# an two different bases be used when calculating the determinant of a linear map from V to V?

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?

• 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. – Lubin Nov 23 at 4:37

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 $$\det M = \det f$$.