Kernel and image of a diagonalizable linear map

Question

I find it difficult to proceed with this proof: Let $f:V \to V$ be a diagonalizable linear map. Prove that $ker(f)=ker(f^m)$ and $im(f)=im(f^m)$ for every positive integer $m$.

Can anyone give me a clue?

Answer 1

For any basis $v_1,\cdots,v_n$ we have $im(f)= span(v_1,\cdots, v_n)$. In particular for a basis consisting of Eigenvector (existence by diagonalizabilty) we get:

$$im(f^k) = span(\lambda_1^kv_1,\cdots,\lambda_n^kv_n)=span(\lambda_1v_1,\cdots,\lambda_nv_n)=im(f).$$

Similar for the kernel.

Answer 2

I'll assume $V$ is finite-dimensional. Pick a basis $\{b_1,\ldots,b_n\}$ such that $f(b_i)=\lambda_i b_i$. Now, for $v\in V$ write $v=\sum_{i=1}^nv_ib_i$ and use linearity to conclude that $f^m(v)=0$ iff $f(v)=0$, so that $\text{ker} f = \text{ker} f^m$. For the image, you should be able to prove $\text{im} f^m\subset \text{im} f$. Use the dimension formula to conclude that equality holds.