How can I determine if the matrix A spans $\displaystyle R^4$?

$\displaystyle A = \begin{bmatrix}-2 & -5 & 8 & 0 & -17\\ 1 & 3 & -5 & 1 & 5\\ 3 & 11 & -19 & 7 & 1\\ 1 & 7 & -13 & 5 & -3 \end{bmatrix}$

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- Mar 21st 2011, 12:11 PMposix_memalignDoes the matrix span R^4?
How can I determine if the matrix A spans $\displaystyle R^4$?

$\displaystyle A = \begin{bmatrix}-2 & -5 & 8 & 0 & -17\\ 1 & 3 & -5 & 1 & 5\\ 3 & 11 & -19 & 7 & 1\\ 1 & 7 & -13 & 5 & -3 \end{bmatrix}$ - Mar 21st 2011, 12:14 PMAckbeet
- Mar 21st 2011, 12:46 PMposix_memalign
- Mar 21st 2011, 05:21 PMAckbeet
Right-ho. Then the way I would go about it is this: let the columns of $\displaystyle A$ be denoted $\displaystyle \mathbf{a}_{1},\mathbf{a}_{2},\dots,\mathbf{a}_{5} .$ They are column vectors in $\displaystyle \mathbb{R}^{4}.$ Let

$\displaystyle \mathbf{r}=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{ 4}\end{bmatrix}$

be an arbitrary vector in $\displaystyle \mathbb{R}^{4}.$ We want to know if there is a set of scalars, $\displaystyle b_{1},b_{2},\dots,b_{5},$ such that

$\displaystyle b_{1}\mathbf{a}_{1}+b_{2}\mathbf{a}_{2}+\dots+b_{5 }\mathbf{a}_{5}=\mathbf{r},$

right? Now, this equation that I've just written down can be translated into a system of equations for the unknown constants $\displaystyle b_{j}.$ Your goal is to determine the rank of the coefficient matrix for that system. If the rank is 4, then you've got yourself a spanning set with those $\displaystyle \mathbf{a}_{j}$'s. Otherwise, if the rank is less than 4, you don't. Make sense? Where would you go next? - Mar 21st 2011, 05:58 PMposix_memalign
- Mar 21st 2011, 06:01 PMAckbeet
- Mar 22nd 2011, 09:44 AMposix_memalign
I did reduce, I had actually already done that when I was fumbling around without knowing if it was the correct thing to do, I got:

$\displaystyle \begin{bmatrix}1 & 0 & 0 & 0 & 27/10\\ 0 & 1 & 0 & 0 & 3/10\\ 0 & 0 & 1 & 0 & -17/10\\ 0 & 0 & 0 & 1 & -5 \end{bmatrix}$

Hence why the first four columns are linearly independent, thus the rank is 4. - Mar 22nd 2011, 09:57 AMAckbeet
Hmm. WolframAlpha disagrees with you. Maybe you should re-check your work?

- Mar 23rd 2011, 11:40 AMposix_memalign
- Mar 23rd 2011, 11:41 AMAckbeet
That's correct, it does not.