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My question is

Does there exist a real matrix $B_{n\times m}$ with $m<n$ for which $BB^T=I_n$?

Why do I need this?

Suppose we are given a real matrix $Q_{m\times n}$ (again, with $m<n$) and we want to show that $\det(QQ^T)\geq 0$. If we find such a $B_{n\times m}$ for which $BB^T=I_n$, then define $$A= \begin{pmatrix} Q_{m\times n} & 0\\ 0 & B_{n \times m} \end{pmatrix}$$ So $$\begin{align}\det(AA^T) &= \det\begin{pmatrix} QQ^T & 0\\ 0 & BB^T \end{pmatrix}_{(m+n)\times(m+n)} \\ &= \det\begin{pmatrix} QQ^T & 0\\ 0 & I_n \end{pmatrix}_{(m+n)\times(m+n)} \\ &=\det(QQ^T). \end{align}$$ Now, $A$ is a square matrix which means its determinant is non-negative, Q.E.D.

I tried to find an example, but couldn't. I just don't know how to start.

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    $\begingroup$ No, because $\text{rank}\left(XY\right)\leq \min\left(\text{rank}(X), \text{rank}(Y)\right)$. $\endgroup$ – Git Gud Jun 1 '14 at 23:19
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As Git Gud said, such $B$ cannot exist.

The natural way (to me) to do this is the following. For $x\in \mathbb C^n$ a unit eigenvector for an eigenvalue $\lambda$ of $QQ^T$, we have (noting that since $Q$ is real, $Q^T=Q^*$) $$ \lambda=\lambda x^*x=x^*(\lambda x)=x^*QQ^*x=(Q^*x)^*Q^*x\geq0. $$ So all eigenvalues of $QQ^T=QQ^*$ are non-negative, which implies that $\det QQ^T\geq0$.

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  • $\begingroup$ Thanks, Martin! That was really nice. However, you need to do some editions as Git Gud said. $\endgroup$ – Amir Hossein Jun 1 '14 at 23:36
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    $\begingroup$ @GitGud: indeed. Edited, thanks. $\endgroup$ – Martin Argerami Jun 1 '14 at 23:43

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