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**Defunkt** Take any transformation that is not invertible; for example, $\displaystyle T:\mathbb{R}^2 \to \mathbb{R}^2$ defined by $\displaystyle T((x,y)) = (x,0)$. Since T is not invertible, $\displaystyle Ker T \neq \{0\}$. In fact, $\displaystyle Ker T = \{(x,y) \in \mathbb{R}^2 : x = 0\}$.

Then, $\displaystyle Ker T$ is spanned by $\displaystyle (0, 1)$, however $\displaystyle T(0,1) = (0,0)$ which is not a base.

This will work for any transformation that is not invertible:

Since it is not invertible, $\displaystyle Ker T \neq \{0\}$ but for any $\displaystyle w \in Ker T$, $\displaystyle Tw = 0$, and therefore the image of any base of $\displaystyle Ker T$ will be mapped to the zero vector.