Let A be an $n*n$ matrix. a). suppose $A^2=0$ (zero matrix).

Question

Let A be an $n*n$ matrix. a). suppose $A^2=0$ (zero matrix). Prove A is not invertible. b). suppose $AB=0$ for some nonzero $n*n$ matrix B. Could A be invertible?

a). proof : suppose A is invertible, then $A^{-1}A^{-1}AA=0*A^{-1}*A^{-1}$

then $(I_n)^2 \neq 0$

Hence not invertible.

b). suppose A is invertible, $A^{-1}*A*B=0*A^{-1}$ $(I_n)^2*B \neq 0$

contradiction Hence it can't.

Does this look ok? Any alternative proof?

Answer 1

Part a) Your's works but here's another- $$A^2=[0]$$ $$\det(A^2)=\det([0])$$ $$\det(A)^2=0$$ $$\det(A)=0$$ As the matrix A is singular, it is also non-invertible.

Part b) Your proof seems to have no problem and it is also how I would have done it.