# How to find the basis of the following vector space?

I'm trying to find the basis of the following vector space but I can't seem to be able to find it:

$W = \{x = (x_1 , x_2 , x_3 , x_4 , x_5 ) ∈ \Bbb R^5 : x_1 − x_3 − x_4 = 0\}$

I understand that the basis $\mathcal B$ should be a linearly independent set of vectors. So I'm assuming I have to solve the equation above to find the vectors for my basis $\mathcal B$. However, I'm not sure how to solve it in the first place.

Could someone please guide me on how to approach this question?

Thanks.

Rouché-Capelli theorem grants us that if $n$ is the number of equations that characterize the subspace $W$, we have $$\dim(W)=\dim(\mathbb{R^5})-n=4$$

Let v be the vector

$$v=\ (x_1,x_2,x_3,x_4,x_5)^T$$

If $v \in W$, it must satisfy $x_1-x_3-x_4=0 \implies x_1=x_3+x_4$, so a vector in $W$ must be of the kind

$$(x_3+x_4,x_2,x_3,x_4,x_5)$$

Can you find $4$ linearly independent vectors that satisfy the above constraint?

• Now this makes more sense. But my question still holds. How do I find the 4 independent vectors? Is it by trial and error? Right now I can see the two vectors $(1,0,1,0,0)$ and $(1,0,0,1,0)$ but I can't seem to get the other 2. I'm not even sure if the 2 I have is right. – nTuply May 29 '15 at 0:11
• Yes, the fastest way is indeed try and error. I would add the two vectors $(0,1,0,0,0)$ and $(0,0,0,0,1)$ which are clearly linearly independent from the two you listed. – Lonidard May 29 '15 at 7:35
• Here you have 4 parameters. I usually fix one to be $1$ and the others to be zero, find a vector, repeat again varying the parameter chosen to be equal to $1$. Once you reach the number of desired vectors, check if they are linearly independent and you are done! – Lonidard May 29 '15 at 8:19

First you should note that there is no unique basis for this space in the same way that there is no unique basis for $\mathbb R^4$. The way I would approach this is to try to find some vectors in $W$. Once you have one, how might you change it. How many degrees of freedom are there in the definition of $W$? This should tell you how many basis vectors to look for. Recall that all you need for a set of basis vectors is that:

• Any two vectors in the set are linearly independent
• The basis vectors span the set
• None of the basis vectors is $\mathbb 0$

Hint: You should expect there to be 4 basis vectors (think why). Now you just need to find 4 linearly independent vectors in $W.$

• By trying to input some values, I found out that $v_1 =(1,0,1,0,0)\: and \: v_2=(0,0,0,0,0)$ could be a basis. I am then not very sure, how to get the rest. I'm guessing it should be 5 since the dimension is 5 but then maybe $(0,0,0,0,0)$ is not counted, so that's why you say 4? – nTuply May 28 '15 at 23:02
• No there should be 4 vectors as the dimension of $W$ is 4. To get more vectors, try changing some numbers in the ones you have. Think about how some other numbers would need to be changed to keep them in $W$. Also, basis vectors need to be nonzero (otherwise you won't have a linearly independent set). – Dan Robertson May 28 '15 at 23:04
• But in my case I'm dealing with $\Bbb R^5$ shouldn't the dimension be 5, thus 5 linearly independent bases? Apart from trial and error(which might take very long) is there a process to actually get the bases by calculation? – nTuply May 28 '15 at 23:06
• There isn't really any process to get all the basis vectors by calculation. Also there will only be 4 of them. Consider this simpler example: Find the basis for the set $X=\{\mathbf x\in\mathbb R^2\,|\,\mathbf x=(x_1,x_2);\;x_1=x_2\}$. We get that $X\subset \mathbb R^2$ and $\mathbb R^2$ is clearly two-dimensional so has two basis vectors but $X$ is clearly a (one-dimensional) line so only has one basis vector. Each (independent) constraint when defining a subset reduces the dimension by 1. – Dan Robertson May 28 '15 at 23:11

The dimension of the space W is 4. We therefore want to find 4 linearly independent vectors.

To do this in a perhaps more systematic way, I recommend maintaining as many zeroes as possible as this will make independence clear.

For example, this is a basis for W:

$\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$ , $\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$ , $\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$ , $\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}$.