigenspace is orthogonal to the first one and a non-symmetric matrix by declaring it is not, or declaring that the matrix is not diagonalizable.

Now, if your given three vectors span a one dimensional vector space, again you have enough room to do whatever you want.

Of course, in each situation, there are compatibility conditions with your given eigenvalues!

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