I understand that the basis of column space A is just the columns of A that correspond to the pivot columns of U. This is because U is just the reduced row echelon form.
However, as mentioned in the title, I am having difficulty grasping why columns of L that correspond to the pivot columns of U form the basis of column space A?
Just to give u guys a concrete example of what does the proof statement mean:
$$A = LU$$
As you can see that the first 2 columns of U are the pivot columns,
$$U = \begin{bmatrix}5&0&3\\0&1&1\\0&0&0\end{bmatrix}$$
Based on the proof statement, the first 2 columns of L form the basis for A $$L = \begin{bmatrix}1&0&0\\2&1&0\\-1&0&1\end{bmatrix}$$
I am supposed to prove why basis b for column space of A is the first 2 columns of L.
$$b = \left\{ \begin{bmatrix}1\\2\\-1\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix} \right\} $$
In short, how to prove that the basis b for column space of A is the columns of L that correspond to the pivot columns of U?
I appreciate your help!
Best Answer
$A=LU \Rightarrow Col (A)=Col (LU)=Col \big(\begin{bmatrix}1&0&0\\2&1&0\\-1&0&1\end{bmatrix}\begin{bmatrix}5&0&3\\0&1&1\\0&0&0\end{bmatrix}\big)=span (5 [1, 2, -1]^t,[0, 1, 0]^t)$
The idea is that non -pivotal columns of U contribute nothing to Col (LU). In the above example, 3rd column of U is non pivotal (doesn't have any pivots) and hence in Col (LU) expression above, 3rd column is missing.
Let's generalize this for a special case of square matrices (where U is such that pivots appear on diagonals):
Let LU factorization of A exist such that A=LU (assume n by n order). U is in row echelon form and L is a lower triangular matrix with 1's on diagonal. Assume U has k ($\lt n$) pivots $a_{ii} \ne 0$ for $i={1,2,3,...,k}$. Then $Col (A)=Col (LU)=span (Lu_1 ,Lu_2, ...,Lu_{n-1},Lu_n))$ ,
where $u_i$ is the ith column of U. Since L is invertible, $Col (A)=Col(LU)=span(Lu_1,Lu_2,...,Lu_k) \tag{A}$
Remarks:
1) Col () is column space.
2) (A) is true because suppose that for some real $c_i$'s where $i={1,2,3,4,...,k}$ such that $c_1 (Lu_1)+c_2 (Lu_2)+...+c_k(Lu_k)=0$$\Rightarrow L(c_1u_1+c_2u_2+...+c_ku_k)=0$$ \Rightarrow c_1u_1+c_2u_2+...+c_ku_k=0 \Rightarrow c_1=c_2=c_3=...=c_k=0$ [Since $u_i$'s are pivotal columns and hence linearly independent.]