Existence of a subspace of the domain of a linear map with specific properties

Let $$V$$, $$W$$ be two finite-dimensional vector spaces over $$\mathbb{K}$$ and $$T \in L(V,W)$$.

Does a subspace $$U$$ of $$V$$ with the following properties exist:

$$U \cap \mathrm{Ker}(T) = \{0\}$$ and $$\mathrm{Im}(T) = \{ T(u)\;|\;u\in U \}$$

My solution:

Let $$B_0 = \{ v_1, \dots, v_m \}$$ be a basis of $$\mathrm{Ker}(T)$$.

Because every basis of a subspace can be expanded to a basis of its vector space, $$B_0$$ can be expanded to a basis $$B$$ of $$V$$:

$$B = \{v_1, \dots, v_m, v_{m+1}, \dots, v_{n}\}$$

Then let $$U = \mathrm{span}\{v_{m+1}, \dots, v_n \}$$ be the span of the extended vectors.

U now fulfills the first property because the intersection of the span of two linearly independent sets only contains the zero vector.

For the second property I don't know how to get to the property, so far I have:

I now know that $$V = U \oplus \mathrm{Ker}(T)$$ and from $$\mathrm{dim}\;V = \mathrm{dim}(\mathrm{Ker}(T)) + \mathrm{dim}\;U$$ and $$\mathrm{dim}\; V = \mathrm{dim}(\mathrm{Ker}(T)) + \mathrm{dim}(\mathrm{Im}(T))$$

that $$\mathrm{dim}\;U = \mathrm{dim}(\mathrm{Im}(T))$$.

And I also know that $$T$$ restricted to $$U$$ is injective.

Do these observation help me solve the problem or could someone provide a hint? Thank you

• For $z = T(v) \in Im(T)$, write (uniquely as you showed) $v = u + a$ with $u \in U$ and $a \in \ker(T)$. So you get $z = T(v) = T(u)$. Jan 17, 2019 at 17:41
• Thank you for your help! Jan 17, 2019 at 17:59

Since $$V=U\oplus \ker(T)$$, any $$v\in V$$ can be written uniquely as $$v=u+w$$ with $$u\in U$$ and $$w\in \ker(T)$$. Now compute $$T(v)=T(u+w)=T(u)+T(w)=T(u)+0=T(u).$$ It now follows that $$T(V)=T(U)$$ as required.