# KerT + ImT

1. Jul 7, 2016

### Dank2

1. The problem statement, all variables and given/known data
Let T be a Linear Transformation defined on R4 ---> R4
Is that true that the following is always true ?
KerT + ImT = R4

2. Relevant equations

3. The attempt at a solution
Since every vector in R4 must be either in KerT or the ImT, so the addition of those subspace contains R. and ofc every vector in ImT or KerT is in R4.

2. Jul 7, 2016

### Staff: Mentor

Looks OK to me other than you omiitted the 4 in R4, which I'm sure was inadvertent.

Given that T is a transformation from a space to itself, your statement is true. If, however, the domain and codomain weren't the same, then the statement would not be true, as Ker(T) and Im(T) would be subspaces of different dimension.

For example, if T is defined as $T : \mathbb{R}^3 \to \mathbb{R}^2$, with $T\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$ then Ker(T) consists of all vectors of the form $\begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix}$, a subspace of $\mathbb{R}^3$ while Im(T) consists of all vectors of the form $\begin{pmatrix} x \\ y \end{pmatrix}$, a subspace of $\mathbb{R}^2$.

The idea behind this question seems to be the Rank-Nullity Theorem. For a linear transformation T:V --> W, it's usually stated as dim(Ker(T)) + dim(Im(t)) = dim(V).