a multiple choice question on non-negative definite matrices A symmetric matrix in $\mathbb{M}_n(\mathbb{R})$ is said to be non-negative definite if
$x^TAx≥0$ for all (column) vectors $x \in \mathbb{R}^n$. Which of the following statements
are true?
a. If a real symmetric $n×n$ matrix is non-negative definite, then all of its
eigenvalues are non-negative.
b. If a real symmetric $n×n$ matrix has all its eigenvalues non-negative, then
it is non-negative definite.
c. If $A\in\mathbb{M}_n(\mathbb{R})$, then $AA^T$ is non-negative definite.

I know that all of the above options are correct but I did not have proof of any one of them.can anybody tell me about the proof of them.
 A: Hints: The basic fact you need to know in order to prove part (b) is that $A$ is real symmetric if and only if it is orthogonally diagonalizable, i.e. $A$ is real symmetric if and only if $A=QDQ^T$ for some real diagonal matrix $D$ and real orthogonal matrix $Q$ (such that $QQ^T=Q^TQ=I$, i.e. $Q^T=Q^{-1}$).
(b) The eigenvalues of $A$ are the diagonal entries of $D$ (why?). So, if all eigenvalues of $A$ are nonnegative, $D$ has nonnegative diagonal entries. Now, for any vector $x$, let $y=Q^Tx$. Then $x^TAx=y^TDy=\sum_i d_iy_i^2$ and so ...
(a) You may use orthogonal diagonalization to finish this part too. If $x^TAx\ge0$ for every vector $x\in\mathbb{R}^n$, then in particular the inequality holds for $x=Q^Te_i$, where $e_i$ is the $i$-th vector in the standard basis of $\mathbb{R}^n$. Now, for such a vector $x$, what is $x^TAx$?
(a) But part (a) can be proved without orthogoanl diagonalization. If $\lambda$ is a real eigenvalue of $A$ and $v$ is an associated eigenvector, by considering $v^TAv$, you can show that $\lambda\ge0$.
(c) Note that $x^TAA^Tx=(A^Tx)^T(A^Tx)$. In general, for a vector $y$, what is $y^Ty$?
A: For part a)
Suppose A is a real symmetric $n$x$n$ non-negative definite matrix.  Let v be an eigenvector of A then $ Av =\lambda v$ where $\lambda$ is real since A is symmetric.  Since A is non-negative definite we know that $v^TAv \ge 0$ and hence $v^T\lambda v \ge 0$ and hence $\lambda \ge 0$ since $v^Tv$ is the square of the norm of the vector $v$ and hence is non-negative.
