A linear map is an isomorphism if and only if the determinant is nonzero.
I would like to see a proof of this statement ?
A linear map is an isomorphism if and only if the determinant is nonzero.
1
$\begingroup$
linear-algebra
vectors
determinant
-
0This is just because a matrix is invertible if and only if $det\neq 0$ – 2017-01-03
-
0How do you define the determinant of a linear map? Via alternating $n$-forms? Or (yuk) by picking a basis and looking at matrices? – 2017-01-03
-
0I'm trying to prove and show A linear map is an not isomorphism if the determinant is zero, can any one show me a pooof ? – 2017-01-03
1 Answers
1
Let $V$ be an $n$ dimensional $F-$vector space. Let $T:V\to V$ be a linear map, and $B$ a basis of $V$. Then let $A$ be the matrix for $T$ with respect to $B$. Then $$A\, \mathrm{adj}(A) = \det(A)I_n$$ If $\det(A)$ is nonzero, then we see that $A$ is invertible, with $A^{-1} = \frac{1}{\det(A)}\mathrm{adj}(A)$. The above formula can be verified via direct computation, using the Laplace expansion to calculate the determinant.
Now suppose $\det(A) = 0$. If $A$ were invertible, then we would have $1=\det (I_n) = \det (AA^{-1}) = \det(A)\det(A^{-1}) = 0\cdot \det(A^{-1})=0$, which is impossible. Thus, $A$ is not invertible.
Now, one just needs to show that the determinant of $A$ is nonzero iff the determinant of $T$ is nonzero, and that $T^{-1}$ exists iff $A^{-1}$ exists.