Diaonalized Matrix of the form $S^2=D$ If $D$ is a diagonal matrix, with non-negative eigenvalues, prove that there is a matrix $S$ such that $S^2 = D$
 A: Since $D$ is a diagonal matrix, its eigenvalues are on the diagonal, which you know are non-negative. You can show, using matrix multiplication that for a diagonal matrix, $S$, $S^2$ is $S$ with the values on the diagonal squared. Since all the diagonal values of $D$ are non-negative, then there must be a matrix $S$ such that $S^2 = D$.
A: You seem to be hung up on the fact that "you weren't given specific values".  Well, we know that the matrix is a diagonal matrix, so it has to look something like this:
$$
D = 
\pmatrix{
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&&\ddots&\vdots\\
0&0&\cdots&d_{nn}
}
$$
Now, we don't know what the values from $d_{11}$ to $d_{nn}$ are, but we know that every other  entry of the matrix is zero.
Now, what can we say about the eigenvalues of $D$? As Christopher said, the eigenvalues of $D$ are exactly the diagonal entries $d_{11}$ to $d_{nn}$.  Why is this the case?  Check the definition of an eigenvalue, and check to see why this has to be true.
Now that we know that $d_{11},\dots,d_{nn}$ are the eigenvalues, we know that they have to be non-negative.  Since they are non-negative numbers, they have a non-negative square root.  Now, look at the matrix
$$
S = 
\pmatrix{
\sqrt {d_{11}}&0&\cdots&0\\
0&\sqrt{d_{22}}&\cdots&0\\
\vdots&&\ddots&\vdots\\
0&0&\cdots& \sqrt{d_{nn}}
}
$$
What happens when you multiply $S$ by $S$, using the rules for scalar multiplication?  If you can't figure it out right away, try it for the $2\times 2$ and $3 \times 3$ version, and see if you can find the pattern.
