Image and kernel of map of direct sum

Question

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 $\ker A\cap V=(\ker A\cap U_1)\oplus(\ker A\cap U_2)$?

Answer 1

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.

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I hope this helps ^_^

Answer 2

Your first question was already answered.

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 $U=(U\cap U_1)\oplus(U\cap U_2)$.