# Does the matrix span R^4?

• Mar 21st 2011, 12:11 PM
posix_memalign
Does 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 PM
Ackbeet
Quote:

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

Do you mean, "How can I determine if the columns of matrix A span $\displaystyle \mathbb{R}^4$?"

Quote:

$\displaystyle A = \begin{bmatrix}-2 & -5 & 8 & 0 & -17\\ 1 & 3 & -5 & 1 & 5\\ 3 & 11 & -19 & 7 & 1\\ 1 & 7 & -13 & 5 & -3 \end{bmatrix}$
• Mar 21st 2011, 12:46 PM
posix_memalign
Quote:

Originally Posted by Ackbeet
Do you mean, "How can I determine if the columns of matrix A span $\displaystyle \mathbb{R}^4$?"

Yes, sorry about that. :-)
• Mar 21st 2011, 05:21 PM
Ackbeet
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 PM
posix_memalign
Quote:

Originally Posted by Ackbeet
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?

Thank you so much. :-) I would row reduce the matrix to row reduced echelon form and obtain that the first four columns are linearly independent -- i.e. the rank is 4. Is this correct at all?
• Mar 21st 2011, 06:01 PM
Ackbeet
Quote:

Originally Posted by posix_memalign
Thank you so much. :-) I would row reduce the matrix to row reduced echelon form and obtain that the first four columns are linearly independent -- i.e. the rank is 4. Is this correct at all?

You would reduce? Or you did reduce?
• Mar 22nd 2011, 09:44 AM
posix_memalign
Quote:

Originally Posted by Ackbeet
You would reduce? Or you did reduce?

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 AM
Ackbeet
Hmm. WolframAlpha disagrees with you. Maybe you should re-check your work?
• Mar 23rd 2011, 11:40 AM
posix_memalign
Quote:

Originally Posted by Ackbeet
Hmm. WolframAlpha disagrees with you. Maybe you should re-check your work?

Oh dear ...
Well I redid it and see now that the rank is 3. I.e. it does not span R^4.
• Mar 23rd 2011, 11:41 AM
Ackbeet
That's correct, it does not.