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Artin - domains don't seem to match

  1. Oct 21, 2014 #1
    This isn't a homework problem, but rather a bit of confusion regarding something in the textbook we're using; if this isn't the right place, feel free to move it.

    From Artin's Algebra pages 422/423 (slightly paraphrased):

    Let ##Q=\begin{bmatrix}1&\\3&1\end{bmatrix}##, ##A=\begin{bmatrix}2&-1\\1&2\end{bmatrix}##, ##P=\begin{bmatrix}1&1\\1&2\end{bmatrix}##, ##A'=Q^{-1}AP=\begin{bmatrix}1&\\&5\end{bmatrix}## (note: blank spaces are to be interpreted as zeroes.)

    Let ##M## be the integer lattice with its standard basis ##{\bf C}=\left(e_1,e_2\right)##, and let ##L## be the lattice with basis ##{\bf B}=\left(v_1,v_2\right)=\left(\left(2,1\right)^t,\left(-1,2\right)^t\right)##. Its coordinate vectors are the columns of ##A##. We interpret ##P## as the matrix of a change of basis in ##L##, and ##Q## as the matrix of change of basis in ##M##.

    My question is, if ##A## is interpreted as a map ##M\rightarrow L##, wouldn't we have to interpret ##P## as a map ##M\rightarrow M## and ##Q## as a map ##L\rightarrow L## to get ##A'##, which is just ##A## with different bases for ##M## and ##L##, to be ##M\rightarrow L##? Therefore ##P## would be interpreted as a change of basis in ##M## instead of ##L## and ##Q## in ##L## instead of ##M##? In fact, if we had set up the problem so ##A## was 2x3, ##Q## 2x2, and ##P## 3x3, ##M## would be ##\mathbb{R}^3## and ##L## would be ##\subseteq\mathbb{R}^2##, so ##P## could not be interpreted as a change of basis in ##L=\mathbb{R}^2## as it's 3x3.

    Is there something incredibly obvious I'm missing?
     
  2. jcsd
  3. Oct 22, 2014 #2
    Ah, derp. There was something obvious. It turns out M is to be interpreted as the codomain of the free module homomorphism corresponding to A.
     
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