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OpenStudy (anonymous):

Find bases for the four fundamental subspaces of the Matrix A. V1=(1,0,2),V2=(1,1,0),V3=(-1,0,-1) Need answer Find bases for the four fundamental subspaces of the Matrix A. V1=(1,0,2),V2=(1,1,0),V3=(-1,0,-1) Need answer @Mathematics

OpenStudy (anonymous):

just so you don't get the matrix messed up, those are column vectors. I'm not sure how to put in a 3x3 matrix

OpenStudy (anonymous):

for Rank, did you basically get the standard basis for R3?

OpenStudy (anonymous):

well you can see they are all dependent so it could be the originals vectors but i like nicer numbers

OpenStudy (zarkon):

I think you mean to say independent

OpenStudy (anonymous):

\[R(A^T)=(1,0,0)^T,(0,1,0)^T,(0,0,1)^T\] and i mean't columnspace

OpenStudy (anonymous):

You cannot write the other vectors in the form of others therefore they're all dependent

OpenStudy (zarkon):

you have dependent and independent backwards

OpenStudy (chriss):

We just went over some of this stuff tonight in Linear Algebra... not far enough along yet to answer your questions, but I can help with the matrix... if they are column vectors, the matrix should be: [ 1 1 -1 ] [ 0 1 0 ] [ 2 0 -1 ]

OpenStudy (anonymous):

ahh yeah i just noticed that

OpenStudy (chriss):

I thought I was going to be able to do that with the editor but it didn't work, but that should work, but I just saw you meant column space, so I guess that's not what you meant anyway :P

OpenStudy (zarkon):

\[\left[\begin{matrix}1 & 1&-1 \\0 & 1&0 \\ 2 & 0&-1 \end{matrix}\right]\]

OpenStudy (zarkon):

right click on my matrix and view source to see what I did.

OpenStudy (zarkon):

'show source'

OpenStudy (anonymous):

so did you get the same for the transpose of the columnspace and the columnspace

OpenStudy (zarkon):

are you looking for a basis for A and A transpose?

OpenStudy (anonymous):

Its Zarkon! awesome :)

OpenStudy (zarkon):

Hi joe :)

OpenStudy (anonymous):

i'm looking for the four fundamentals, so the Column spaces and nullspaces of Matrix A and it's transpose

OpenStudy (zarkon):

it is me

OpenStudy (anonymous):

awesome!

OpenStudy (zarkon):

the matrix has full rank (and it is square) so the null space is just the zero vector

OpenStudy (anonymous):

yeah that's what i was assuming but you got the standards for the column spaces?

OpenStudy (anonymous):

The the null space of A transpose will also be just the 0 vector, as the rank of A is the same as the rank of A transpose.

OpenStudy (zarkon):

use the originals if you want

OpenStudy (zarkon):

they are L.I.

OpenStudy (anonymous):

So this problem is that bad. All the columns form a basis for the Col Space, all the rows form a basis for the Row Space, and the Null Space and Left Null Space are trivial.

OpenStudy (anonymous):

isnt* i meant isnt >.> lol

OpenStudy (anonymous):

That's why i asked for the answers lolz XD it's on my pretest that my teacher gives no answers

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