Answer to Question #146814 in Linear Algebra for Sourav Mondal

Question

Find the range space and a basis for the
kernel of the linear transformation
T : R4 ->R4 defined by
T(x1, x2, x3, x4) = (x1 - x2, x2 - x3, x3 - x4, x4 - x1).

Answer 1

$1.\text{ Kernel of a transformation}.\\ T: \mathbb{R}^4\longrightarrow \mathbb{R}^4\\ T(x_1,x_2,x_3,x_4)=(x_1-x_2,x_2-x_3,x_3-x_4,x_4-x_1)\\ \begin{cases} x_1-x_2=0\\ x_2-x_3=0\\ x_3-x_4=0\\ x_4-x_1=0 \end{cases}\\ \begin{pmatrix} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1\\ -1&0&0&1 \end{pmatrix}\\ \text{We add the first row to the fourth row:}\\ \begin{pmatrix} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1\\ 0&-1&0&1 \end{pmatrix}\\ \text{We add the second row to the fourth row:}\\ \begin{pmatrix} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1\\ 0&0&-1&1 \end{pmatrix}\\ \text{We add the third row to the fourth row:}\\ \begin{pmatrix} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1\\ 0&0&0&0 \end{pmatrix}\\ \begin{cases} x_1-x_2=0\\ x_2-x_3=0\\ x_3-x_4=0 \end{cases}\\ \begin{cases} x_1=x_2\\ x_2=x_3\\ x_3=x_4 \end{cases}\\ x_1=x_2=x_3=x_4\\ \text{ker}(T)=\{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4|x_1=x_2=x_3=x_4\}.\\ \text{Hence } \{(1,1,1,1)\}\text{ is a basis of ker(T)}.\\ 2.\text{ Range space}.\\ \text{Let } (a,b,c,d)\in \mathbb{R}^4.\\ \mathbb{R}^4\text{ is a codomain of T}.\\ \text{Then:}\\ \begin{cases} x_1-x_2=a\\ x_2-x_3=b\\ x_3-x_4=c\\ x_4-x_1=d \end{cases}\\ \text{We have shown above that the rank of the coefficient matrix equals 3}.\\ \text{Using Rouché–Capelli theorem:}\\ \text{rank} \begin{pmatrix} 1&-1&0&0&a\\ 0&1&-1&0&b\\ 0&0&1&-1&c\\ -1&0&0&1&d \end{pmatrix}=3\\ \text{So the range space of T consists of all the vectors }(a,b,c,d)\in\mathbb{R}^4\\ \text{such that the last equality is true.} \\ \begin{pmatrix} 1&-1&0&0&a\\ 0&1&-1&0&b\\ 0&0&1&-1&c\\ -1&0&0&1&d \end{pmatrix}\sim \begin{pmatrix} 1&-1&0&0&a\\ 0&1&-1&0&b\\ 0&0&1&-1&c\\ 0&-1&0&1&a+d \end{pmatrix}\sim \begin{pmatrix} 1&-1&0&0&a\\ 0&1&-1&0&b\\ 0&0&1&-1&c\\ 0&0&-1&1&a+b+d \end{pmatrix}\sim \begin{pmatrix} 1&-1&0&0&a\\ 0&1&-1&0&b\\ 0&0&1&-1&c\\ 0&0&0&0&a+b+c+d \end{pmatrix}\\ \text{Then } a+b+c+d=0.\\ \text{The range space of T consists of all the vectors }(a,b,c,d)\in\mathbb{R}^4\\ \text{such that }a+b+c+d=0.$