Interrelation between Kernel and Range space.

Question

Suppose $T$ is a linear transformation from $V$ to $V$, where $V$ is a vector space. Now suppose $\ker T$ is its kernel and $\operatorname{im} T$ is its image space. Both are subspaces of $V$. Now while doing problems involving $\ker T$ and $\mathrm{T}$ in $\mathbb{R}^3$. I have found some examples where $\ker T$ and $\operatorname{im} T$ are direct sum complements in the entire $\mathbb{R}^3$. So I got a question in my mind – can it be generalized?(An intuition also comes from Rank-Nullity theorem).So my question is that if $T$ is a linear transformation on $V$ then do $\ker T$ and $\operatorname{im}T$ necessarily form complements to each other? If the statement is false then I would be interested in a counterexample and if this statement is true: why? I also would be interested, in case the statement is false, what additional conditions are required to be imposed so that the same becomes true.

Answer 1

No, this will fail in any dimension. You can always use an $n\times n$ matrix with a $1$ in the upper-right corner and zeroes everywhere else.

The issue is that nothing stops $\operatorname{im}(T)\cap \ker(T)\ne 0$. For $V$ to be an internal direct sum of $W$ and $W'$, we need $W+W'=V$ and $W\cap W'=0$. Of course, we know by rank-nullity that $$\dim (\operatorname{im}(T))+\dim (\ker(T))=\dim V.$$ By a standard result, if $W\cap W'=0$, then $$ \dim (W+W')=\dim W+\dim W'-\dim(W\cap W')=\dim W+\dim W'.$$ So, if $\operatorname{im}(T)\cap \ker(T)=0$, we get $\dim (\operatorname{im}(T)+\ker(T))=n$ and hence $V=\operatorname{im}(T)\oplus \ker(T)$. So, a necessary and sufficient condition is triviality of the intersection.

Answer 2

Actually, there a condition for such endomorphisms:

(i) If $U_1,U_2$ are complementary subspaces of a vector space $V$, the projections $p_1, p_2$ from $V$ onto each factor satisfy the equality $\; p_i^2=p_i$.

(ii) Conversely, if an endomorphism $T$ of $V$ satisfies $T^2=T$, then we have a direct sum decomposition: $$V=\ker T\oplus \operatorname{im}T.$$