Diaonalized Matrix of the form $S^2=D$

Question

If $D$ is a diagonal matrix, with non-negative eigenvalues, prove that there is a matrix $S$ such that $S^2 = D$

Answer 1

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 $S^2 = D$.

Answer 2

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.