• Support PF! Buy your school textbooks, materials and every day products Here!

[Linear Algebra] Maximal linear independent subset

  • Thread starter iJake
  • Start date
  • #1
41
0

Homework Statement



In the follow cases find a maximal linearly independent subset of set ##A##:
(a) ##A = \{(1,0,1,0),(1,1,1,1),(0,1,0,1),(2,0,-1,)\} \in \mathbb{R}^4##
(b) ##A = \{x^2, x^2-x+1, 2x-2, 3\} \in \mathbb{k}[x]##

The Attempt at a Solution



The first part of the exercise is trivial, as it is easy to observe that the second vector is a linear combination of the first and third vectors. My question is one of mechanics. Should I be writing the elements of each vector as the rows or columns of a matrix?

In a previous exercise, I had to determine whether or not a vector space was spanned by a set of vectors. In that case, it was:

Determine if ## V = \mathbb{k}[x]_3## is spanned by ##A = \{1, 1+x^2, 1-x+x^2+x^3, 4-x+2x^2+x^3\}##

I reordered the elements and presented them as the columns of a matrix:

##
\begin{matrix}
0 & 0 & 1 & 1\\
0 & 1 & 1 & 2\\
0 & 0 & -1 & -1\\
1 & 1 & 1 & 4\\
\end{matrix}
##

I did row operations (switch R4 with R1, then with R3, add R3 to R4) and found a row of 0s, thus telling me the rank of the matrix is 3 and that A does not span V.

However, for my current problem, I wrote the matrix using the vector elements as rows, not columns. When I write them as columns I do not reach the same conclusion, which confuses me.

##
\begin{matrix}
1 & 0 & 1 & 0\\
1 & 1 & 1 & 1\\
0 & 1 & 0 & 1\\
2 & 0 & -1 & 0\\
\end{matrix}
##

Here I clearly observe that the second row is the sum of the first and third rows. However, if I write it where the vector elements are columns, I see something else:

##
\begin{matrix}
1 & 1 & 0 & 2\\
0 & 1 & 1 & 0\\
1 & 1 & 0 & -1\\
0 & 1 & 1 & 0\\
\end{matrix}
##

Can someone explain the difference to me? Thanks.
 
Last edited:

Answers and Replies

  • #2
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,792
1,390
It might be because the last matrix has an error in it. The bottom left element should be 0, not 1.

If all you are trying to do is test whether n vectors in a n-dimensional space are independent, by calculating the rank of a matrix, you should get the same result regardless of whether you make the vectors rows or columns. I think it is more usual to make the vectors columns, but that's just convention.
 
  • #3
41
0
Sorry, that was just a typo. It's fixed now, and I find that R4 - R2 gives me a row of 0s and that the rank of that matrix is 3. Is it irrelevant that the linearly dependent vector corresponds to the second column of the matrix, in this case?
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement



In the follow cases find a maximal linearly independent subset of set ##A##:
(a) ##A = \{(1,0,1,0),(1,1,1,1),(0,1,0,1),(2,0,-1,)\} \in \mathbb{R}^4##
(b) ##A = \{x^2, x^2-x+1, 2x-2, 3\} \in \mathbb{k}[x]##

The Attempt at a Solution



The first part of the exercise is trivial, as it is easy to observe that the second vector is a linear combination of the first and third vectors. My question is one of mechanics. Should I be writing the elements of each vector as the rows or columns of a matrix?

In a previous exercise, I had to determine whether or not a vector space was spanned by a set of vectors. In that case, it was:

Determine if ## V = \mathbb{k}[x]_3## is spanned by ##A = \{1, 1+x^2, 1-x+x^2+x^3, 4-x+2x^2+x^3\}##

I reordered the elements and presented them as the columns of a matrix:

##
\begin{matrix}
0 & 0 & 1 & 1\\
0 & 1 & 1 & 2\\
0 & 0 & -1 & -1\\
1 & 1 & 1 & 4\\
\end{matrix}
##

I did row operations (switch R4 with R1, then with R3, add R3 to R4) and found a row of 0s, thus telling me the rank of the matrix is 3 and that A does not span V.

However, for my current problem, I wrote the matrix using the vector elements as rows, not columns. When I write them as columns I do not reach the same conclusion, which confuses me.

##
\begin{matrix}
1 & 0 & 1 & 0\\
1 & 1 & 1 & 1\\
0 & 1 & 0 & 1\\
2 & 0 & -1 & 0\\
\end{matrix}
##

Here I clearly observe that the second row is the sum of the first and third rows. However, if I write it where the vector elements are columns, I see something else:

##
\begin{matrix}
1 & 1 & 0 & 2\\
0 & 1 & 1 & 0\\
1 & 1 & 0 & -1\\
0 & 1 & 1 & 0\\
\end{matrix}
##

Can someone explain the difference to me? Thanks.
Both
$$A = \begin{bmatrix}
1 & 0 & 1 & 0\\
1 & 1 & 1 & 1\\
0 & 1 & 0 & 1\\
2 & 0 & -1 & 0\\
\end{bmatrix}
$$
and its transpose
$$B = \begin{bmatrix}
1 & 1 & 0 & 2\\
0 & 1 & 1 & 0\\
1 & 1 & 0 & -1\\
0 & 1 & 1 & 0\\
\end{bmatrix}
$$
have the same row and column ranks---3 in this case.

If you perform row-operations on ##A## you will produce a new matrix whose rows are linear combinations of the row-vectors ##\vec{a}_i## in ##A##, so will be manufacturing directly some linear combination ##c_1 \vec{a}_1 + c_2 \vec{a}_2 + c_3 \vec{a}_3 + c_4 \vec{a}_4## that is the zero (row) vector; that will be the "all-zero" row of the transformed matrix. When you perform row operations on ##B## that is not what you are doing: the new columns are not linear combinations of the old columns, and row operations that produced a zero-row in ##A## will not necessarily produce a zero row in ##B##---although, of course, some other row operations will do so. To find out what linear combinations of the column vectors ##\vec{b}_i## in ##B## are needed to get the zero (column) vector, we need to find ##k_i## that give ##k_1 \vec{b}_1 + k_2 \vec{b}_2 + k_3 +\vec{b}_3 + k_4 \vec{b}_4 = \vec{0}##. That can be done by solving the equation system
$$B \vec{K} = \vec{0},$$
where
$$\vec{K} = \pmatrix{k_1\\ k_2 \\ k_3 \\ k_4}.$$
Row operations on ##B## are just helping you to solve that system of equations.
 

Related Threads on [Linear Algebra] Maximal linear independent subset

  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
21
Views
5K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
6
Views
1K
Replies
4
Views
6K
Replies
3
Views
3K
Top