Let $\mathbb{F}^{m \times n}$ be the vector space of all $m \times n$ matrices and let $\mathcal{L}(V, W)$ be the vector space of all linear maps from a vector space $V$ to a vector space $W$ ($V$ and $W$ both over $\mathbb{F}$) with $\dim V = n$, $\dim W = m$. Is $M_{B_V, B_W}: \mathcal{L}(V, W) \rightarrow \mathbb{F}^{m \times n}$ an isomorphism for each choice of bases $B_V$ of $V$ and $B_W$ of $W$?

EDIT: Sorry if this question did not seem motivated - my main reason for asking is that I am confused by matrices in $\mathbb{F}^{m \times n}$ whose entries are all the same or in which there is a repetition of columns: in other words, how can such a map be surjective? How can an $a_{ij} \in \mathbb{F}^{m \times n}$ whose entries are all the same or in which there is a repetition of columns correspond to $B_V$ and $B_W$?

  • $\begingroup$ Yes, yes it is. If you'd like more details, you should begin by adding more details to your question. In particular, check out the information here about How to ask a good question. $\endgroup$ – Michael Albanese Oct 21 '13 at 5:11
  • $\begingroup$ Yes. ${}{}{}{}{}$ $\endgroup$ – copper.hat Oct 21 '13 at 5:11
  • $\begingroup$ @MichaelAlbanese: Thanks for the advice - I edited my question. I will make sure to take a look at that guide you linked me! $\endgroup$ – user102237 Oct 21 '13 at 5:37

$\mathcal{L}(V, W)$ is the set of all linear maps $V \to W$, they need not be injective or surjective. The map $M_{B_VB_W}$ takes a linear map $\varphi : V \to W$ and associates to it the unique matrix $A_{\varphi} \in \mathbb{F}^{m\times n}$ such that $[\varphi(v)]_{B_W} = A_{\varphi}[v]_{B_V}$; here $[\,\cdot\,]_{B_W}$ and $[\,\cdot\,]_{B_V}$ mean the components with respect to the bases $B_W$ and $B_V$ respectively. That is, $M_{B_VB_W}(\varphi) = A_{\varphi}$. If the matrix $A_{\varphi}$ has repeated columns, that just means the map $\varphi$ is not injective.

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  • $\begingroup$ I understand this - I am not talking about the maps in $\mathcal{L}(V, W)$. I am talking about the map $M_{B_V,B_W}$ $\endgroup$ – user102237 Oct 21 '13 at 5:47
  • $\begingroup$ Okay...this makes wonderful sense. I should have known that this is exactly what you meant in your original answer. Thanks! $\endgroup$ – user102237 Oct 21 '13 at 5:52

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