I am trying to find the column space of a matrix

I am working on a lab for my class and am having issues finding the column soace for the matrix.

A = [1,1,0,2,0;0,1,1,3,0;2,0,0,0,1;3,1,0,2,1;2,1,1,3,0;1,0,0,2,1]

A = 6×5

1 1 0 2 0 0 1 1 3 0 2 0 0 0 1 3 1 0 2 1 2 1 1 3 0 1 0 0 2 1

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

This is the matrix that I am using. Tried the code

rrefA = rref(A)

rrefA = 6×5

1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

pivotColumns = rrefA(:,1:end-1)

pivotColumns = 6×4

1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

basisColumnSpace = A(:,pivotColumns)

Index in position 2 is invalid. Array indices must be positive integers or logical values.

but i cant get it to work I keep getting a error that i cant clear. I am wondering if there is another way to find the column space with the basic MatLab package. I would apperciate any help.

Thank You

Answer 1

[ProblemSolver]

@Taylor -- You just need to use this:

pivotColumns = find(any(rrefA, 1));

I hope this works!

[Torsten]

Counterexample:

A = [1,1,0,2,4;0,1,1,3,6;2,0,0,4,8;3,1,0,5,10;2,1,1,6,12;1,0,0,7,14]

A = 6×5

1 1 0 2 4 0 1 1 3 6 2 0 0 4 8 3 1 0 5 10 2 1 1 6 12 1 0 0 7 14

rrefA = rref(A)

rrefA = 6×5

1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0

pivotColumns = find(any(rrefA, 1))

pivotColumns = 1×5

1 2 3 4 5

basisColumnSpace = A(:,pivotColumns)

basisColumnSpace = 6×5

1 1 0 2 4 0 1 1 3 6 2 0 0 4 8 3 1 0 5 10 2 1 1 6 12 1 0 0 7 14

[Walter Roberson]

Carter Stahly comments to @Torsten

This does not work FYI. It does not properly find the column space vector (try with A = [1,3,2;0,1,0])

[Walter Roberson]

I believe that @Torsten is providing a counter-example to illustrate the failure of @ProblemSolver approach. I do not believe that Torsten is attempting to illustrate a correct general method.

[Carter]

That would make sense, thank you for the clarification.