How to prove the equality of determinant?

Question

Let A be a square matrix, and B a nilpotent. (size of A and B is same)

Assume AB=BA

Show that, det(A+B)=det(A)

The matrices have coefficients in an arbitrary field? – Martin Brandenburg Nov 13 '14 at 16:31 — Martin Brandenburg, Nov 13 '14 at 16:31

score 3 · Accepted Answer · answered Nov 13 '14 at 16:30

3

Hint: Since they commute, they are simultaneously upper triangularizable. For some unitary $U$, we can write $$ UAU^* = \pmatrix{\lambda_1&&&&*\\&\lambda_2\\&&\ddots\\&&&&\lambda_n}: = T_A\\ UBU^* = \pmatrix{0&&&&*\\&0\\&&\ddots\\&&&&0} := T_B $$ Note that $\det(A + B) = \det(T_A + T_B)$

answered Nov 13 '14 at 16:30

Ben Grossmann

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Better replace "unitary" by "invertible" and $U^T$ by $U^{-1}$ if the matrices have coefficients in an arbitrary field, wlog algebraically closed. – Martin Brandenburg Nov 13 '14 at 16:32
thank you!! your solution is simple and wonderful!! – user192276 Nov 13 '14 at 16:33
@MartinBrandenburg does Schur-decomposition work in arbitrary fields? I'm not so sure this solution method extends. In fact, I'm not even sure the statement is true in fields of finite characteristic. – Ben Grossmann Nov 13 '14 at 16:36
Well as I've indicated, don't think about such complicated things special to $\mathbb{C}$. It is a general fact about algebraically closed fields that (as you say) commuting matrices are simultaneously upper triangulizable. The latter means that $U A U^{-1}$ is triangular for some invertible matrix $U$. – Martin Brandenburg Nov 13 '14 at 16:38
@MartinBrandenburg Ah, so simultaneous upper-triangularization does extend, even if the concept of a "unitary matrix" is no longer in play. The only complicated thing that's added with Schur is that over $\Bbb C$, we can choose our similarity (to an upper-triangular matrix) to be unitary. – Ben Grossmann Nov 13 '14 at 16:42

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

Here is an alternative proof which avoids triangularization. The proof works over any field $k$.

Case 1: $\det(A)=0$.

Then $\det(A+B)=0$, since otherwise $A+B$ is invertible, and then $A=(A+B)-B$ is invertible too (SE/119904), namely via the geometric series. This is a contradiction.

Case 2: $\det(A) \neq 0$.

Then $A$ is invertible. Then $C:=A^{-1} B$ is nilpotent with $\det(A+B)=\det(A) \det(1+C)$, so that it suffices to prove $\det(1+C)=1$. But this is clear since $0$ is the only eigenvalue of $C$ and therefore $1$ is the only eigenvalue of $1+C$. (In general, eigenvalues add when we add constants.)

How to prove the equality of determinant?

2 Answers2