MHB Question on rank-nullity theorem

SarahER · Nov 9, 2014

View attachment 3493

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

Opalg · Nov 9, 2014

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.

SarahER · Nov 9, 2014

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?

MHB Question on rank-nullity theorem

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