0
$\begingroup$

I was watching a video about linear algebra and computing the four fundamental subspaces. The problem that was given in the video was the following: Suppose $$ B=\left[\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -1 & 0 & 1 \\ \end{matrix}\right] \left[\begin{matrix} 5 & 0 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{matrix}\right] $$ Find a basis for and compute the dimension of each of the 4 fundamental subspaces. Note: the matrix B is given in the B=LU form, if you have watched Gilbert Strang Lectures on Linear Algebra this form will make more sense.

They gave the solution: Dimension of column space C(B)=2 (since there are two pivots) A basis for C(B) is : $$\left[\begin{matrix} 1 \\ 2 \\ -1\end{matrix}\right] \left[\begin{matrix} 0 \\ 1 \\ 0 \\ \end{matrix}\right] $$

Dimension of row space $C(B^T)=2$

$$\left[\begin{matrix} 5 \\ 0 \\ 3 \\ \end{matrix}\right] \left[\begin{matrix} 0 \\ 1 \\ 1 \\ \end{matrix}\right] $$ Dimension of null space N(B) is 1 ( since there are 3 columns- number of independent columns which is 2) $$\left[\begin{matrix} -5/3 \\ -1 \\ 1 \\ \end{matrix}\right] $$ Dimension of left null space $N(B^T)=1$ $$\left[\begin{matrix} 1 \\ 0 \\ 1 \\ \end{matrix}\right] $$

I can't understand how they got the values for the basis, and why they use matrix L to find the basis for the column space. If someone can clarify this to me. Thank you!

$\endgroup$

1 Answer 1

1
$\begingroup$

The point is that matrix L is invertible while matrix U, with the last row all 0s, isn't. It is U, not L, that keeps B from have a column space of dimension 3. The product, B= LU, has column space of dimension 2.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .