$\ker\left(T\right)\cap\text{Im}\left(T\right)=\left\{ 0\right\}$ iff $\mathbb{R}^{n}=\ker\left(T\right)+\text{Im}\left(T\right)$ .

I'm trying to prove the following claim: Let $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ be a linear transformation. Then $\ker\left(T\right)\cap\text{Im}\left(T\right)=\left\{ 0\right\}$ iff $\mathbb{R}^{n}=\ker\left(T\right)+\text{Im}\left(T\right)$. The if part was easy but for some reason I can't manage to show the only if. Help would be appreciated.

Proof of if direction: Proof. Let $U\subseteq\ker\left(T\right)\subseteq\mathbb{R}^{n}$ and $V\subseteq\text{Im}\left(T\right)\subseteq\mathbb{R}^{n}$ be a basis for $\ker\left(T\right)$ and $\text{Im}\left(T\right)$ respectively. Since $\ker\left(A\right)\cap\text{Im}\left(T\right)=\left\{ 0\right\}$ no non-trivial vector in $\text{Im}\left(A\right)$ can be expressed as a linear combination of vector in $U$ and likewise no vector in $\text{Im}\left(T\right)$ can be expressed as a linear combination of vectors in $V$ and thus $U$ is linearly independent of $V$ . Furthermore by the rank-nullity theorem $$\left|U\right|+\left|V\right|=\text{dim}\left(\ker\left(T\right)\right)+\text{dim}\left(\text{Im}\left(T\right)\right)=\dim\left(\mathbb{R}^{n}\right)$$ So $U\cup V$ is a set of $n$ linearly independent vectors and thus a basis of $\mathbb{R}^{n}$ . This of course implies that every $x\in\mathbb{R}^{n}$ can be expressed as a unique linear combination of vectors in $U\cup V$ and in particular as a unique sum of two vectors in $\text{span}U=\ker\left(T\right)$ and $\text{span}V=\text{Im}\left(T\right)$ proving that $\mathbb{R}^{n}=\ker\left(T\right)\oplus\text{Im}\left(T\right)$ .

• Perhaps include your proof for the only-if direction in your question? It might itself contain a hint for the if direction. – Travis Oct 8 '15 at 14:34

If you know the Grassmann formula $$\dim(U_1+U_2)=\dim U_1+\dim U_2+\dim(U_1\cap U_2)$$ and the rank nullity theorem $$\dim\ker T+\dim\operatorname{Im}T = n$$ then you're done.