I have a linear map $T:\mathbb R^n \rightarrow \mathbb R$. and this map is not a constant map that get all vectors into $0$ vector. if I have a vector $|u\rangle \in R^n$ and $|u\rangle \notin \operatorname{Ker}(T)$, how to prove that for any vector $|v\rangle$ in $\mathbb R^n$ there is a vector $|a\rangle$ in $\operatorname{Ker}(T)$ such that:
$|v\rangle = |a\rangle + ( \frac{T|v\rangle}{T|u\rangle})|u\rangle$
and what does this mean?
note: $\frac{T|v\rangle}{T|u\rangle}$ means vector $T|v\rangle$ is divided by $T|u\rangle$ and this is could happen when one vector is a coefficient of the another vector.
Hint: If we work backwards from the expression given, we see that for such a vector $|a \rangle$ to exist, it would have to be true that $$ |a \rangle = |v \rangle - \frac{T|v\rangle}{T|u\rangle} |u \rangle. $$ Show that this vector $|a\rangle$ will always be an element of $\ker T$. That is, show that $T|a\rangle = 0$.