I was given the matrix:
$$A=\left(\begin{matrix}-4 & -7 & 1 & 2\\ 0 & 0 & 3 & 8\\5 & -1 &1 & -4\end{matrix}\right)$$
I arranged them into an augmented matrix and added a 5th column of zeroes. I row reduced until I got
$$B=\left(\begin{matrix}1 & 0 & 0 & -2 & 0\\ 0 & 1 & 0 & 0 & 0\\0 & 0 &1 & 8 & 0\end{matrix}\right)$$
I let $t$ denote a free variable: $x2=0, x3=-8, x1=2, x4=t$.
I know that the pivot columns must stay in the matrix. I also know that if the amount of vectors is greater than or equal to the span we are still unsure if the vectors are in the given span.
I want to know if I am on the right track and what step I should take next. Can I take out the 4th column and row reduce again to verify that the leftover vectors are in the span?
$\endgroup$ 72 Answers
$\begingroup$You don't need an augmented matrix in this case.
Starting with $$A=\left(\begin{matrix}-4 & -7 & 1 & 2\\ 0 & 0 & 3 & 8\\5 & -1 &1 & -4\end{matrix}\right)$$ just row reduce like you did. I'll assume you're correct and that the RREF of this matrix is $$\operatorname{RREF}(A)=\left(\begin{matrix}1 & 0 & 0 & -2\\ 0 & 1 & 0 & 0\\0 & 0 &1 & 8\end{matrix}\right)$$
Then you just count the pivots:
$$\require{enclose}\left(\begin{matrix}\enclose{circle}{1} & 0 & 0 & -2\\ 0 & \enclose{circle}{1} & 0 & 0\\0 & 0 &\enclose{circle}{1} & 8\end{matrix}\right)$$
There are $3$ pivots in this case, meaning the row rank is $3$. By the theorem which tells us the row rank = the column rank of a matrix, we also know that the column rank of $A$ is $3$. Thus there are $3$ linearly independent columns of $A$.
$\Bbb R^3$ has a dimension of $3$ (can you prove this?), thus any $3$ linearly independent vectors will span it. Thus the columns of $A$ do indeed span $\Bbb R^3$.
Appendix: Theorems needed for the above
- Row reduction (by elementary row operations) preserves the row space of a matrix
- The number of pivots of a matrix in RREF equals the row rank of a matrix.
- The row rank = the column rank of a matrix (and thus you can just refer to it as the rank of the matrix).
- The dimension of a vector space $V$ is well-defined. I.e. if the dimension of $V$ is $n$, then $(1)$ any $n$ linearly independent vectors $v_1, v_2, \dots, v_n \in V$ is a spanning set, $(2)$ any set of less than $n$ vectors will NOT be a spanning set, and $(3)$ any set of more than $n$ vectors will necessarily be linearly dependent. (So this one is really $3$ different theorems.)
If you haven't encountered one or more of these theorems, you may have to do the problem a different way (or prove these theorems yourself). ;)
$\endgroup$ 6 $\begingroup$Perhaps changing the format makes it easier to see. The given matrix provides you with 4 vectors with 3 components each. Thus this matrix can also be seen as
$\begin{bmatrix}-4 & 0 & 5 \\-7 & 0 & -1 \\ 1 & 3 & 1 \\ 2 & 8 & -4 \end{bmatrix}$
You can row reduce this matrix and find that it equals
$\begin{bmatrix}1 & 0 & -(5/4) \\0 & 1 & (3/4) \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ In this format it is clear that of the 4 given vectors, 3 are linearly independent with one linearly dependent and thus it indeed spans the vector space.
$\endgroup$
