Given the Matrix A = [[1,3,-4,2,-1,6],[0,0,1,-3,6,0],[0,0,0,1,4,-3],[0,0,0,0,0,0]] (this matrix is 4x6);

1.) Find a basis for col(A^T)

2.) Show that nul(A) is orthogonal to col(A^T)

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- Nov 30th 2006, 01:58 PMBlodenBasis for Col(A)/Nul(A)
Given the Matrix A = [[1,3,-4,2,-1,6],[0,0,1,-3,6,0],[0,0,0,1,4,-3],[0,0,0,0,0,0]] (this matrix is 4x6);

1.) Find a basis for col(A^T)

2.) Show that nul(A) is orthogonal to col(A^T) - Dec 3rd 2006, 04:39 PMBloden
Ok so my hint for this was:

**For 1.) Reduce (A^T)^T = A to echelon form (for your problem it already**

is in echelon form) then transpose the non-zero rows. Or, reduce A^T to

echelon form then choose the columns of A^T which contain leading row

entries (i.e. pivots). There is, in this case, a third way: eyeball

A^T.

For 2.) Let w be in null(A) and v be in col(A^T). Then Aw = 0 and

v = A^Tx for some x. Compute w dot v = w^Tv = ...

For #1, I found the transpose of A to be:

[[1,0,0,0],[3,0,0,0],[-4,1,0,0],[2,-3,1,0],[-1,7,4,0],[6,0,-3,0]]

(6x4 matrix)

When I rref the transpose of A I get:

[1,0,0,0], [0,1,0,0], [0,0,1,0], [0,0,0,0], [0,0,0,0], [0,0,0,0]

The first 3 columns of pivots, if I did that correctly...but whats the basis? The first 3 columns in the orig?

And for #2

What's the nul(A) and when I multiply them I should get 0. So I will have some m * 3 matrix times another matrix? Dot product them? - Dec 3rd 2006, 07:43 PMBloden
Well I figured it out.

I'm just confused if col(A^T) is:

the vectors: [1,3,-4,2,-1,6], [0,0,1,-3,7,0],[0,0,0,1,4,-3] <--- 3 6x1 vectors

OR: if it is [1,3,-4,2,-1,6], [0,0,1,-3,7,0],[0,0,0,1,4,-3] <-- 3 1x6 vectors...

For #2 I was able to find the nul(A) and when multiplying each of the vectors from nul(A) to col(A) I got 0, so I am positive I did this right...which leads me to think the 6x1 vectors are right for col(A) but I want to make sure...

but...col(A^T) is the same as row(A) which makes me think of the 2nd...so im confused.