Is $\text{Im}(\sigma)\cap \text{Ker}(\sigma)=\{0\}$ true for any linear transformation?

Question

For a given linear transformation $\sigma:V\to V$. We know $\dim V= \dim (Im(\sigma))+\dim(ker(\sigma))$. Does this can imply $Im(\sigma)\cap\ker(\sigma)=\{0\}$ ?. I am confused when calculate some given linear transformation's kernel and image. And find their intersection alwasys be $\{0\}$. I can't prove it's true for any linear transformation . Also I couldn't find a counterexample.

Answer 1

For a given linear transformation $\sigma:V\to V$. We know $\dim V= \dim (Im(\sigma))+\dim(ker(\sigma))$

Which is true only when $V$ is finite dimensional!

In general , $Im(\sigma)\cap\ker(\sigma)=\{0\} $ cannot hold

For example, consider $T: \Bbb{R}^2 \rightarrow \Bbb{R}^2$ defined by $$T(x,y)=(y,0)$$Then $\text{Im}(T)=\text{ker}(T)$ and so $\text{Im}(T) \cap\text{ker}(T)=\text{ker}(T)=\text{Im}(T) \neq \{0\}$