Find a Basis for U with $\text{rref}(A)$

[Dethrone]

Find a basis for $U=\text{span}{}\left\{\begin{bmatrix}1\\1\\0\\0\end{bmatrix}\begin{bmatrix}0\\0\\1\\1\end{bmatrix}\begin{bmatrix}1\\0\\1\\0\end{bmatrix}\begin{bmatrix}0\\1\\0\\1\end{bmatrix}\right\}$

Let $U=\text{col}(A)$, and applying row reduction on A, we obtain $\text{rref}(A)=\left[\begin{array}{1,0,0,1,5,3,2}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right]$.

Since the first, second, and third columns are columns with leading ones, then those corresponding columns of $A$ should form a basis for $A$, namely:

${}\left\{\begin{bmatrix}1\\1\\0\\0\end{bmatrix}\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\begin{bmatrix}1\\0\\1\\0\end{bmatrix}\right\}$

But the answer uses the first, second, and fourth column, why?

 

[I like Serena]

Hey Rido! (Smile)

Row reduction doesn't work here - you need column reduction. (Wasntme)

The matrix is just a short hand notation for the 4 vectors that are represented in its columns. To figure out the proper basis, you can add a multiple of any vector (aka column) to any of the other vectors (aka columns). This does not apply to the rows.

Row reduction is for solving a set of equations in which each row in the matrix represents an equation. (Nerd)

 

[Dethrone]

Hi ILS! :D

That is interesting. In my notes, as a lemma, my prof writes "The set of columns with leading 1's of rref(A), constitutes a basis for col(rref(A)), and the set of corresponding columns of A constitutes a basis for col(A)." Why does that not apply here? (Wondering)
Since, clearly, the column space of A is clearly $U$, what we are trying to find a basis for. Applying the lemma above, then what I did in my first post follows...

We were not formally introduced to column reduction, although I've heard of it. Can we take the transpose of the four vectors and perform row reduction on them, then take the transpose of the resulting basis? :D

 

[I like Serena]

Hi ILS! :D

That is interesting. In my notes, as a lemma, my prof writes "The set of columns with leading 1's of rref(A), constitutes a basis for col(rref(A)), and the set of corresponding columns of A constitutes a basis for col(A)." Why does that not apply here? (Wondering)
Since, clearly, the column space of A is clearly $U$, what we are trying to find a basis for. Applying the lemma above, then what I did in my first post follows...

Can't help that. Sorry. (Crying)

To use row reduction, the vectors need to be inserted in the matrix as rows instead of columns.
I'm afraid that transposing the matrix, as you did, will mess up the results.

We were not formally introduced to column reduction, although I've heard of it. Can we take the transpose of the four vectors and perform row reduction on them, then take the transpose of the resulting basis? :D

Yep. That would work! (Happy)

 

[Jameson]

I agree with your professor that you can take the pivot columns of $A$ to form a basis for the column space of $A$.

I didn't do the row reduction by hand, but using this site I got a different rref form than the one you did. I also checked it in R and got that all the columns are pivot columns. Is there a typo here?

 

[Dethrone]

Hi Jameson,

Sorry, there was a typo :(
$U=\text{span}{}\left\{\begin{bmatrix}1\\1\\0\\0\end{bmatrix}\begin{bmatrix}0\\0\\1\\1\end{bmatrix}\begin{bmatrix}1\\0\\1\\0\end{bmatrix}\begin{bmatrix}0\\1\\0\\1\end{bmatrix}\right\}$

I checked it before posting, but guess I overlooked that one. I have edited the first post.

While I agree with ILS, I still don't see why my method, which follows from my Professor's lemma, doesn't work.

Let $A=\left[\begin{array}{c}1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{array}\right]$. Then we want to find a basis for $U$, or a basis for $\text{col}(A)$ since $U=\text{col}(A)$. By the lemma, we can perform row reduction and then a basis can be found by noting the columns in $\text{rref}(A)$ that have leading 1's and find its corresponding columns in $A$.
I get that to be the first, 2nd, 3rd columns; not the first, 2nd, and 4th columns in the answer key.

 

[Jameson]

I would also use the 1st, 2nd and 3rd columns however after playing around with this matrix it seems that $c_4$ can be written in terms of $c_1,c_2,c_3$ and similarly $c_3$ can be written in terms of $c_1,c_2,c_4$. The rank of this matrix is 3 and there can be more than one basis of course so my thought is that your answer is correct as well as the book's.

This is something we just discussed in one of my own classes and I feel pretty confident with the systematic way to find a basis for the column space is to select the pivot columns of the matrix.

 

[Dethrone]

Thanks very much, ILS and Jameson! It is all clear now :D

 

[Jameson]

Thanks very much, ILS and Jameson! It is all clear now :D

Just one last thing (I'm always expecting someone to point out an error with my linear algebra comments but I think I'm actually correct here ...

Row reduction will help you find the rank of the matrix, which is equivalent to the number of linearly independent columns, equivalent to the number of pivot columns. Once you know the rank to form a basis you need $r$ linearly independent columns from $A$ to make a basis for the column space, so any 3 linearly independent ones from $A$ will do actually. It might make extra work though to take 3 random vectors and make sure they are linearly independent, so the 3 from the pivot positions are easier because we know they are linearly independent.

For example, consider the matrix: $E=\left(\begin{array}{ccc}1 & 1 & 2 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \end{array}\right)$. The rank of $E$ is 2 and there are two pivot positions once you get to rref form. Column 3 is clearly just double column 2, so we can see that columns 2 and 3 won't form a basis for $E$ but columns 1,2 and 1,3 will.

Lastly, say we have $F=\left(\begin{array}{ccc}1 & 1 & 1 \\ 2 & 2 & 2\\ 3 & 3 & 3 \end{array}\right)$. The rank is clearly 1 and by convention we denote the first column to be a pivot column but actually any of the three columns form a basis for the column space by themselves.

 

[Dethrone]

I see, so matrix $E$ is analogous to $A$ in the sense that either column 1 and 3, or column 1 and 2 will work as a basis. [typo edited by Jameson]

I never knew of the term "pivot columns", but it certainly is much easier to say than the two sentences I used to describe it... :D Thanks again!

 

[Jameson]

I made a typo that I fixed. For $E$, columns 1,2 or 1,3 will form a basis for the column space of $E$. 2,3 will not since they are linearly dependent.