Let us find the kernel:
ker(T)={(x1,x2,x3)∈R3 ∣ T(x1,x2,x3)=(0,0,0,0)}=
={(x1,x2,x3)∈R3 ∣ (x1+x2,x2+x3,x1−x3,2x1+x2−x3)=(0,0,0,0)}
It follows that we must solve the following system:
⎩⎨⎧x1+x2=0x2+x3=0x1−x3=02x1+x2−x3=0 which is equivalent to the system
⎩⎨⎧x1=−x2x2=−x3x1=x32x3−x3−x3=0 and to the system
⎩⎨⎧x2=−x3x1=x3x3∈R
Therefore, ker(T)={(x1,x2,x3)∈R3 ∣ x2=−x3,x1=x3}={(x3,−x3,x3) ∣ x3∈R}={x3(1,−1,1) ∣ x3∈R}.
It follows that dim(ker(T))=1.
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