# Decomposition of a positive semi -definite matrix

Prove the following: "If $$A$$ is a positive semi-definite matrix then, $$A=BB^{t}$$ for some $$B$$. In particular, we can take $$B$$ as full column rank."

Since, $$A$$ is a symmetric matrix, rank$$(A)$$=number of nonzero eigenvalues. All the eigenvalues of A is more or equal to $$0$$.

I can take the diagonal matrix $$D=diag(\lambda _{1},\lambda _{2}, ... ,\lambda _{r},0, ... ,0)$$, where $$\lambda _{i}> 0$$ $$(i=1,2, ... ,r)$$.

Let $$P$$ be an orthogonal matrix such that $$P^{t}AP=D$$. Then, $$A=PDP^{t}$$.

I can factor $$A$$ as $$A=PD^{1/2}D^{1/2}P^{t}$$.

But in this case, I cannot say that $$B=PD^{1/2}$$ has full column rank.

I know $$A$$ can be expressed as $$A=BB^{t}$$ for some $$B$$, but I don't know whether I can take a matrix $$B$$ as full column rank.

If $$D$$ has zeros on the diagonal, you want to let $$B$$ have fewer columns than $$n$$. Thus
$$D = diag(\lambda_1, \ldots, \lambda_r, 0, \ldots, 0) = C^t C$$ where $$C$$ is $$n \times r$$ with $$C_{ii} = \lambda_i^{1/2}$$ for $$1 \le i \le r$$, otherwise $$C_{ij} = 0$$. You can then take $$B = PC$$.
However, the statement does not work for $$A = 0$$, unless you allow $$n \times 0$$ matrices.