Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a question that has stumped me. It asks "Without computing A, find bases for the four fundamental subspaces." The $LU$ equation is as follows:

$$\begin{bmatrix} 1 & 0 & 0 \\ 6 & 1 & 0 \\ 9 & 8 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 2 \end{bmatrix}$$

I understand that $$N(A) = \begin{bmatrix} 0 \\ 2 \\ -1 \\ 1 \end{bmatrix} $$ and that there is no $N(A^T)$ because there are no zero rows in $U$, but I do not know what to do for the column/row spaces without computing $A$. I could compute columns 1, 2, and 3 because it's not "$A$" but I know that's not what they want.

share|improve this question
add comment

2 Answers

up vote 0 down vote accepted

The row space of $A=LU$ is the same as the rowspace of $U$ since $L$ is not singular. $L$ represents only a change of basis on the rows of $U$. The column space is the full span, since the null space is nothing.

Also, I think this is the null space: $$N(A) = \begin{bmatrix} 0 \\ 1 \\ -2 \\ 1 \end{bmatrix}$$

share|improve this answer
    
Yes, you are correct. I copied my answer down incorrectly. I completely forgot about how there were three pivots and the vectors were in R^3. Thank you for pointing that out! –  Goyatuzo Oct 4 '12 at 18:30
add comment

If we are given two matrices $A$ and $B$, then the columns of $AB$ are linear combinations of the columns of $A$ and the rows of $AB$ are the linear combinations of the rows of $B$. This follows immediately from block multiplication $$AB = \begin{pmatrix} A\mathbf{b_1} & \cdots & A\mathbf{b_n}\end{pmatrix} = \begin{pmatrix} \mathbf{a_1}^\mathrm{T}B \\ \vdots \\ \mathbf{a_m}^\mathrm{T}B\end{pmatrix}$$ where $a_i$ and $b_i$ denote the row/column vectors of $A$ and $B$ respectively. From this, what can you conclude?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.