Let $V$ be a finite-dimensional vector space and $T:V\rightarrow V$ be linear.
(a) Suppose that $V = R(T) + N(T)$. Prove that $V = R(T)\oplus N(T)$.
(b) Suppose that $R(T)\cap N(T) = \{0\}$. Prove that $V = R(T)\oplus N(T)$.
MY ATTEMPT
(a) Let us take a vector $v\in R(T)\cap N(T)$. Thus $v = T(w)$ for some $w\in V$ and $T(v) = 0$.
Consequently, considering that $w = w_{1} + w_{2}$, one has
\begin{align*}
T(T(w)) = T(T(w_{1} + w_{2})) = T(T(w_{1})) + T(T(w_{2})) = T(T(w_{1})) = 0
\end{align*}
Hence $T(w_{1})\in N(T)$. Then I get stuck.
(b) Let $\mathcal{B}_{N} = \{v_{1},v_{2},\ldots,v_{m}\}$ be a basis for $N(T)$. Then we can extend it to $\mathcal{B}_{V} = \{v_{1},v_{2},\ldots,v_{n}\}$, which is a basis for $V$ ($\dim V = n$). Since $\mathcal{B}_{R} = \{T(v_{m+1}),T(v_{m+2}),\ldots,T(v_{n})\}$ spans $R(T)$ and $\dim R(T) = \dim V – \dim N(T) = n – m$, it results that $\mathcal{B}_{R}$ is a basis for $R(T)$ indeed.
If we prove that $\mathcal{B}_{N}\cup\mathcal{B}_{R}$ is LI, we are done. Indeed, this is the case.
Suppose otherwise, that is to say, assume the set $\{v_{1},v_{2},\ldots,v_{m},T(v_{m+1}),T(v_{m+2}),\ldots,T(v_{n})\}$ is LD.
Without loss of generality, we may assume that $T(v_{n})$ is a linear combination of the vectors $v_{1},v_{2},\ldots,v_{m}$.
Consequently, we would have that $T(v_{n})\in R(T)\cap N(T)$, which contradicts the fact the $R(T)\cap N(T) = \{0\}$.
Hence the proposed proposition holds.
Could someone help me with this?
Best Answer
Are you familiar with the rank-nullity theorem? if so, you can combine it with the following fact: if $A$ and $B$ are finite-dimensional subspaces, then $\operatorname{dim}(A+B)=\operatorname{dim}(A)+\operatorname{dim}(B) - \operatorname{dim}(A \cap B)$.