Question

I have completed this LINEAR algebra problem, and I was wondering it everything is correct.

Here is the problem.

Find a basis for rowsp, collsp, nullsp, and compute ranks of

A= [1 2 -1 2 , 2 3 1 2 , 4 5 5 2]


I first got A into row echelon form

[1 2 -1 2 , 0 1 -3 2, 0 0 0 0 ]

I then saw there were two leading ones, so the basis for the row space was

[ 1 2 -1 2]^t , [0 1 -3 2]^t

it has a rank of 2, since it is 2 vectors

then, to find the basis for colsp, I took the first 2 COLUMNS of a, to get

[1 2 4]^t and [2 3 5]^t

I'll post the rest in the comments

Answer 1

I then found the null space by augmenting the 0 matrix with my row echelon form. I noticed that there was a row of 0's, so I knew that there had to be 2 vectors in the null space.


I got

[ 1 0 5 -2 | 0
0 1 -3 2 | 0
0 0 0 0 | 0 ]

I then set x3=1, x4=0 and vice versa

and I got N(A) = { [-5 3 1 0]^t and [2 -2 0 1]^t}

nullity is 2.

And that is good, because nullity + rank = 4 = # of columns

Does everything seem right?

Answer 2

It seems right to me.

Answer 3

Sweet. Thanks!

Answer 4

you're welcome.