Given the following set of vectors, determine which forms a basis for $\mathbb{R}^3$. For those that do not, which span $\mathbb{R}^3$?

\begin{Bmatrix} \begin{pmatrix}1\\0\\-2\end{pmatrix},& \begin{pmatrix}-1\\1\\0\end{pmatrix},& \begin{pmatrix}-1\\2\\-2\end{pmatrix},& \begin{pmatrix}0\\0\\1\end{pmatrix} \end{Bmatrix}

Attempted solution:

By constructing a matrix with these vectors as columns and following the row operations R3 + 2R1 then R3+2R2, and finally R1+R2, I obtain:


Which gives me columns 1, 2, and 3 as pivot columns. Consequently, column 3 is linearly dependent. Thus the following set of vectors forms the basis for $\mathbb{R}^3$:

\begin{Bmatrix} \begin{pmatrix}1\\0\\0\end{pmatrix},& \begin{pmatrix}0\\1\\0\end{pmatrix},& \begin{pmatrix}0\\0\\1\end{pmatrix} \end{Bmatrix}

For the second part of the question: for those that do not, which span $\mathbb{R}^3$?

Since the only vector that does not form a basis for $\mathbb{R}^3$ is:


and this is only one vector, it cannot span $\mathbb{R}^3$ (since a set of vectors spanning $\mathbb{R}^3$ would have to contain at least three vectors).

Have I solved this correctly?

  • $\begingroup$ Also - for the first question - would the vectors that form a basis for R3 be {1, 0 -2}, {-1, 1, 0}, and {0, 0, 1} - or would they be the vectors that I got when I reduced the matrix into RREF, which were {1, 0, 0}, {-1, 1, 0}, and {0, 0, 1}? $\endgroup$ – Hazim Mar 2 '16 at 4:23
  • $\begingroup$ Both of those sets form a basis for $\mathbb{R}^3$. But in the context of the question, the correct answer would be the first set you gave (original vectors from the given matrix). The set of vectors you got after reducing the matrix to reduced row-echelon form is just the standard basis for $\mathbb{R}^3$. $\endgroup$ – SplitInfinity Mar 2 '16 at 4:54

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