Question on rank-nullity theorem

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The discussion centers on the rank-nullity theorem, specifically addressing the basis of the kernel of a linear transformation. The vector $(0, 1, -1)$ serves as the sole basis for the kernel of the transformation \( g \), confirming that the dimension is 1. This conclusion is drawn from the condition that vectors of the form $(0, b, -b)$ are scalar multiples of $(0, 1, -1)$, establishing the dimensionality based on the relationship \( b + c = 0 \).

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View attachment 3493

So my question on this is: Why is the basis (0, 1, -1) because the dim is 1?
 

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SarahER said:
View attachment 3493

So my question on this is: Why is the basis (0, 1, -1) because the dim is 1?
Hi Sarah, and welcome to MHB!

If the vector $(a,b,c)$ satisfies $a=0$ and $b+c=0$, then $c=-b$ and the vector is equal to $(0,b,-b) = b(0,1,-1).$ So all such vectors are scalar multiples of $(0,1,-1)$. Thus the single vector $(0,1,-1)$ forms a basis for the kernel of $g$, which therefore has dimension 1.
 
Opalg said:
Hi Sarah, and welcome to MHB!

If the vector $(a,b,c)$ satisfies $a=0$ and $b+c=0$, then $c=-b$ and the vector is equal to $(0,b,-b) = b(0,1,-1).$ So all such vectors are scalar multiples of $(0,1,-1)$. Thus the single vector $(0,1,-1)$ forms a basis for the kernel of $g$, which therefore has dimension 1.

So, is it 1 dimensional because b = -c?
 

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