# does the set of vectors $S =\{u,v,w\} = \{(1,2,3),(2,2,1),(0,4,-5)\}$ span $\mathbb R^3$? If not, What does it span? Describe it geometrically.

does the set of vectors $S=\{u,v,w\}=\{(1,2,3),(2,2,1),(0,4,-5)\}$ span $\mathbb R^3$? If not, What does it span? Describe it geometrically.

I already know I'm supposed to turn this into a $3\times3$ matrix and take the determinent, so I got $14 -14 = 0$ so it does not span $\mathbb R^3$. What I'm lost on is how to find what it does span, and describe it geometrically. the example of this type of problem only left off on finding wether it spans or not...

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One vector spans a line. Two linearly independent vectors span a plane. And $\geq 3$ linearly independent vectors span a hyper-plane. Therefore, we need to figure out how many linearly independent vectors are in this set.

The determinant will only tell you whether your set of vectors as a whole are linearly independent. If your determinant turns out to be zero, as it did in this case, then we cannot deduce how many independent vectors are in the set - only that the number is less than $3$. So at this point, all we know is that $Span(u, v, w)$ will turn out to be either a line or a plane.

Now, an easy way to proceed from here is to plug your vectors into a matrix as its rows. Then, perform Gauss-Jordan elimination to get the matrix into row-echelon form. From here, the number of nonzero rows in row echelon form will give the dimension of the row space of the original matrix. That is, it will give the number of linearly independent row vectors in the original matrix.

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ohhh that explains a lot, thanks! –  J L Jul 5 '14 at 10:15
Glad I could help! –  Kaj Hansen Jul 5 '14 at 10:15
so i guess it's pointless to ever approach this kind of problem by taking the determinant.. –  J L Jul 5 '14 at 10:19
@Kaj: Nice answer - straight to the point. –  johannesvalks Jul 5 '14 at 10:20
@JL, A big problem is that calculating determinants can be time consuming for larger matrices. Gauss-Jordan elimination is potentially faster for matrices larger than $2 \times 2$, and it will give you more information in the long run. –  Kaj Hansen Jul 5 '14 at 10:24

It looks like there's some bugs in computing the determinant.

By taking a cofactor expansion along the first row \begin{align*} \det \begin{bmatrix} 1 & 2 & 3 \\ 2 & 2 & 1 \\ 0 & 4 & -5 \\ \end{bmatrix} &= 1 \det \begin{bmatrix} 2 & 1 \\ 4 & -5 \\ \end{bmatrix} - 2 \det \begin{bmatrix} 2 & 1 \\ 0 & -5 \\ \end{bmatrix}+3\det \begin{bmatrix} 2 & 2 \\ 0 & 4 \\ \end{bmatrix} \\ &= 1 \times (2 \times (-5) - 1 \times 4)-2 \times (2 \times (-5) - 1 \times 0)+3 \times (2 \times 4 - 2 \times 0) \\ &=-14-(-20)+24 \\ &=30. \end{align*}

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Ah! JL will want to double-check his work. I was taking that for granted. –  Kaj Hansen Jul 5 '14 at 10:26

For a large matrix, instead of computing the determinant using a general formula, simpler to do (partial) Gaussian elimination to find out whether it has full rank. In this case, we start with $$\begin{bmatrix}1 & 2& 3 \\ 2 & 2 & 1 \\ 0 & 4 & -5 \end{bmatrix}$$ and subtract twice the first row from the second, yielding $$\begin{bmatrix}1 & 2& 3 \\ 0 & -2 & -5 \\ 0 & 4 & -5 \end{bmatrix}$$ Then adding twice the second row to the third gives $$\begin{bmatrix}1 & 2& 3 \\ 0 & -2 & -5 \\ 0 & 0 & -15 \end{bmatrix}$$ which is upper triangular with no zeroes on the diagonal, so the matrix has full rank and the original vectors span $\mathbb R^3$.

(We can also easily get the determinant here as the product of the diagonal elements, if we need it.)

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so full rank = same number of row space as rows in original matrix? –  J L Jul 5 '14 at 10:59
@JL: Yes. "Full rank" here is just shorthand for "the largest rank possible", i.e. the length of the shortest side of the matrix. –  Henning Makholm Jul 5 '14 at 11:00