0

The will be linear transofmation $T:V \rightarrow V$

The formula says that $\dim V=\dim(\operatorname{Ker}(T))+\dim(\operatorname{Im}(T))$

and $\operatorname{Ker}(T)\subseteq V$ and $\operatorname{Im}(T) \subseteq V$

Does it implies that $V=\operatorname{Ker}(T)\oplus \operatorname{Im}(T)$ iff $T$ is a homomorphism?

gbox
  • 12,867

2 Answers2

1

Actually, I just noticed that this is a duplicate. :'(

I'll leave my answer be for now.

@ScroogeMcDuck already gave an appropriate counterexample, but let me just elaborate a bit:

The following is a good thing to know:

Fact: $V = W\oplus W' \iff W+W' = V \text{ and } W\cap W' = \{0\}$

So, in our case, all we need to find as a counterexample to your "guess" is a $T: V\to V$ such that $\ker T \cap \operatorname{Im} T \neq \{0\}$. In fact, consider the example of "shift right" linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ given by $T(1,0)= (0,1), \ T(0,1) = (0,0)$. Well, clearly, $\ker T \cap \operatorname{Im} T = \operatorname{span}\{(0,1)\}\neq \{(0,0)\}$

A more higher view of this is the following (if you care):

Given a linear transformation $T: V\to V$, we have the following sequence of linear maps: $$\ker T \overset{\text{inclusion}}\to V \overset{T}\to \operatorname{Im} T$$

Let's limit ourselves to $V$ being finite dimensional for simplicity. By just choosing basis for $\ker T$ and then extending it to the basis for $V$, you can see that $\ker T\oplus \operatorname{Im} T = V$ exactly means that the above sequence is a "short exact sequence" $0\to \ker T \to V\to \operatorname{Im} T\to 0$, which just means that image of each arrow is exactly the kernel of the next arrow. Your question is about splitting exact sequence of vector spaces, which has nice answers.

chriseur
  • 1,037
0

No. Let $V=U=\mathbb{R}^2$, define $T=\begin{bmatrix} 0&1 \\\\ 0&0\end{bmatrix}$. $T$ is linear hence a homomorphism of vector spaces. Then $\ker (T)=\{(x, y)\in\mathbb{R}^2 | y=0\}= \text{im}(T)$. Clearly then $V=\mathbb{R}^2\neq \ker(T)\oplus\text{im}(T)$.

ScroogeMcDuck
  • 921
  • I don't think that is correct. Ker(T)={(x,0)} and Im(T)={(0,y)} hence $\mathbb{R}^2$ is a direct sum of Ker(T) and Im(T). – Sangeeta Mar 26 '14 at 21:20
  • False. Working with columns: $\ker(T)=\left{ \begin{pmatrix} x \ 0 \end{pmatrix} ~|~ x\in\mathbb{R}\right}$ and $\mathrm{im}(T)=\left{ \begin{pmatrix}x\0\end{pmatrix} ~|~ x \in\mathbb{R}\right}$. Working with row vectors: $\ker(T)=\mathrm{im}(T)=\mathrm{span}\big( (0,1) \big)$. This counterexample is valid. The formula $\dim(V)=\dim(\ker(T))+\dim(\mathrm{im}(T))$ only makes a statement about dimensions. – Oliver Braun Mar 26 '14 at 21:41