Question

T(x₁,x₂,x₃) = (x₁₊x_2, x₂₊x_3, x_1-x_3, 2x₁+x₂-x₃) find he dimension of the kernel

Answer 1

Let us find the kernel:

$\ker (T)=\{(x_1,x_2,x_3)\in\mathbb R^3\ |\ T(x_1,x_2,x_3)=(0,0,0,0)\}=$

$=\{(x_1,x_2,x_3)\in\mathbb R^3\ |\ (x_1+x_2, x_2+x_3, x_1-x_3, 2x_1+x_2-x_3)=(0,0,0,0)\}$

It follows that we must solve the following system:

$\begin{cases} x_1+x_2=0\\ x_2+x_3=0\\ x_1-x_3=0\\ 2x_1+x_2-x_3=0 \end{cases}$ which is equivalent to the system

$\begin{cases} x_1=-x_2\\ x_2=-x_3\\ x_1=x_3\\ 2x_3-x_3-x_3=0 \end{cases}$ and to the system

$\begin{cases} x_2=-x_3\\ x_1=x_3\\x_3\in\mathbb R \end{cases}$

Therefore, $\ker (T)=\{(x_1,x_2,x_3)\in\mathbb R^3\ |\ x_2=-x_3,x_1=x_3\}=\{(x_3,-x_3,x_3)\ |\ x_3\in\mathbb R\}=\{x_3(1,-1,1)\ |\ x_3\in\mathbb R\}.$

It follows that $\dim(\ker(T))=1.$