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It is known that the set of all $m \times n$ matrices has a vector space isomorphism with the set of all linear transformations between two vector spaces $V$ and $W$ of finite dimensions $m$ and $n$ respectively. Does such a result hold when either $m$ or $n$ is equal to zero?

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what is a zero-dimension vector space? –  dineshdileep Dec 3 '12 at 9:15
    
@dineshdileep: The question is not "What is a zero-dimension vector space?". Rather: "What is an $m\times 0$-matrix?" –  Rasmus Dec 3 '12 at 9:30
    
@dineshdileep: Vector space generated by the empty set. –  Sugata Adhya Dec 3 '12 at 13:31
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A linear transformation must be zero whenever the domain has dimension $0$ (since linear transformations must map $0$ to $0$) and also whenever the codomain has dimension $0$ (here there is no choice, even without linearity). If follows that the vector spaces of $n\times0$ and of $0\times m$ matrices are both $0$-dimensional. This is in accordance with the fact that such matrices have $n0=0m=0$ entries (they are called empty matrices). So much is easy.

The harder part is to realise* that although neither the unique $n\times0$ matrix nor the unique $0\times m$ matrix have any entries, they are not the same. Indeed the $n\times0$ has no columns, but it has $n$ rows (each of length $0$, and therefore a bit hard to discern), while the $0\times m$ matrix has no rows but $m$ columns of length $0$. And one can multiply the $n\times0$ matrix by the $0\times m$ matrix, giving a $n\times m$ matrix which is not empty: it has $mn$ entries, all equal to $0$, corresponding to the composition of the linear maps $K^n\to K^0\to K^m$. In the opposite order one can only multiply the $0\times m$ matrix by the $n\times0$ matrix if $n=m$, in which case it gives the (very) empty $0\times0$ matrix, corresponding to the composition of the linear maps $K^0\to K^n\to K^0$ (which, being not only the zero map but also the identity map of $K^0$, has determinant $1$).

*A notorious blunder of Bourbaki is to state that all empty matrices are equal.

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If $m$ or $n$ is zero, then the vector space of linear transformations between $V$ and $W$ is the zero-dimensional vector space $\{0\}$.

To make the isomorphism work, one could argue that there is exactly one $m\times 0$ (or $0\times n$)-matrix, namely, the zero matrix all of whose entries vanish.

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Vanish in the sense of not existing at all. Entries of a zero matrix always have the property of being $0$, which is the usual mathematical sens of "to vanish", but in the case considered here there are no entries to begin with. –  Marc van Leeuwen Dec 3 '12 at 10:07
    
@MarcvanLeeuwen: Right, if there are no entries then all entries vanish. –  Rasmus Dec 3 '12 at 17:04
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