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I am trying to create a matrix for a linear map from V to V and calculate the determinant. This would be way easier if I could use two different bases on the horizontal and vertical directions, but I'm not sure if this is correct. Can I do this?

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    $\begingroup$ It depends. You’re looking at a linear transformation $f:V\to V$. For some purposes you may use different bases in the domain and the codomain. Are you interested only in the zeroness/nonzeroness of the determinant? In this case, yes. If, on the other hand, you are interested in the specific nonzero numerical value of the determinant, the answer is no. $\endgroup$ – Lubin Nov 23 at 4:37
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To choose a basis is to choose an isomorphism $\phi : V \to \Bbb R^n$. Then the matrix $M$ is the linear map defined by $M = \phi\circ f\circ \phi^{-1}$, and $$\begin{align}\det M &= \det(\phi\circ f\circ \phi^{-1})\\ &=(\det \phi)(\det f)(\det \phi^{-1})\\ &=(\det f)(\det \phi)(\det \phi)^{-1}\\ &=\det f\end{align}$$

But suppose you use a different isomorphism $\psi$ for one of the conversions: $$\begin{align}\det M &= \det(\psi\circ f\circ \phi^{-1})\\ &=(\det \psi)(\det f)(\det \phi^{-1})\\ &=(\det f)(\det \psi)(\det \phi)^{-1}\end{align}$$

But there is no guarantee that $(\det \psi)(\det \phi)^{-1} = 1$. Both values are non-zero, since the maps are isomorphisms, but that is all we can say. Because they are non-zero, $\det M = 0 \iff \det f = 0$, but we cannot conclude that $\det M = \det f$.

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