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enter image description here

I have attached what A is in a picture. I'm wondering how I can find the bases of A without computing A (i.e. by just looking at the LU factorisation of A, how can we work out the bases for the column, row, null and left nullspace?)

So far I think rank of A is 3, because rank of L and U are both 3. And because L and U both have 3 linearly independent columns, A (=LU) has 3 independent columns as well? Which leads to dimension of column space of A being 3?

I also know that the dimension of the nullspace of A is 4-3 = 1. But how do I work out some general bases of the 4 subspaces?

Thank you for your time!

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  • $\begingroup$ You want to be careful when you say, "I think rank of $A$ is $3$, because rank of $L$ and $U$ are both $3$". The rank of a matrix product cannot be established that easily. For example, consider $A_1=(1,1)$ and $A_2=(-1,1)^T$. Then $A_1$ and $A_2$ both have rank $1$ but $A_1A_2$ has rank $0$. $\endgroup$
    – Matthew H.
    Feb 11, 2021 at 18:24
  • $\begingroup$ @MatthewPilling I see - thanks for clearing that up for me! $\endgroup$
    – Antimatics
    Feb 12, 2021 at 8:36

1 Answer 1

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It turns out $A=LU$ has rank $3$. To see this, we'll show that $R(A)=\mathbb{R}^3$. Choose $b\in \mathbb{R}^3$ arbitrary. Note $Ax=b$ has a solution iff $\Big[U\Big|L^{-1}b\Big]$ is consistent. The latter augmented system is consisent since $U$ has $3$ pivot columns, so $A$ has rank $3$; any basis for $\mathbb{R}^3$ will suffice as a basis for $R(A)$. Next, observe how $Ax=0$ iff $Ux=0$ so $N(A)=N(U)$. You can find $N(A^T)$ using the fact $\Big[R(A)\Big]^{\perp}=N(A^T)$. Lastly, to find $R(A^T)$, use the facts that $$\text{rank}(A^T)=\text{rank}(A)=3$$ $$\text{rank}(U^T)=3$$ $$R(A^T)\subseteq R(U^T)$$ to conclude $R(A^T)=R(U^T)$.

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  • $\begingroup$ Thank you very much for a clear explanation! $\endgroup$
    – Antimatics
    Feb 12, 2021 at 8:39
  • $\begingroup$ Just another quick question - why is the latter augmented system consistent since U has 3 pivot columns? Is it because since there are more unknowns than equations so there has to be a solution (4 unknowns but only 3 rows of equations)? $\endgroup$
    – Antimatics
    Feb 12, 2021 at 8:49
  • $\begingroup$ It's because every row contains a pivot so the columns of $U$ span $\mathbb{R}^3$. Just because we ave more unknowns than equations doesn't guarantee there has to be a solution. $\endgroup$
    – Matthew H.
    Feb 12, 2021 at 17:22
  • $\begingroup$ ah I see, thank you! $\endgroup$
    – Antimatics
    Feb 13, 2021 at 11:59

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