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)
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?
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.