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If $X$ is an $N\times M$ real valued matrix (with $N < M$), the product of its transpose with itself ($X^t\cdot X$) results in a square $M\times M$ matrix.

Is there some simple property that $X$ has to obey so the the product ($X^t\cdot X$) is the identity matrix?

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You might find this interesting: en.wikipedia.org/wiki/Orthogonal_matrix#Rectangular_matrices –  Adam Saltz Jan 2 '13 at 4:55
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It can't happen. Rank of $X^T X$ is at most the rank of $X$, so at most N. That's less than M. –  hardmath Jan 2 '13 at 4:55
    
Thanks @AdamSaltz! That is indeed interesting. –  mathhead Jan 2 '13 at 4:57
    
I would say that the simple property you want is orthogonality, which is defined as $X^TX=I$. But as @hardmath says, this is impossible if $N<M$. –  Mario Carneiro Jan 2 '13 at 5:00
    
@hardmath Ah yes, I see. Thanks! –  mathhead Jan 2 '13 at 5:07

1 Answer 1

up vote 1 down vote accepted

If $X$ can be obtained by removing columns from a square orthogonal matrix, then $X^TX=I$. (Note that this imposes the condition $N\geq M$.)

Proof: Let $X_{ij}=A_{ik_j}$, where $1\leq k_0<k_1<\dots<k_M\leq N$ selects the columns from the square matrix $A$ satisfying $A^TA=I$. In full sum form,

$$\begin{align} A^TA=I\Rightarrow\sum_aA_{ai}A_{aj}&=\delta_{ij} \\ \sum_aX_{ai}X_{aj}=\sum_aA_{ak_i}A_{ak_j}&=\delta_{k_ik_j}=\delta_{ij}\Rightarrow X^TX=I, \\ \end{align}$$

where $\delta_{k_ik_j}=\delta_{ij}$ because $k_i=k_j$ iff $i=j$.

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Thanks @mario-carneiro. Would upvote as well but i don't have enough reputation yet. –  mathhead Jan 2 '13 at 23:48

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