Vtyjkyuk

I have a matrix $A_3x3$ whose basis for the column space are $a_1=(2,2,5)$, $a_2=(9,5,3)$ and $a_3=(3,6,1)$. If these are the basis for the column space, then column space can always be generated by linear combination of these basis vectors. Right?

Now, here is what I did.

I went for a non-linear combination of these basis vectors given by $a_1^2+a_2+a_3^3$ which is equal to $(40,225,29)$. Now since, $(40,225,29)$ is a non-linear combination of basis vectors, I should not be able to get this by linear combination of basis vectors. To verify this I tried to get the coefficients $x_1,x_2$ and $x_3$ such that $x_1(2,2,5)+x_2(9,5,3)+x_3(3,6,1) = (40,225,29)$. Surprisingly, I am getting a consistent solution.

Why is this discrepancy ? By definition the above system should have been inconsistent.

Any suggestions ?

Let me do a quick computation to check what does $a_1, a_2, a_3$ span

octave:1> A = [2 2 5; 9 5 3; 3 6 1]
A =

 2 2 5
 9 5 3
 3 6 1

octave:2> rref(A)
ans =

 1 0 0
 0 1 0
 0 0 1

It turns out that it spans $mathbbR^3$, hence anything in $mathbbR^3$ can be written as linear combination of those vectors.

Being a basis of a vector space doesn't say anything about whether it can span any non-linear transform of those three vectors. It can certainly cover its linear combination, but it doesn't guarantee whether it can do so for a particular non-linear combination. After all, a vector that is obtained by a non-linear transform could have been obtained by a linear combination as well.