Error in my argument that $A^m=0$ implies $A^{m-1}=0$ for matrix $A$?

Question

So nilpotent is the matrix where $A^m =0$ but $A^{m-1}$ is not equal to $0$. Well here's my logic-

$$A^{m-1} = I \cdot A^{m-1} = A^{-1} \cdot A \cdot A^{m-1} = A^{-1} \cdot A^m = A^{-1} \cdot 0 = 0$$

This gives $A^{m-1}$ always zero when $A^m=0$! Now I am sure there's some glitch in my logic so can someone point that out? (Note that $A^{-1}$ here is inverse of matrix $A$)

Sorry for the weird symbols but I am not good at coding those math symbols.

Answer 1

Yes, but you don't know if $A^{-1}$ actually exist. In fact it does not.

Say $$A =\pmatrix{0&1&0\\0&0&1\\0&0&0}\implies A^2 = \pmatrix{0&0&1\\0&0&0\\0&0&0}$$

and $A^3=0$. Notice that $A^{-1}$ does not exist since $\det A =0$.

---

Why it newer exists? If $\lambda$ is it eigenvalue and $v$ eigenvector, then we have $$Av = \lambda v\implies 0=A^{n}v =\lambda^n v \implies \lambda =0$$

So all it eigenvalues are $0$ and thus $\det A =0$ and so $A^{-1}$ does not exists.

Answer 2

You're assuming that $A^{-1}$ exists!

You'll find that nilpotent matrices cannot have inverses. In fact, your contradiction proves this!

---

I hope this helps ^_^

Answer 3

Since $A^{m-1} \ne 0$, there is $x \ne 0$ such that $y:=A^{m-1}x \ne 0.$ Then we get

$$Ay=A^mx=0.$$

This shows that $0$ is an eigenvalue of $A$. Therefore $A$ is not invertible.