A matrix is positive semidefinite iff it can be written in the form $X'X$ 
I know the fact that a matrix is positive semidefinite if and only if it can be written in the form $X'X$. But how to prove it? 

Thanks in advance.
 A: OK, first off, I am assuming the notation $X'$ our OP Ian uses in his question is in fact, as I asked in my comment, the same as the conventional $X^T$, which I shall deploy throughout.
It is easy to see that a matrix $A$ such $A = X^TX$ is positive definite; we have
$\langle v, Av \rangle = \langle v, X^TXv \rangle = \langle Xv, Xv \rangle \ge 0. \tag{1}$
However, going the other way is much more difficult.  And the reason it is much more difficult is:
IT IS FALSE that every positive semi-definite matrix $A$ can be written $A = X^TX$.
However,
IT IS TRUE that every symmetric positive semi-definite matrix $A$ can be so written.
To see this, suppose $A = A^T$; then $A$ may be diagonalized by some orthogonal matrix $O$:  $O^TAO = \Lambda = \mathrm{diag} (\lambda_1, \lambda_2, . . . , \lambda_n)$, where $n$ is the size of $A$.  Then for any vector $x$, set $y = O^Tx$, so that $Oy = OO^Tx = x$; then
$$
\langle x, Ax \rangle = \langle Oy, AOy \rangle = \langle y, O^TAOy \rangle = \langle y, \Lambda y \rangle, \tag{2}
$$
so that taking $y = (y_1, y_2, . . . , y_n)^T$ yields
$$
\langle x, Ax \rangle = \sum_1^n \lambda_i y_i^2; \tag{3}
$$
this shows we must have all $\lambda_i \ge 0$, hence we can take 
$$
\Lambda^{\frac{1}{2}} = \mathrm{diag}\left(\lambda_1^{\frac{1}{2}}, \lambda_2^{\frac{1}{2}},\ldots, \lambda_n^{\frac{1}{2}}\right), \tag{4}
$$
and setting $X = O\Lambda^{\frac{1}{2}} O^T$, we see that 
$$
X^TX = O^{TT} \left(\Lambda^{\frac{1}{2}T} O^T O \Lambda^{\frac{1}{2}}\right) O^T = O \Lambda O^T = A, \tag{5}
$$
establishing the desired result.  Note we have exploited the symmetry of $\Lambda^{\frac{1}{2}}$ in performing this calculation.    
It should be observed that $A = X^TX$ implies $A$ symmetric: $A^T  = (X^TX)^T = X^TX^{TT} = X^TX = A$; thus a positive semi-definite $A$ would automatically be symmetric if the assertion $A= X^TX$ were true.
To see that $A$ positive semi-definite is not sufficient to force $A = X^TX$, consider the $2 \times 2$ matrix
$$A = 
\begin{bmatrix} \lambda & \epsilon \\
 0 & \lambda \end{bmatrix}, \tag{6}
$$
and let $v = (x,y)^T$.  Then 
$
\langle v, Av \rangle = (x, y)
\begin{bmatrix} \lambda & \epsilon \\
0 & \lambda \end{bmatrix}
\begin{pmatrix} x \\ y \end{pmatrix} = \lambda (x^2 + y^2) + \epsilon xy \tag{7}
$
I claim that, at least for $\lambda > 0$ and $\vert \epsilon \vert \le \lambda$,
$\lambda (x^2 + y^2) + \epsilon xy \ge 0 \tag{8}$
for all $(x, y)^T$.  To see this, note that we can, without loss of generality, take $\vert x \vert \le \vert y \vert$, since the quadratic form $\lambda (x^2 + y^2) + \epsilon x y$ is symmetric in $x$ and $y$.  Then we have
$\vert \epsilon x y \vert  = \vert \epsilon  \vert \vert x \vert \vert y \vert  \le  \lambda \vert y \vert ^2 = \lambda y^2, \tag{9}$
or
$- \lambda y^2 \le \epsilon x y  \le \lambda y^2, \tag {10}$
whence
$0 \le \lambda y ^2 + \epsilon x y \le \lambda (x^2 + y^2) + \epsilon x y, \tag {11}$
showing that $A$ is positive semi-definite.  But there is no similarity transformation which diagonalizes $A$; indeed, if for some nonsingular $W$, $WAW^{-1}$ were diagonal, we would have to have $WAW^{-1} = \lambda I$,
since the characteristic polynomial of a matrix is a similarity invariant.  But then we would have $W(A - \lambda I)W^{-1} = 0$; but this is impossible since $A - \lambda I = \epsilon N$,  where $N = [\delta_{i, i + 1}]$ is a non-vanishing, albeit nilpotent, matrix; it cannot be "similaritized" into vanishing.
In closing:  certain generaizations may be possible:  if $A$ is an $n \times n$ matrix of the form 
$A = \lambda I + \epsilon N, \tag{12}$
then  $A$ is positive semi-definite provided $\vert \epsilon \vert \le \frac{\lambda}{n - 1}$ for $\lambda > 0$; furthermore $A$ can't be diagonalized; the arguments are essentially the same as those given above.
Sorry about all the false starts, false alarms, and repetitive editing.   
Hope this helps.  Cheers.
A: Let me provide a rather elementary method for real symmetric matrix $A$ (since OP didn't mention in which field the entries are):
First, suppose that a real symmetric matrix $A$ is positive semidefinite, then we can find an invertible matrix $P$, such that $$P^{T}AP=\operatorname{diag}\{\underbrace{1,1,...,1}_r,\underbrace{0,0,...,0}_{n-r}\}.$$
Multiply $(P^{T})^{-1}=(P^{-1})^T$ and $P^{-1}$ on the left and right side of the RHS, we have
$$ A=(P^{-1})^{T}\operatorname{diag}\{\underbrace{1,1,...,1}_r,\underbrace{0,0,...,0}_{n-r}\}P^{-1} .$$
If we force the last $(n-r)$ rows of $P^{-1}$ to be zero, and denote the new matrix by $X$, then
$$ A=X^{T}X $$ and we are done.
To show the converse, it suffices to note that if $ A=X^{T}X $, then for every $x=(x_1,x_2,...,x_n)\ne 0$ we have $$xAx^{T}=xX^{T}Xx^{T}=(xX^{T})(xX^{T})^{T}\ge 0.$$
