Positive semidefinite cone is generated by all rank-$1$ matrices.

Solution 1:

$C=\{xx^\top:x\in\mathbb{R}^n\}$ is a cone, because for any $c\in\mathbb{R}^+$ we have $$cxx^\top=(\sqrt{c}x)(\sqrt{c}x)^\top\in C.$$ Also observe that $-cxx^\top\not\in C$, since $\sqrt{-c}\not\in\mathbb{R}$. And $C$ is a pointed cone since $x=0\Rightarrow 0\in C$. (These aren't necessary, only facts about the cone.)

$xx^\top$ is rank one because each row is a scalar multiple of any other (nonzero) row. (Similarly for columns.)

$xx^\top$ is positive semi-definite for any $x\in\mathbb{R}^n$ because for any $y\in\mathbb{R}^n$ you have $$y^\top xx^\top y=(y^\top x)(y^\top x)^\top=\left(\sum_{k=1}^n y_k x_k\right)^2 \ge 0.$$

For some $xx^\top$ pick any nonzero $y$ such that $y^\top xx^\top y=0$, and any small positive definite matrix $M$. Then $$y^\top(xx^\top+M)y=y^\top xx^\top y+y^\top My=y^\top My>0.$$ Hence also $y^\top (xx^\top-M) y<0$, and $xx^\top-M$ is not positive semidefinite. This shows that $xx^\top$ is on the boundary.

Use the same argument for PSD matrices with one zero eigenvector. (When $M$ has a zero eigenvalue there is always a nonzero $y$ such that $y^\top M y=0$.)

We have that the space is convex: for any two $x_0,x_1\in\mathbb{R}^n$ and $0\le \lambda\le 1$ we know $$y^\top \big(\lambda x_0x_0^\top + (1-\lambda)x_1x_1^\top\big) y = \lambda y^\top x_0x_0^\top y + (1-\lambda) y^\top x_1x_1^\top y \ge 0.$$

Now you need to show how you can create an affine combination of matrices in the form $xx^\top$ to create any positive definite matrix: this will show they are all in the interior.