Suppose $T^2$ is diagonalizable and $\ker{T}=\{0\}$, and every eigenvalue of $T^2$ is nonnegative. Show that $T$ is diagonalizable. Suppose $T^2$ is diagonalizable and $\ker{T}=\{0\}$, and every eigenvalue of $T^2$ is nonnegative. Show that $T$ is diagonalizable.
Of course $T$ is an operator on $V$.
It seems to me that if I take a basis of eigenvectors of $T^2$ I can define:
$$S(e_j)=\sqrt{\lambda_j}e_j$$
So by definition $S$ is diagonalizable since every eigenvector of $T^2$ is an eigenvector of $S$ and so there's a basis of eigenvectors of $S$, and obviously $S^2=T^2$. I want to say that $T$ has to be of this form (perhaps $T(e_j)=\pm\sqrt{\lambda_j}e_j$), and therefore has to be diagonalizable. I feel that this explanation is flawed somehow, and obviously strange since I didn't use the given about $\ker{T}$.
Many thanks!
EdiT: Okay, I found a different proof(by someone else) that uses the minimal polynomial and @julien 's algebraic note. I'll give it a little more thought but I think I got this down. Thank you.
 A: These things become very easy with the following fact, related to the notion of minimal polynomial.
Fact: a linear operator $S\in L(V)$ over a finite-dimensional $K$ vector space $V$ is diagonalizable if and only if there exists a polynomial $p\in K[X]$ such that $p(X)=\lambda (X-\mu_1)\cdots (X-\mu_s)$ with $\mu_j$ pairwise distinct in $K$ and $p(S)=0$ (i.e. there exists an annihilating polynomial which splits into pairwise distinct linear factors over $K$).
Proof: I believe this is done  in most textbooks. The only if is easy: take $p(X)=\prod (X-\mu_j)$ with $\mu_j$ the pairwise distinct eigenvalues of $S$. The if follows from the fact that if $p(X)=\prod p_j(X)$ with $p_j$ relatively prime, then $\ker p(S)=\bigoplus \ker p_j(S)$. $\Box$
Application: with your assumptions, denote $\mu_j$ the pairwise distinct (nonnegative) eigenvalues of $T^2$. Note that they are all positive, since $\ker T=\{0\}$. Now $T^2$ is annihilated by $p(X)=\prod_{j=1}^s(X-\mu_j)$. Which means that $T$ is annihilated by
$$
q(X)=\prod_{j=1}^s(X^2-\mu_j)=\prod_{j=1}^s(X-\sqrt{\mu_j})(X+\sqrt{\mu_j})
$$
which is a product of pairwise distinct linear factors. So $T$ is diagonalizable by the fact above.
Note: Your argument does not work because, if $T^2e_j=\mu_je_j$, $e_j$ need not be an eigenvector for $T$. Consider $T=\pmatrix{1&0\\0&-1}$ and the vector $(1,1)$ for instance. Finally, you see why there is a problem if some $\mu_j$ is $0$, that is $\ker T\neq \{0\}$. This yields a factor $X^2$ which does not split into pairwise distinct linear factors. So you can't guarantee that $T$ be diagonalizable. Sometimes it will, other times it won't (cf. Gerry's example).
A: Hint: Jordan normal form. Since $\ker\ T = \{0\}$, it is nonsingular. Can nonsingular Jordan block vanish or otherwise become diagonalizable when computing $T^2$?
Expansion of the idea: Let $T = S J S^{-1}$ be a Jordan decomposition of $T$ (or, if seen as an operator, a matrix representation in a basis defined by $S$). Then
$$T^2 = T \cdot T = S J S^{-1} S J S^{-1} = S J^2 S^{-1}.$$
Now, $J = \mathop{\rm diag}(J_1, J_2, \dots, J_k)$ is a block-diagonal matrix, with $J_1,\dots,J_k$ being Jordan blocks. So,
$$J^2 = \mathop{\rm diag}(J_1^2, J_2^2, \dots, J_k^2).$$
Obviously, $T^2$ is diagonalizable if and only if $J^2$ is diagonalizable, which is possible if and only if $J_i^2$ are diagonalizable for all $i=1,\dots,k$.
Observe one such block:
$$J_i = \begin{bmatrix}
\lambda_i & 1 & 0 & \dots & 0 & 0 \\
& \lambda_i & 1 & \dots & 0 & 0 \\
& & \ddots & \ddots & & \vdots \\
& & & \lambda_i & 1 & 0 \\
& & & & \lambda_i & 1 \\
& & & & & \lambda_i \\
\end{bmatrix}.$$
Now, try to see what is $J_i^2$. You should get a nondiagonalizable matrix (with diagonals $\lambda_i^2$, $2\lambda_i$, $1$) whenever $J_i$ is of order greater than $1$.
