# Doubt with vectorial spaces (Basis and dimension)

Good night, i'm working in a problem, i need an basis and the dimension of the space.

$a_{1}=(1,0,0,-1),\:a_{2}=(2,1,1,0),\:a_{3}=(1,1,1,1),\:a_{4}=(1,2,3,4),\:a_{5}=(0,1,2,3)$

I make this:

$\left[ \begin {array}{cccc} 1&0&0&-1\\ 2&1&1&0 \\ 1&1&1&1\\ 1&2&3&4 \\ 0&1&2&3\end {array} \right]$

and i apply gauss for reduce the matrix:

$\left[ \begin {array}{cccc} 1&0&0&-1\\ 0&1&0&1 \\ 0&0&1&1\\ 0&0&0&0 \\ 0&0&0&0\end {array} \right]$

now, i have 3 linearly independent vectors and my dimensions is 3. but, i can take any vector and they go linearly independent? for example: $a_{1}=(1,0,0,-1),\:a_{2}=(2,1,1,0),\:a_{3}=(1,1,1,1)$

or

$a_{3}=(1,1,1,1),\:a_{4}=(1,2,3,4),\:a_{5}=(0,1,2,3)$

and going to be linearly independent? thanks!!

• what exactly are you looking for? A basis for the image of your linear function? or for the kernel? or...? – user190080 May 27 '16 at 23:29

In this case you are finding a basis for the rowspace of the matrix you set up. The nonzero rows of your row-reduction matrix span the row space of the original matrix. Only the nonzero rows are linearly independent.
You can take a look at this site for details: http://www.math.tamu.edu/~fnarc/psfiles/find_bases.pdf

Since you constructed a matrix from the set of vectors as row vectors, you are essentially looking for a basis of the row space of the matrix (the span of the row vectors), let's call this matrix $A$.

Row operations do not change the row space of a matrix, so to construct a basis for the row space, what you must look at are the pivot rows in the reduced row-echelon form of $A$.

We can choose either the pivot rows of $\mathrm{RREF}(A)$ or the corresponding rows in $A$ as a basis for the vector space spanned by those vectors. So in this case $\left\{(1,0,0,-1), (2,1,1,0), (1,1,1,1)\right\}$ and $\left\{(1,0,0,-1), (0,1,0,1), (0,0,1,1)\right\}$ are both bases for the vector space spanned by the given vectors.

One way I prefer to do it, is construct the matrix from the vectors as columns. In this case we are trying to determine a basis for the column space of $A$:

$$A=\begin{bmatrix}1 & 2 & 1 & 1 &0 \\ 0 & 1 & 1 & 2 & 1 \\ 0 & 1 & 1 & 3 & 2 \\ -1 & 0 & 1 & 4 & 3\end{bmatrix},$$

$$\mathrm{RREF}(A)=\begin{bmatrix}1 & 0 & -1 & 0 & 1 \\ 0 & 1 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0\end{bmatrix}.$$

What $\mathrm{RREF}(A)$ is telling us is that columns $1$, $2$, and $4$ of matrix $A$ form a basis for the column space of $A$ (because these are the pivot columns). i.e. $\left\{(1,0,0,-1), (2,1,1,0), (1, 2, 3, 4)\right\}$ is a basis for the space spanned by the given set of vectors.

What $\mathrm{RREF}(A)$ also tells us is that $\mathbf{c}_{3}=-1\mathbf{c}_{1}+1\mathbf{c}_{2}$, or $$(1,1,1,1)=-1(1,0,0,-1)+1(2,1,1,0),$$ and also, $\mathbf{c}_{5}=1\mathbf{c}_{1}-1\mathbf{c}_{2}+1\mathbf{c}_{4}$, or $$(0,1,2,3)=1(1,0,0,-1)-1(2,1,1,0)+1(1,2,3,4).$$

The vector which is expressed as said linear combination is of course the column in $A$ which corresponds to a non-pivot column of $\mathrm{RREF}(A)$. The vectors in the linear combination are those column vectors of $A$ which correspond to the pivot in that row, and its corresponding coefficient in the linear combination is the entry in that same row, in the current non-pivot column. It is kind of confusing to explain but hopefully you can see by the example I gave.

This is perhaps not what you expect but I give it anyway so you see another way to solve your problem. Since given vector components are simple you can try to see at first sight with a little attention about linear relations among them. You have $$a_2-a_1=a_3\\a_4-a_3=a_5$$ hence $$a_4+a_1-a_2=a_5$$ Putting now, in order to know if $a_1,a_2,a_4$ are linearly independent, $$\alpha a_1+\beta a_2+\gamma a_4=0\iff \begin{cases} \alpha+2\beta+\gamma=0\\\beta+2\gamma=0\\\beta+3\gamma=0\\-\alpha +4\gamma=0\end{cases}\iff \alpha=\beta=\gamma=0$$ Consequently you have dimension $3$ and you can take the base $$\{a_1,a_2,a_4\}$$

• How is it that we are in 3 dimensions even though each vector has 4 components? I'm having trouble reconciling the idea of a vector with 4 components in a 3-dimensional space (took Linear a while ago now, so I know I'm missing something but I can't figure out what). I do understand though (I think...) the idea that $x$ linearly independent vectors define a basis in $x$ dimensions, just not how we can have 4 components per vector in a 3D space. – jeremy radcliff May 28 '16 at 6:49
• Ah...I think I'm being dumb. Is it just that they define a 3D subspace in their original 4D space? – jeremy radcliff May 28 '16 at 6:51
• Even in dimension $n$ the subspace generated by a vector (which obviously has $n$ components because it belongs to the space with $n$- dimension) has dimension 1. And two linearly independant vectors generate a subspace of 2-dimension, etc. – Piquito May 29 '16 at 3:35
• Yes, say $\left\{(1,0,0,0),(0,1,0,0)\right\}$ is a basis for a $2$-dimensional subspace of the $4$-dimensional space $\mathbb{R}^{4}$. Vectors in this subspace take the form $k(1,0,0,0)+l(0,1,0,0)=(k,l,0,0),\ k, l\in \mathbb{R}$. So while the vectors in this subspace have $4$ coordinates, the 3rd and 4th coordinates must be zero, hence, a $2$-dimensional subspace. – yung_Pabs May 30 '16 at 4:37