column space having non-linear combinations of basis vectors

Question

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 ?

Answer 1

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 $\mathbb{R}^3$, hence anything in $\mathbb{R}^3$ 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.