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This is a problem from the textbook Hoffman and Kunze (Sec 3.2, problem 12, page 84)

Let p, m and n be positive integers and F a field. Let V be the space of m by n matrices over F and W be the space of p by n matrices over F. Let B be a fixed p by m matrix and let T be the linear transformation, T(A)= BA. Prove that T is invertible if and only if p=m and B is an invertible m by m matrix.

I could prove the left pointing implication and I am trying to prove the right pointing implication. Suppose T is invertible, then this means that T is 1-1 and onto and thus, by the rank-nullity theorem we have that dim V = dim W and thus p=m (Is this a correct argument? ). I would be grateful if there is a slight hint about how to prove that B is invertible.


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up vote 2 down vote accepted

As to your first question: yes, since $\dim(V)=mn$ and $\dim(W)=np$, if $\dim(V)=\dim(W)$, then $mn=np$, so $m=p$ (since $m,n,p\gt 0$). So your argument that $p=m$ is correct.

To show that $B$ is invertible, let $\mathbf{x}\in\mathbf{N}(B)$, and consider the matrix $A$ that consists of a copy of $\mathbf{x}$ in each of its columns. What is $T(A)$? Since $T$ is invertible, what does that tell you about $A$?

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Dear Professor Magidin, Thank you so much for the answer. Here is my answer sketch following your ideas. If x is in the null space of B we have that Bx=0 and if we make A be the matrix with x in each of its columns, we have that T(A)=BA=0. Since T is invertible this implies that A=0 and thus x=0. This thus implies that B is invertible. Best Regards. – Shibi Vasudevan Jun 10 '11 at 22:43
@Shibi: Yes; to make it a a bit more explicit, this implies that $\mathbf{N}(B)=\{0\}$, and since $B$ is a square matrix, this implies that $B$ is invertible. – Arturo Magidin Jun 11 '11 at 2:25

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