Solving Nilpotent Matrices: Show det(I-A)=det(I+A)=1

[kmjonec]

Homework Statement

Show that if A is nilpotent then det(I-A)=det(I+A)=1.

Homework Equations

I know that det(A)=0 if A is nilpotent and det(I)=1, so this seems like it follows logically. I also know that the tr(A)=0 and that tr(I-A)=tr(I+A)=n, and that the characteristic polynomial of A is xⁿ.

However, I am lost as to how to start this solution because I don't know what I can say about the det(I-A)? Any hints about where I could start?

 

[fzero]

Use

$$\det M = e^{\text{Tr} \ln M}$$

and the fact that the power series for $$\ln (I\pm A)$$ is finite.

 

[tiny-tim]

welcome to pf!

hi kmjonec! welcome to pf!

hint: (I + A)² ​