# Finding bases for a matrix's four subspaces without computing the matrix

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?

• 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 $0$. Feb 11, 2021 at 18:24
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 $$R(A^T)=R(U^T)$$.
• 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. Feb 12, 2021 at 17:22