MHB Question on rank-nullity theorem

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The discussion centers on the rank-nullity theorem and the basis of a kernel in linear algebra. The vector (a, b, c) is constrained by the conditions a=0 and b+c=0, leading to the conclusion that c=-b. This results in the representation of any vector in the kernel as a scalar multiple of (0, 1, -1). Consequently, the kernel has a dimension of 1, as it is spanned by the single vector (0, 1, -1). The dimension is confirmed to be 1 due to the relationship b = -c.
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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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