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Let $V_k(\mathbb{R}^n)$ be the set of all orthonormal $k$-tuples of vectors $v_1,\ldots,v_k\in\mathbb{R}^n$. Let $M_{k,n}$ be the set of all $k\times n$ matrices. Let $W=\{A\in M_{k,n}\mid AA^t=I_k\}$, where $I_k$ is the identity $k\times k$ matrix. Prove that $V_k(\mathbb{R^n})$ can be identified with $W$.

(I'm not sure what "can be identified with" means. Maybe it just means showing there is a bijection.)

We just put in the vectors $v_1,\ldots,v_k$ as rows of $A$. Then these vectors are columns of $A_t$, so that $AA^t=I_k$ by definition of orthonormality. Conversely, given any $A$, we can let $v_1,\ldots,v_k$ be its rows.

Is there anything more to say about this?

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It is exactly as you say. One says "identified" because you are not defining a bijection at random (after all, both sets have the same cardinality as $\mathbb R$), but rather you are creating your bijection in terms of the properties of the objects you are using.

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  • $\begingroup$ Thank you, Martin. May I know why both sets have the same cardinality as $\mathbb{R}$? It does not seem obvious to me. $\endgroup$
    – Mika H.
    Commented Nov 9, 2013 at 3:44
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    $\begingroup$ You are welcome. The cardinality of $\mathbb R^n$ is the same as that of $\mathbb R$, so that's the upper limit. So all we need to show is that there are uncountably many orthogonal tuples. And that already happens for the $2$-tuples: $(\cos t,\sin t)$, $(-\sin t, \cos t)$ for $t\in[0,2\pi)$. $\endgroup$ Commented Nov 9, 2013 at 3:55

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