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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.

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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.

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