Proving $\overline{\ker(A-\lambda {\rm Id})^{r}} = \ker(A-\overline{\lambda}{\rm Id})^{r}.$

Question

Is there a simple proof of $\overline{\ker(A-\lambda {\rm Id})^{r}} = \ker(A-\overline{\lambda}{\rm Id})^{r}$, with $A \in M(n,\mathbb{R})$?

I think induction might work, but if there are other methods I'd rather accept them.

Any help or tip would be appreciated.

Answer 1

You just need to remember the definitions. $x\in \overline{\text{Ker}(A-\lambda)}$ means $$ (A-\lambda)\bar x=0 $$ Then if you take complex conjugate and use $\bar A=A$ $$ (A-\bar\lambda)x=0 $$ and $x\in\text{Ker}(A-\bar\lambda)$

Nothing really changes if you add a exponent $r$ because the conjugate of a product is the product of the conjugates.