Question

8. Let
A =

1 -3 2 -4
0 3 0 4
2 3 4 4
3 -24 6 -32
-1 -3 -1 3


(a) Specify the row space, row(A), of the matrix A.
(b) Determine the dimension of row(A) and the dimension of the column space, col(A) and hence rank(A).
(c) Find a basis of row(A).
(d) Find a basis of col(A).

Please help :(

Answer 1

row reduce the matrix. that will give you the information to answer all of those questions.

Answer 2

thanks ill have a go at it

Answer 3

post what you get back on here. the row reduction process is the most tedious part. after that its just pointing out things.

Answer 4

might take a while

Answer 5

right this is what i got

1 -3 2 -4
0 1 0 4/3
0 0 1 7
0 0 0 0
0 0 0 0

Answer 6

what do i do now :/

Answer 7

ack sry bout that.

So there are 3 pivots (those 1's along what i'll call a pseudo diagonal). That means the rank of the matrix is 3. This is also the dimension of the row space and columns space.

If you didnt switch any of the rows around, then the first 3 rows of the original matrix span the row space of A. They would form a basis for the row space.

Likewise, the first three columns form a basis for the columns space of A.

Answer 8

To find the basis for column space of A, do I not have to transpose the original matrix the do row operations to reduce again to echelon form? which will give my row basis?

Answer 9

and also i did switch the rows around so the span would be just original rows of A?
i.e. span {{1,-3,2,-4), (0, 3, 0, 4) ...etc}

Answer 10

the dimension of the row space is 3, so you would only need 3 of the rows to span the row space. Now that i am thinking about it, you might need to row reduce the transpose of A to see which rows (which would be columns in A transpose) are the 3 that form a basis.

The row reduced form of A is good for letting you know which columns form a basis for the column space for sure though. since the first 3 columns have the pivots in the row reduced form, that means the first 3 columns of A are linearly independent and form the basis of the columns space.

Answer 11

thank you ever so much :) appreciate your help