Please help me to write a step by step solution to this problem:

Let $T$ be a linear operator defined on a finite dimensional vector space $V$. If $W$ is a $T$ invariant subspace with $V = \mathrm{im}(T) \oplus W$, then show that W must be $N(T)$ (the null space of $T$).

I am getting an intuition but am at a loss as how to write that formally...

  • $\begingroup$ What is $(+)W$? $\endgroup$ Jan 6, 2016 at 11:30
  • $\begingroup$ @ElChapo I guess it means $\oplus$, and maybe $R(T)=\text{im}(T)$? $\endgroup$
    – user160738
    Jan 6, 2016 at 11:30
  • $\begingroup$ @ElChapo :user160738 is absolutely right. $\endgroup$
    – Qwerty
    Jan 6, 2016 at 11:32
  • $\begingroup$ @Qwerty What does $\oplus$ mean in this context? Is the finite dimensional vector space equipped with a scalar product, and it means the orthogonal sum, or is it the direct sum? $\endgroup$
    – Roland
    Jan 6, 2016 at 11:34
  • $\begingroup$ @Qwerty, what have you tried so far, and where are you stuck? $\endgroup$ Jan 6, 2016 at 11:42

2 Answers 2


Suppose $w\in W$; then $T(w)\in\operatorname{im}(T)$ by definition and $T(w)\in W$ since $W$ is $T$-invariant. Thus $T(w)=0$; therefore $W\subseteq N(T)$.

Can it be $W\subsetneq N(T)$? Use the fact that, by assumption, $$ \dim V=\dim \operatorname{im}(T)+\dim W $$ and that, by the rank-nullity theorem, $$ \dim V=\dim \operatorname{im}(T)+\dim N(T) $$

  • $\begingroup$ please add a proof of $N(T)\subseteq W$ .I am at a loss as how to to that.. $\endgroup$
    – Qwerty
    Jan 6, 2016 at 15:20
  • $\begingroup$ @Qwerty The dimension of $W$ is equal to $\dim V-\dim\operatorname{im}(T)$ by assumption, which is the same as the dimension of $N(T)$ by the rank-nullity theorem. $\endgroup$
    – egreg
    Jan 6, 2016 at 16:24
  • $\begingroup$ I'm sorry, but I can't accept two answers. The one I accepted is slightly better explained. $\endgroup$
    – Qwerty
    Jan 7, 2016 at 12:21

One way to go about this is by using the following fact:

For a vector space $V$, if $U$ and $W$ are subspaces of $V$, then their sum $U+V$ is a direct sum if and only if $U \cap W=\{0\}$.

Now since $T\in \mathcal{L}(V)$ we have that $\text{im}(T)$ is a subspace of $V$, and we are given that $W$ is a T-invariant subspace of $V$, so we can make the following conclusion:

$\text{im}(T) \cap W=\{0\}$

Given that $W$ is T-invariant, we have that for any $w\in W$, $T(w)\in W$. But we also know that $T(w)\in \text{im}(T)$. Given that we know their intersection is $\{0\}$ this means that $T(w)=0$, which by definition, means that each $w\in W$ is sent to $0$ by T, and therefore $W\subset N(T)$.

By the rank-nullity theorem we know that $\text{dim}(V)= \text{dim(im}(T))+\text{dim}(N(T))$. Since $V=\text{im}(T) \oplus W$, we also know that $\text{dim}(V)=\text{dim(im}(T)) +\text{dim}(W)$. This tells us that $\text{dim}(W)=\text{dim}(N(T))$.

Since $W\subset N(T)$ and their dimensions are equal, we must have that $W=N(T)$.

  • $\begingroup$ @Qwerty hopefully this clarifies everything. $\endgroup$ Jan 7, 2016 at 6:51
  • $\begingroup$ yeah it does ... Thanks $\endgroup$
    – Qwerty
    Jan 7, 2016 at 12:19

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