Linear Transformation R4 to R4: KerT + ImT = R4

[Dank2]

Homework Statement

Let T be a Linear Transformation defined on R⁴ ---> R⁴
Is that true that the following is always true ?
KerT + ImT = R⁴

Homework Equations

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.

 

[Mark44]

Homework Statement

Let T be a Linear Transformation defined on R⁴ ---> R⁴
Is that true that the following is always true ?
KerT + ImT = R⁴

Homework Equations

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.

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).