(a) Prove that a $2 \times 2$ scalar matrix $A$ over a field $F$ has a square root (i.e., a matrix $B$ satisfying $B^2 = A$).
(b) Prove that a real symmetric matrix having the property that every negative eigenvalue occurs with even multiplicity has a square root. [Hint: Use (a).]
(a) Since $A$ is a $2 \times 2$ scalar matrix it is of the form $ \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} $. If $\lambda^{1/2} \in F$ then we can use the matrix $B = \begin{pmatrix} \lambda^{1/2} & 0 \\ 0 & \lambda^{1/2} \end{pmatrix} $. Otherwise take $B = \begin{pmatrix} 0 & \lambda \\ 1 & 0 \end{pmatrix} $
(b) Since A is symmetric we have that $A = QDQ^{-1}$ for where $D$ is a diagonal matrix with the eigen values of $A$ on the main diagonal. It suffices to find a matrix $B$ such that $B^2 = D$ since then $(QBQ^{-1})^2 = QBQ^{-1}QBQ^{-1} = QB^2Q^{-1} = QDQ^{-1} = A$.
I know that every non negative real number is a square of another real number, so I think that can be used to some how prove that the block on the main diagonal of $D$ with all the non negative eigen values is the square of another matrix? Also since every negative eigenvalue occurs with even multiplicity I can group the negative eigen values together to create $2 \times 2$ scalar matrix blocks along the main diagonal and by part (a) I know those blocks are the square of another matrix.
Is it true that if I can show that each block along the main Diagonal of $D$ is the square of another matrix then $D$ is the square of another matrix?