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!

  • 1
    $\begingroup$ Try to form any vector (a, b, c) from a linear combination of your vectors. You will get 4 algebraic equations with 3 unknowns. Then you can figure out if they span. $\endgroup$ – jdods Mar 10 '15 at 4:07
  • $\begingroup$ What's the dimension of $\mathbb{R^3}$? $\endgroup$ – science Mar 10 '15 at 4:14

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 $x$.

  • $\begingroup$ Thanks a lot for the help! $\endgroup$ – Kenshin Mar 10 '15 at 20:20

$$ \begin{pmatrix} x\\ y\\ z\end{pmatrix}= a\begin{pmatrix} 1\\ -2\\ 3\end{pmatrix} +b\begin{pmatrix} 0\\ 3\\ -1\\\end{pmatrix} +c\begin{pmatrix} 2\\ -1\\ 5\\\end{pmatrix} +d\begin{pmatrix} 0\\ 0\\ 1\\\end{pmatrix} $$

So we get $$ \begin{aligned} x&= \quad a\quad \quad \quad+2c\\ y&=-2a \ +3b \ \ -c\\ z&=\quad 3a \ \ -b \ +5c+d\\ \end{aligned} $$

Now you can solve for $a,b,c$, and $d$ using standard algebra or matrix methods. I get the feeling that they do span all of $\mathbb{R}^3$ assuming you did your row reduction correctly.


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