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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?

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    $\begingroup$ Use elementary column operations, not elementary row operations to get it into RCEF (or at least CEF). $\endgroup$ – user137731 Jul 20 '16 at 0:15
  • $\begingroup$ @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... $\endgroup$ – imranfat Jul 20 '16 at 0:20
  • $\begingroup$ 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? $\endgroup$ – user137731 Jul 20 '16 at 0:31
  • $\begingroup$ 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 $(0,1,1/6)$ $\endgroup$ – user23 Jul 20 '16 at 0:38
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    $\begingroup$ $(-8,-2,5)$ 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. $\endgroup$ – user137731 Jul 20 '16 at 0:45
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First: if you already know that space is two dimensional then check any two of the given vectors are linearly independent (as none of them is a scalar multiple of the other one) and then you're done.

Otherwise, set up a matrix with your vectors as rows (since you're going to reduce it by rows)...and reduce the matrix:

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

Thus, the vector representing the third row is lin. dependent in the other two so you can chose these two first ones as basis of the space...or you can choose the vectors representing the two non-zero rows in the rightmost matrix as basis: it's the same.

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