Thread: Quick question on colomun space

I have the matrix A:
1 2 3 4 5
2 5 -1 0 -2
5 11 8 12 13
2 3 13 13 16
I found the basis for the row space:
1 2 2
2 5 3
3 -1 13
4 0 13
5, -2, 16

and a basis for the column space:
1 2 4
2 5 0
5 11 12
2, 3, 13

How do you go about expressing each row of A and column of A as a linear combination of vectors from the basis for row space and column space respectively.

It is difficult to explain through the internet but let me try.

1)First you need to bring the matrix into reduced echelon form.

2)Then from this row equivalent matrix you can tell by just looking (because it is in reduced form) how to express the non leading 1 coloums as linear combinations of the leading ones coloums.

3)The same linear combination you used in this row equivalent you use in the original matrix.

The reduced row echelon form of the augmented matrix A is:
1 0 -17 0 -11
0 1 -7 0 4
0 0 0 1 2
0 0 0 0 0

The reduced row echelon form of the transpose of A is:
1 0 3 0
0 1 1 0
0 0 0 1
0 0 0 0
0 0 0 0

So what are you saying?

Originally Posted by USCGuy

The reduced row echelon form of the augmented matrix A is:
1 0 -17 0 -11
0 1 -7 0 4
0 0 0 1 2
0 0 0 0 0

The leading one are the respective linearly independent coloum vectors. Good you found that.

Look at the non leading ones.
That is the 3rd and 5th Coloums.

The 3rd coloum vector can be expressed as,

Where "First" and "Second" are the coloum vectors.
Look how easy it is when it is in reduced echelon form.
Now that means in the original matrix the same relationaship holds. Meaning -17 times the first coloum vector and -7 times the second colomun vector express the Thrid Coloum vector as a linear combination.

And the 5th one is also easy to find,

Ok. Thank you.