Question

Verify rank nullity theorem for the linear 

transformation T : R

3 → R

3

 defined by : 9 

T(x, y, z) = (x + 2y – z, y + z, x + y – 2z)

Answer 1

$T:\mathbb{R}^3\rightarrow \mathbb{R}^3$ , $T(x,y,z)=(x+2y-z,\ y+z,\ x+y-2z)$

Let us find the dimension of the image:

$T(x,y,z)=\begin{pmatrix} 1 & 2&-1 \\ 0 & 1&1\\ 1&1&-2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=x\begin{pmatrix} 1\\0\\1 \end{pmatrix}+ y\begin{pmatrix} 2\\1\\1 \end{pmatrix}+z\begin{pmatrix} -1\\1\\-2 \end{pmatrix}$

$\text{Im} \ T=\text{Span}\begin{Bmatrix} \begin{pmatrix} 1\\0\\1 \end{pmatrix}, \begin{pmatrix} 2\\1\\1 \end{pmatrix}, \begin{pmatrix} - 1\\1\\-2 \end{pmatrix} \end{Bmatrix}= \text{Span}\begin{Bmatrix} \begin{pmatrix} 1\\0\\1 \end{pmatrix}, \begin{pmatrix} 2\\1\\1 \end{pmatrix}, \begin{pmatrix} 2\\1\\1 \end{pmatrix}-3 \begin{pmatrix} 1 \\0\\ 1 \end{pmatrix} \end{Bmatrix}= \text{Span}\begin{Bmatrix} \begin{pmatrix} 1\\0\\1 \end{pmatrix}, \begin{pmatrix}2\\1\\1 \end{pmatrix} \end{Bmatrix}$

$\text{dim } \text{Im} \ T=2$

Now we find the dimension of the kernel:

$T(x,y,z)=(x+2y-z,\ y+z,\ x+y-2z)=(0,0,0)$

We have that $y=-z, \ \ x=z-2y=3z$ .

So, $\text{Ker} \ T=\text{Span} \begin{Bmatrix} \begin{pmatrix} 3\\-1\\1\end{pmatrix} \end{Bmatrix}$

$\text{dim Ker}\ T=1$

The rank-nullity theorem:

The dimension of the kernel plus the dimension of the image equals the dimension of the domain.

In this case we have $\text{dim Ker} \ T+\text{dim Im}\ T=\text{dim} \ \mathbb{R}^3$ and $1+2=3$ .