$\dim(V) = \dim T(V) + \dim T^{-1}(0)$

Question

Let $T\colon V \rightarrow W$ a linear transformation between the real vector spaces $V$ and $W$ both with finite dimension.

How can i prove that $\dim(V) = \dim T(V) + \dim T^{-1}(0)$.

I can't understand this problem and how to solve it , if you can help me please.

Answer 1

Consider a space $F$ such as $V = F\oplus T^{-1}(0)$. Consider the restriction $T'$ of $T$ from $F$ to $T(V)$.

1. Let $x,y\in F.$

$$ T'(x) = T'(y)\implies T(x-y) = T(x)-T(y) = 0\implies x-y\in T^{-1}(0) $$but as $x,y\in F$: $$ x-y\in T^{-1}(0)\cap F= \{0\}\implies x=y. $$2. Let $y\in T(V)$. $\exists x\in V\,\,y=T(x)$. Let us write $x = x_0 + x_1$, $x_0\in T^{-1}(0) $ and $x_1\in F$.

$$ y = T(x) = T(x_0+x_1) = T(x_0)+T(x_1) = 0+T(x_1) = T(x_1) = T'(x_1) $$hence $T'$ is onto.

Conclusion: this proves the second equality in $$ \dim V - \dim T^{-1}(0) = \dim F = \dim T'(F) = \dim T(V) $$

Answer 2

Hint:

1. consider $\mathcal{A}=\{v_1,v_2,\dots,v_k\}$ such that $\{T(v_1),T(v_2),\dots, T(v_k)\}$ is a basis of $T(V)$;

2. consider a basis $\mathcal{B}=\{u_1,u_2,\dots,u_h\}$ of $\ker T=T^{-1}(0)$;

3. prove that $\mathcal{A}$ and $\mathcal{B}$ are disjoint;

4. prove that $\mathcal{A}\cup\mathcal{B}$ is a basis of $V$.