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Let $M$ be a matrix of size $n$x$m$ over the reals. The following statement is clearly right:

$(\exists x,y \in R^{n} ,x \neq y \; \land M*x =M*y) \rightarrow$ $M$ is non unitarary matrix

Does the opposite direction hold as well? i.e. does:

$M$ is non unitarary matrix $\rightarrow (\exists x,y \in R^{n} ,x \neq y \; \land M*x =M*y)$

?

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up vote 1 down vote accepted

No, the opposite direction does not hold. Consider, for example, $M = \begin{bmatrix}1 & 0\\\ 0&2\end{bmatrix}$.

$M$ is not unitary because $MM^\ast = \begin{bmatrix} 1& 0 \\ 0& 4\end{bmatrix} \neq I$.

On the other hand $M(v_1,v_2)^t = (v_1,2v_2)^t$ so $Mx = My$ iff $x = y$.

More generally, unitary means $MM^\ast = I$ while your condition only means that $M$ is $1-1$. More colloquially, unitary means "keeps all distances the same" while 1-1 means "doesn't change any nonzero distance to $0$", so from this perspective it's clear that there should be a difference between the two types of matrices.

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