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

  • 1
    $\begingroup$ 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 $z = T(v) = T(u)$. $\endgroup$
    – C. Zhihao
    Jan 17, 2019 at 17:41
  • $\begingroup$ Thank you for your help! $\endgroup$
    – strelsol
    Jan 17, 2019 at 17:59

1 Answer 1


Try this:

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.


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