Let's say I fix a list of two real numbers $\sigma = (\sigma_1, \sigma_2)$, and I want to show that there exists a real, entrywise-nonnegative matrix $A$ with $\sigma$ as its spectrum.
How could I construct such a matrix?
My work:
I try to back out the matrix, knowing that I have some constraints to work with, namely, if $\sigma$ were to be the spectrum of $A$, then we must have that
$$\sigma_1 + \sigma_2 = tr(A)$$ $$\sigma_1 \sigma_2 = det(A)$$
Now, looking at the matrix
$$ A= \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} $$
with $a,b,c,d \ge 0$, I compute its eigenvalues to be
$$ \lambda_1, \lambda_2 = \frac{(a+d) \pm \sqrt{(a-d)^2+4bc}}{2} $$
So, in working the problem in reverse, I know that $\sigma_1, \sigma_2$ must have the above form.
And that the matrix's entries are to be expressed in terms of $\sigma_1, \sigma_2$.
But the problem is: how can I actually solve for $a,b,c,d$ ? I don't see how the trace and determinant constraint equations are enough to help me construct the matrix.
Any help is appreciated.
Thanks.
If $\sigma_1,\sigma_2\ge0$ then $$ \begin{pmatrix}\sigma_1 & b\\0 & \sigma_2\end{pmatrix},\quad b\ge0 $$ does it. Suppose $\sigma_1<0$. Since the trace has to be nonnegative, it must be that $\sigma_2\ge-\sigma_1$. You can choose any nonnegative $a,b,c,d$ such that $$ a+d=\sigma_1+\sigma_2,\quad b\,c=a\,d-\sigma_1\,\sigma_2. $$ A possible choice is $$ a=d=\frac{\sigma_1+\sigma_2}{2},\quad b=c=\frac{-\sigma_1+\sigma_2}{2}. $$