# Is it true that if $T$ is a linear operator on a finite-dimensional vector space $V$ then $V=\ker T\oplus \operatorname{im}T$?

$\newcommand{\im}{\operatorname{im}}$I am trying to prove or disprove the following statement:

Let $V$ and $W$ be finite-dimensional vector spaces. If $T:V\rightarrow W$ is a linear transformation then $V=\ker T\oplus \im T$. (By the symbol $\oplus$ I mean the direct sum of two vector spaces.)

This statement cannot be true if $V\neq W$ because a vector space can only be a direct sum of its subspaces. However, I am not sure about the case when $V=W$, i.e., when $T$ is a linear operator.

I want to use the following proposition:

$\textbf{Proposition.}$ Let $V$ be a finite-dimensional vector space and let $U$ and $W$ be subspaces of $V$. Then $V=U\oplus W$ if and only if $V=U+W$ and $U\cap W=\left\{ 0 \right\}$.

First I want to show that $V=\ker T + \im T$. I just don't have a clue how to possibly do this, which leads me to believe there must be a counterexample. I believe that $\ker T\cap \im T=\left\{0\right\}$ since $T(0)=0$ for any linear transformation and it is not possible for $Tv\neq 0$ if $v\in \ker T$. Some help?

• If you knew that $\ker T \cap \operatorname{im} T= \emptyset$, then you'd have a proof. But this isn't true, and you can easily find an example in small dimensional spaces. – hardmath Jul 16 '16 at 21:02
• Aside: the statement you're studying is only claiming that $V$ is the displayed direct sum -- in particular, it is not claiming that the direct sum is the internal direct sum of two subspaces, so you're going about the problem wrongly. (the proposition you want to use is about the internal direct sum) – user14972 Jul 16 '16 at 21:18
• Actually, this statement is true iff $T$ is a projection operator.Note that, in order for $T$ to be a projection operator, you need $W \subseteq V$. – onurcanbektas Jul 29 '17 at 7:31
• @onurcanbektas Yes if T is a projection then this is true. But it's also true for example if T is a rotation (which is not a projection) or a reflection: So being a projection is not a necessary condition. – Tom Collinge Apr 3 '18 at 10:38
• @TomCollinge That is correct; The necessary condition for that condition to be hold for $T$, we need $T$ to be normal, i.e it needs to commute with its adjoint (the operator whose matrix is the transpose of the matrix of $T$). – onurcanbektas Apr 3 '18 at 17:57

You have the right ideas. Indeed, your claim is not true. Consider, for example, the transformation $$T = \pmatrix{0&1\\0&0}$$ Verify that im$(T) = \ker(T)$, and that both of these are one-dimensional subspaces of $\Bbb R^2$.

Notably, however, the statement will hold for any self-adjoint (symmetric) operator $T:V \to V$.

• Thank you for providing me with this counterexample. I see that kerT=imT={(0,y): y \in R}. – aimedaca Jul 16 '16 at 21:05
• In fact, @aimedaca, it is quite difficult to find an example of matrices which do satisfy the condition you wanted! Did you try some examples before embarking in the proof? – Mariano Suárez-Álvarez Jul 16 '16 at 21:10
• @MarianoSuárez-Alvarez Not enough, apparently! – aimedaca Jul 16 '16 at 21:13
• Did you find an example which does have the property? – Mariano Suárez-Álvarez Jul 16 '16 at 21:18
• Geometrically this $T$ is an orthogonal projection onto the $y$-axis followed by a right-angle clockwise rotation that takes the oriented $y$-axis to the oriented $x$-axis. (In case anybody wants to use geometry to verify ${\rm img}(T)=\ker(T)$.) – arctic tern Jul 16 '16 at 21:21

This is not true in general. I will restrict the question to the case where $V = W$, i.e. $T$ is a linear operator on $V$.

Consider the linear operator over $V = \mathbb{R}^3$ where the matrix is written with respect to the standard basis: $(e_1, e_2, e_3)$

$\begin{bmatrix} 0 & 1& 1 \\ 0 & 0& 0\\ 0 & 0& 0 \\ \end{bmatrix}$

It is easy to see that $\mathrm{Null}(T) = span(e_1, e_2 - e_3)$, and $\mathrm{Im}(T) = span(e_1)$. Clearly $V \neq \mathrm{Null}(T) \oplus \mathrm{Im}(T)$.

However, this is an interesting question, because by adding a few additional hypothesis, we can get a few true statements that similar to this, and these become very important in linear algebra.

$1.$ For a finite dimensional complex vector space (or over an algebraically closed field), $V$, where the dimension of $V$ is $n$, for any linear operator $T$, $V = \mathrm{Null}(T^n) \oplus \mathrm{Im}(T^n)$.

$2.$ If we assume $V$ is complex (or over an algebraically closed field), finite dimensional, and we have a linear transformation such that $T^2 = T$, we do indeed have: $V = \mathrm{Null}(T) \oplus \mathrm{Im}(T)$.

• Thank you for this counterexample. By $T^n$ do you mean $T$ composed with itself $n$ times or are you treating $n$ as an exponent? – aimedaca Jul 16 '16 at 21:11
• In linear algebra, we usually use $T^n$ to denote T composed with itself $n$ times, which is what I am referring to. Equivalently, you could think of this as raising the matrix associated with $T$ to the $n$th power (multiplying the matrix by itself $n$ times), as the composition of operators is the same as the multiplication of their matrix representations. – Christian Jul 16 '16 at 21:14
• $Null(T)$ will also contain $e_2-e_3$. – Berci Jul 16 '16 at 21:14
• @aimedaca Those are the same thing. – arctic tern Jul 16 '16 at 21:15
• (3) The rank-nullity thm: $\dim V=\dim\ker T+\dim {\rm img}\,T$ – arctic tern Jul 16 '16 at 21:15