I came across this problem:
I have successfully found the bases of the null space but I can't seem to understand the second part. I looked around online and found nothing useful. I would appreciate some help or a link to this.
Update: Here is the answer to the first part, the bases of the null spaces:
We have $$M(x_{1}-a)=0$$ $$M(x_2-a)=0$$ so that $$(x_1-a){\in}K_{1}$$ $$(x_2-a){\in}K_2$$ You have a basis for $K_1$ and $K_2$ so you know that: $x_1=sv+a$ where v is your 1 dimensional basis vector for $K_1$ and similarily $x_2=tu+a$ where u is your basis vector for $K_2$ (s,t are scalars). $$x_1-x_2=sv-tu$$ Using what you know about u,v and $x_1-x_2$, solve for s and t then p and q. Looking coordinate-wise, you see: $$p=s-t$$ $$5=s-2t$$ $$7=2s-t$$ $$q=-s+t$$ Note: It seems like the vector a does not really matter.