# Find a linearly independent set of vectors that spans the same subspace of $\mathbb{R}^3$

Find a linearly independent set of vectors that spans the same subspace of $\mathbb{R}^3$ as that spanned by the vectors

$\begin{bmatrix}2\\2\\-1\end{bmatrix}, \begin{bmatrix}-8\\-2\\5\end{bmatrix}, \begin{bmatrix}-3\\0\\2\end{bmatrix}$

I got the RREF form of $\begin{bmatrix}2&0&1\\0&1&1/2\\0&0&0\end{bmatrix}$ can't I just omit the last row?

Why is $(2,0,1)$ and $(0,1,1/2)$ not correct as a linearly independent set?

I know that I will have two free variables so that I will have two vectors, I also see that $(-8,-2,5)$ and $(-3,0,2)$ could be a linearly independent set. I'm confused on when I actually worked out why my set isn't working?

• Use elementary column operations, not elementary row operations to get it into RCEF (or at least CEF). – user137731 Jul 20 '16 at 0:15
• @Arthur, no it doesn't, I was hinting as to how to do this problem from the given vectors. There's nothing more to it in this question, in my view. Nonetheless I deleted my comment... – imranfat Jul 20 '16 at 0:20
• Pista número dos: any two of these vectors are clearly linearly independent. Can you represent one vector as a linear combination of the other 2? – user137731 Jul 20 '16 at 0:31
• The first two are not Linearly Independent, they are multiples of each other @Bye_World but I Did $A^T$ and then $rref(A^T)$ and then did $(A^T)^T$ and got the correct Linearly Independent set: $(1,0,-2/3)$ and $(0,1,1/6)$ – user23 Jul 20 '16 at 0:38
• $(-8,-2,5)$ is a multiple of $(2,2,-1)$ if and only if $(-8,-2,5) = k(2,2,-1) = (2k,2k,-k)$. That is you need $$\begin{cases}-8=2k \\ -2=2k \\ 5=-k\end{cases}$$ for one single $k$. No such $k$ exists so these two vectors are not multiples of each other. – user137731 Jul 20 '16 at 0:45

$$\begin{pmatrix}2&2&-1\\-8&-2&5\\-3&0&2\end{pmatrix}\stackrel{\begin{cases}R_2+4R_1\\R_3+\frac32R_1\end{cases}}\longrightarrow\begin{pmatrix}2&2&-1\\0&6&1\\0&3&\frac12\end{pmatrix}\stackrel{R_3-\frac12R_2}\longrightarrow\begin{pmatrix}2&2&-1\\0&6&1\\0&0&0\end{pmatrix}$$