Question states: "Describe the possible echelon forms of the matrix.
$A$ is a $4×2$ matrix, $A = [ a_1 \ a_2 ]$, and $a_2$ is not a multiple of $a_1$."
I have found one fo the solution, which is of the form
$$
P \ \ C
$$
$$
0 \ \ P
$$
$$
0\ \ 0
$$
Where P is the leading entry in echelon form and C is any real number.
I thought that this was the only solution. Here's my argument: since $a_1$ is not equal to $c•a_2$, for any scalar $c$, the two vectors are linearly independent and thus cannot have any free variables.
However, the solution manual states that another acceptable echelon form is
$$
0 \ \ P
$$
$$
0 \ \ 0
$$
$$
0\ \ 0
$$
Other than just finding an example of this echelon form, whu is this acceptable? Obviously, the vector with all zeroes is not a multiple of the second vector with a nonzero leading entry. But is my argument for my answer invalid? In the second acceptable form of $A$, the $x_1$ term is free, which cannot happen in linearly independent set of vectors…
P.S. I apologize for not being able to put things into matrix… I looked hard on this site on how to make the brackets [ ] bigger, but I couldn't find how.
Best Answer
There's a subtlety in the question: $a_2$ is not a multiple of $a_1$. This is not the same as saying that $a_2$ and $a_1$ are independent, however. What if $a_1$ is the zero vector, and $a_2$ is a nonzero vector? They are linearly dependent, but $a_2$ is not a multiple of $a_1$.
Actually, the zero vector is a multiple of all vectors. The second vector is the one that is not a multiple of the first.
In sum: the vectors are not independent, so there can (and is) a free variable in the accepted echelon forms.
P.S. \bigg[ will render as $\bigg[$ in LaTeX.