I am solving problem (Linear Algebra by Hoffman, Excercise 5.4.8) :
Let $V$ be the vector space of $n\times n$ matrices over the field $F$. Let $B$ be a fixed element of $V$ and let $T_B$ be the linear operator on $V$ defined by $T_B(A)=AB-BA$. Show that $\det T_B=0$.
But I can not find the meaning of the determinant of linear operator, not of matrix, in my textbook.
I assume $\det T=0$ means $T$ is not a bijection, but this does not provide the information about nonzero values of determinant. Moreover, I can not convince my assuming.
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$\begingroup$From section 5.6, p. 172, of Hoffman and Kunze:
...we see that the determinant of $T$ is the determinant of the matrix which represents $T$ in any ordered basis $\{\beta_1,\ldots,\beta_n\}$.
I haven't been able to find a definition of the determinant of a linear operator that appears prior to problem 5.4.8 in Hoffman and Kunze. However, the definition is hinted at in problem 5.3.11.
$\endgroup$ $\begingroup$The definition of the determinant can be done this way:
- define a determinant on matrices. Prove that $\det (A B) = \det (B A)$ or, alternatively, that $\det (AB) = \det A \times \det B.$
- define a pseudo-determinant on $E$ (depending on a choice of a certain basis $B$ of $E$), $$\det f = \det M(f,B) $$ is the determinant of the matrix of $f$ in the base $B$.
- prove that this application does not depend on the choixe of the basis, so that the ad hoc choice in step 2. does not matter, using the lemma proved in 1.
Determinant is the factor by which volume change. The new volume is the norm of the transformed n-blade, n-parallelotope, is $\Vert Av_1\wedge\ldots\wedge Av_n\Vert=\vert\det A\vert\cdot \Vert v_1\wedge\ldots\wedge v_n\Vert$. See also here. A zero determinant means that the mapping "collapses" some dimensions, so the map is not an injection.
$\endgroup$ $\begingroup$As said in the comment of filmor, the basis is irrelevant. Your "$T$ is not a bijection" is the key: $$T_B(I)= IB − BI = B - B = 0,$$ so $\ker T_B\ne\{0\}$ and $\det T_B = 0.$
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