So I have a homework question which I am not sure if I am answering correctly. The questions is as follows.
Determine whether the set is a basis for $\mathcal{R}^3$. If the set isn't a basis, determine if it's linearly independent or spans $\mathcal{R}^3$.
So I have 4 column vectors
$$\begin{pmatrix} 1\\ -2\\ 3\end{pmatrix},\begin{pmatrix} 0\\ 3\\ -1\\\end{pmatrix}, \begin{pmatrix} 2\\ -1\\ 5\\\end{pmatrix},\begin{pmatrix} 0\\ 0\\ 1\\\end{pmatrix}$$
I combined the vectors into a $3\times4$ matrix and got the RREF of that matrix which is
$$\begin{pmatrix} 1 \hspace{5pt}0 \hspace{5pt}2 \hspace{5pt}0\\ 0 \hspace{5pt}1 \hspace{5pt}1 \hspace{5pt}0\\ 0 \hspace{5pt}0 \hspace{5pt}0 \hspace{5pt}1\\\end{pmatrix}$$
So what I got from the RREF is that the vectors are not a basis for $\mathcal{R}^3$ since they're not linearly independent. But how do I know if they span $\mathcal{R}^3$?
I'm having trouble finding good examples where the combined column matrices are of size $n\times m$ instead of size $n \times n$ and that tell me what does the RREF of the matrix actually means.
Thanks for all the help in advance!
Best Answer
When you perform the reduction operations on a matrix $A$, you are premultiplying by invertible matrices, so you end up with $R = O A$, where $O$ is an invertible matrix and $R$ is the RREF.
Since $O$ is invertible, you can show that $\dim {\cal R}R = \dim {\cal R}A$ (in fact you have ${\cal R}R = O ({\cal R}A)$, and you have $\ker R = \ker A$.
In your matrix above, you can see that Columns 1,2 & 4 of $R$ spans $\mathbb{R}^3$, so we have $\dim {\cal R}R = 3$ from which we can deduce that $\dim {\cal R}A = 3$ and so ${\cal R}A = \mathbb{R}^3$ (since $\dim \mathbb{R}^3 = 3$).
You can also see that the 3rd column can be written as a linear combination of the first two (you have $A(-2,-1,1,0)^T = 0$), so they are not linearly independent. This means that $\ker A \neq \{0\}$.
Note: To solve $y=Ax$ for a specific $y$ (assuming that $y$ is in the range of $A$), then you would need to perform exactly the same operations on $y$ first to get $Oy$ and then solve $Oy = OAx = Rx$ for $x$.