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I am trying to do the following question, which is Exercise 2 on page 241 of Linear Algebra by Hoffman and Kunze:

Let $T$ be a linear operator on the finite-dimensional space $V$, and let $R$ be the range [image] of $T$.

(a) Prove that $R$ has a complementary $T$-invariant subspace if and only if $R$ is independent of the null space $N$ of $T$ [$R \cap N = 0$].

(b) If $R$ and $N$ are independent, prove that $N$ is the unique $T$-invariant subspace complementary to $R$.

I am having trouble with even starting this problem. I do not have any intuition as why this would be true.

LinearGuy
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    You can use that the sum of the dimension of the range and of the null space is equal to the dimension of $V$. You then have two subspaces which by assumption do not intersect and whose sum of dimensions equals the dimension of the space. – Eric Dec 03 '18 at 21:24
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    By rank-nullity, $\dim R + \dim N = \dim V$ and the question is asking about when we have a direct sum decomposition $V = R \oplus N$. – Trevor Gunn Dec 03 '18 at 21:24

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Suppose $\exists W$ subspace of $V$ such that $V=R\oplus W$ and $W$ is invariant under $T$. Note $V=R\oplus W$ simply means $V=R+W$ and $R\cap W=\{0\}$. By theorem 6 section 2.3, $$\dim (V)=\dim (R+W)=\dim (R)+\dim (W)-\dim (R\cap W)= \dim (R)+\dim (W).$$ So $\dim (V)=\dim (R)+\dim (W)$. By rank-nullity theorem, $\dim (V)=\dim (R)+\dim (N)$. So $\dim (W)=\dim (N)$. It suffice to show $W$ is contained in $N$ or vice versa, to conclude $W=N$. Let $\alpha \in W$. Since $W$ is invariant under $T$, we have $T(\alpha)\in W\cap R=\{0\}$. So $T(\alpha)=0$. Thus $W\subseteq N$. Hence $W=N$.

user264745
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