# Image and kernel of map of direct sum

If $$A$$ is a matrix and $$V=U_1\oplus U_2$$ (finite dimensional). Then in general it's not true that $$AV=AU_1\oplus AU_2$$, because for example for $$U_1=(\mathbb{R},0)^T, U_2=(0,\mathbb{R})^T$$ and $$A=\begin{pmatrix}1&1\\1&1\end{pmatrix},$$ it holds that $$AU_1=AU_2$$ so obviously $$AU_1\cap AU_2\neq\{0\}$$. But what if $$A$$ is invertible on $$V$$? Is there any condition under which the statement is true?

Maybe a more direct question for what I want to do: I have two vectors $$u_1\in U_1$$ and $$u_2\in U_2$$ where $$U_1\cap U_2=\{0\}$$ such that $$V=U_1\oplus U_2$$. I want to check if $$A$$ is injective on $$V$$. Can I check if $$A$$ is injective on $$U_1$$ and $$U_2$$ separately? Is $$\ker A\cap V=(\ker A\cap U_1)\oplus(\ker A\cap U_2)$$?

If $$A$$ is invertible, then it's injective (indeed, bijective). So this might trivialize the problem you're interested in. But yes, if $$A$$ is invertible then it preserves direct sums in the sense that $$V = U_1 \oplus U_2$$ gets sent to $$AV = AU_1 \oplus A U_2$$. To see why,

1. Note images of subspaces are subspaces, so $$A U_1$$ and $$A U_2$$ are subspaces of $$AV$$.
2. Then let $$w \in AV$$. Then $$A^{-1}w = u_1 + u_2$$, so $$w = Au_1 + Au_2$$ and $$AV = AU_1 + AU_2$$
3. Finally, say $$w \in AU_1 \cap AU_2$$. Then $$A^{-1}w \in U_1 \cap U_2 = \{0\}$$ so $$w = A0 = 0$$

Altogether, this means that $$AV = AU_1 \oplus AU_2$$, as desired.

I hope this helps ^_^

Regarding your second question: no, you can't in general check injectivity on $$U_1$$ and $$U_2$$ separately. For a counterexample, consider $$V=\mathbb{R}^2$$ and $$A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$$ with $$U_1=\{\,(x,x)^T\mid x\in\mathbb{R}\,\}\qquad\text{and}\qquad U_2=\{\,(x,-x)^T\mid x\in\mathbb{R}\,\}$$ Then $$V=U_1\oplus U_2$$ but $$\ker A=\{\,(0,y)^T\mid y\in\mathbb{R}\,\}\ne0$$ while $$\ker A\cap U_1=\ker A\cap U_2=0$$.
This shows that it's not true in general for a subspace $$U$$ that $$U=(U\cap U_1)\oplus(U\cap U_2)$$.
However, this is true in some cases. For example if $$U_1\subseteq U$$, then for $$u\in U$$ we can write $$u=u_1+u_2$$ with $$u_1\in U_1$$ and $$u_2\in U_2$$ and we know $$u_1\in U$$, so $$u_2=u-u_1\in U$$ also and we have $$u_1\in U\cap U_1$$ and $$u_2\in U\cap U_2$$. It follows that $$U=(U\cap U_1)\oplus(U\cap U_2)$$.