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Prove that the vector space V

$V = \begin{bmatrix} a& 0& 0& 0 \\ b& 0& 0& 0 \\ c& d& e& f \\ \end{bmatrix}$

is isomorphic to the vector space $M_{2,3}$ of all 2 x 3 matrices under usual addition of matrices and scalar multiplication.

My prof didn't really explain as to how to go about this, but I was thinking that I could transform this into a one-dimensional vector, such that:

$V = (a, 0, 0, 0, b, 0, 0, 0, c, d, e, f)$

$M_{2,3}$ = (g, h, i, j, k, l)

I was thinking maybe I could remove the 0 values in V so they will have the same dimension? I'm not really sure. Please help.

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Both are obviously 6-dimensional vector spaces isomorphic to $\mathbb{R}^6,$ ie vectors with 6 components.

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    $\begingroup$ So is this the way to go for addition proving of isomorphic matrices? $\begin{pmatrix} a& b& c& d& e& f \\ \end{pmatrix} + \begin{pmatrix} g& h& i& j& k& l \\ \end{pmatrix} = \begin{pmatrix} a+g& b+h& c+i& d+j& e+k& f+l \\ \end{pmatrix}$ $\leftrightarrow \begin{pmatrix} a \\ b \\ c \\ d \\ e \\ f \end{pmatrix} + \begin{pmatrix} g \\ h \\ i \\ j \\ k \\ l \end{pmatrix} = \begin{pmatrix} a+g \\ b+h \\ c+i\\ d+j \\ e+k \\ f+l \end{pmatrix}$ $\endgroup$ – ss1919 May 15 at 17:35
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The map $\begin{pmatrix} a & 0 & 0 & 0 \\ b & 0 & 0 & 0 \\ c & d & e & f \\ \end{pmatrix} \rightarrow \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}$

Is an isomorphism.


In general if you have clear bases for two vector spaces and they have the same dimension then you can give an explicit isomorfism in this way.

What I mean is that if $V$ has base $v_1,\dots, v_n$ and $U$ has base $u_1,\dots,u_n$ then the bijection $f(\sum\limits_{i=1}^n a_iv_i) = \sum\limits_{i=1}^na_iu_i$ is an isomorfism.

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Any two finite dimensional vector spaces of the same dimension over the same field are isomorphic.

In this case there are $6$ free variables for each, giving common dimension $6$.

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A basis, by definition, allows a vector space to be completely determined by linear combinations of the vectors in that basis, and this in turn means that any linear action is completely determined by the action on a basis. That is, given any basis $e_1, e_2, ...$ and a vector $v$, there is a unique tuple $c_1, c_2, ...$ such that $v = \sum c_ie_i$, and given any linear function $f$, $f(v) = f( \sum c_ie_i) = \sum f(c_ie_i) =\sum c_if(c_i)$. It follows that if $f(e_1), f(e_2), ...$ is a basis for the destination vector space, then $f$ is an isomorphism between the spaces. So to construct an isomorphism between two vector spaces, we need only pick bases of the spaces, and pick a bijection between those bases.

There are issues with infinite-dimensional spaces, but we have finite dimensional spaces here, so we don't have to worry about those caveats.

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