# $V = \operatorname{Im} T + \ker T$ then $\operatorname{Im} T \cap \ker T = \{0\}$

Let $F$ be a field, let $V$ be a vector space with finite dimension over $F$ and let $T$ be a linear operator on $V$. Prove that:

a) If $V = \operatorname{Im} T + \ker T$ then $\operatorname{Im} T \cap \ker T = \{0\}$;

b) If $\operatorname{Im} T \cap \ker T = \{0\}$ then $V = \operatorname{Im} T \bigoplus \ker T$

I'm really stuck with this problem, some help to solve this please.

I will just provide the details to (i) using the hints suggested by vadim123. You can work on the other part yourself.

i) Suppose that $\def\Im{\operatorname{Im}}V = \Im(T) + \ker(T)$, then by the dimension theorem, $\dim(V) = \dim(\Im(T)) + \dim(\ker(T))$ and so $\dim(V) = \dim(\Im(T) + \ker(T))$. Hence, by the second hint we have that $\dim(\Im(T) \cap \ker(T)) = 0 \implies \Im(T) \cap \ker(T) = \{0\}$, as needed.

Hint: $\dim V = \dim(Im T) + \dim(Ker T)$

Second hint: $\dim(U+W)+\dim(U\cap W)=\dim(U)+\dim(W)$

• but I need another condition ? for example i need to suppose too that $rank(T) = rank (T^2)$ ? – Rachel Jun 29 '14 at 2:36
• This is known as Rank-Nullity. And note that rank(T)$\ge$rank(T$^2$) as you can only remove rank by sending more to the kernal. – Chris C Jun 29 '14 at 2:52

These results hold only in finite dimensional vector spaces, so you must use a dimension argument somewhere. The rank-nullity theorem tells you that $\dim(V)=\operatorname{rk}(T)+\dim(\ker T)$, where $\operatorname{rk}(T)$ is by definition the dimension of the image of $T$. This reduces the question to the following (after which you should forget $T$)

If $U,W$ are subspaces of a space $V$ of finite dimension$~n$, and if $\dim U+\dim W=n$ then show that

1. if $U+W=V$ then $U\cap W=\{0\}$, and

2. if $U\cap W=\{0\}$ then $U\oplus W=V$.

If you know that in finite dimension one has $\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$ for any subspaces $U,W$ (a formula notable for not extending in the obvious way to more than $2$ subspaces) then things become easy. For point 1. the formula gives you $\dim(U\cap W)=n-(\dim U+\dim W)=0$ which allows you to conclude, and for point 2. it similarly gives you that $\dim(U+W)=\dim U+\dim W=n$.

In case you need a proof for the formula, the basic method is to choose a basis for $U\cap W$, extend this basis separately to bases of $U$ and of $W$, and then prove that the $\dim U+\dim W-\dim(U\cap W)$ vectors in the union of those bases together form a basis of $V$; this proof is easy though not trivial. A more slick proof consists of forming the vector space $X=U\times W=\{\,(u,w)\mid u\in U,w\in W\,\}$, which has $\dim X=\dim U+\dim V$, and then apply the rank-nullity theorem to $f: X\to V$ defined by $f(u,w)=u-w$, using that $\ker(f)=\{\, (v,v)\mid v\in U\cap W\,\}$ whence $\dim(\ker f)=\dim U\cap W$.

• I'm looking for a counterexample for the infinite dimensional case - can you help with an accessible reference ? – Tom Collinge Apr 3 '18 at 11:20
• @TomCollinge An obvious counterexample for a) in infinite dimension is to take $V=\Bbb RX]$, and $T$ the operation of (formal) differentiation (or alternatively taking the quotient in the Euclidian division by $X$, or indeed by any non-constant polynomial; this works over any field). Then $T$ is surjective but has a non-trivial kernel, so a) fails. For b), take multiplication by any non-constant polynomial. – Marc van Leeuwen Apr 3 '18 at 11:50
• Many thanks, but I don't know what the space $V=\Bbb RX]$ is referring to (and Google doesn't deal with expanding these notations very well) - could you please elaborate a little ? – Tom Collinge Apr 3 '18 at 12:01
• I think I figured it - $\Bbb R[X]$ is the space of polynomials in a single variable with real coefficients ? – Tom Collinge Apr 3 '18 at 12:11
• Sorry, I meant $\Bbb R[X]$, and indeed that is the space of univariate polynomials. – Marc van Leeuwen Apr 3 '18 at 13:15