Answer to Question #155053 in Linear Algebra for kamya rawat

Question #155053

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


1
Expert's answer
2021-01-12T16:06:46-0500

Let us find the kernel:


ker(T)={(x1,x2,x3)R3  T(x1,x2,x3)=(0,0,0,0)}=\ker (T)=\{(x_1,x_2,x_3)\in\mathbb R^3\ |\ T(x_1,x_2,x_3)=(0,0,0,0)\}=

={(x1,x2,x3)R3  (x1+x2,x2+x3,x1x3,2x1+x2x3)=(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:


{x1+x2=0x2+x3=0x1x3=02x1+x2x3=0\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


{x1=x2x2=x3x1=x32x3x3x3=0\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


{x2=x3x1=x3x3R\begin{cases} x_2=-x_3\\ x_1=x_3\\x_3\in\mathbb R \end{cases}


Therefore, ker(T)={(x1,x2,x3)R3  x2=x3,x1=x3}={(x3,x3,x3)  x3R}={x3(1,1,1)  x3R}.\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.\dim(\ker(T))=1.


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