# Linear Algebra: finding a basis for a subspace of $\mathbb{R}^4$

So I am stuck on this example from my Into. Linear Algerbra book

I'm not exactly sure how I'm supposed to find the basis in this case. Am I just supposed to use a random t and s value and call the single vector a basis? (There were no previous examples in the book that were similar)

Thank you

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What do you think the dimension of W is? – John M Jul 2 '11 at 4:13
How much information do you need in order to determine a vector in $W$? How can you specify that information? That will tell you what a basis for $W$ is. – Arturo Magidin Jul 2 '11 at 4:17
Split your vector into a sum of two vectors; one with only s-components and the other with only t-components. – gary Jul 2 '11 at 4:26
In other words, write $(2s-t,s,t,s) = s (\cdot,\cdot,\cdot,\cdot) + t (\cdot,\cdot,\cdot,\cdot)$. The two vectors are obviously a basis. – anon Jul 2 '11 at 5:04
A basis for a subspace is a $linearly$ $independent$ set which $spans$ the space. So you need to check these two conditions. The clues give you a pretty obvious spanning set, and every finite spanning set contains a basis - so you need to check linear independence. Next, since the size of the basis is the dimension of the space, any set of linearly independent vectors of that size will be a basis - and a set of the right size chosen from the subspace at random will probably be independent. But best to go with the obvious ones which make it easy to check. – Mark Bennet Jul 2 '11 at 6:23

Virtuoso, you have two free variables to choose from in forming a vector in a subspace $\mathbb{W}$, namely $s$ and $t$, they determine your vector. So, we know that we can represent any vector in $\mathbb{W}$ as $\begin{bmatrix} 2s-t\\ s\\ t\\ s \end{bmatrix}$ = $\begin{bmatrix} 2s\\ s\\ 0\\ s \end{bmatrix} + \begin{bmatrix} -t\\ 0\\ t\\ 0 \end{bmatrix}$. Here, I am just decomposing any generic vector in our subspace into the independent components -- you can't break it down any further, as it is determined by these two variables. Now, you can pull out $s$ and $t$, to get that any vector in $\mathbb{W}$ can be represented as $s \begin{bmatrix} 2\\ 1\\ 0\\ 1 \end{bmatrix} + t \begin{bmatrix} -1\\ 0\\ 1\\ 0 \end{bmatrix}$, where you can pick any $s, t \in \mathbb{R}$. In other words, vectors $\begin{bmatrix} 2\\ 1\\ 0\\ 1 \end{bmatrix}$ and $\begin{bmatrix} -1\\ 0\\ 1\\ 0 \end{bmatrix}$ span the subspace $\mathbb{W}$, and since they are linearly independent, they form a basis for your subspace. Since the dimension of the subspace is equal to the number of linearly independent vectors needed to span the subspace (or, alternatively speaking, the number of basis vectors), you can infer that the dimension of $\mathbb{W}$ is equal to 2.

Note, that this is not the basis, as there is no such thing -- there are infinitely many combinations of two vectors that would form a basis for $\mathbb{W}$. All you need is for these two vectors to be (1) linearly independent and (2) contained in the subspace $\mathbb{W}$. So, you could, for example, multiply our basis vectors by any real numbers you want, as one way of getting a new couple of basis vectors.

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I think it is clearer if you write if out like this (à la gary above): The set, namely $W$ consists of all points $(w,x,y,z)$ in $\mathbb{R}^4$ such that

$\left[\begin{array}{c} w\\ x\\ y\\ z\end{array}\right]=s\left[\begin{array}{c} 2\\ 1\\ 0\\ 1\end{array}\right]+t\left[\begin{array}{c} -1\\ 0\\ 1\\ 0\end{array}\right]$.

À la Mark Bennet, you need to check that this linear system above is consistent for every L.H.S. Big hint: consider the matrix formed by the two vectors above as columns, namely the matrix:

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

At most how many pivot positions can I have when the matrix above is in reduced row echelon form? What does this mean?

Also, what's so special about the vectors in the system, they have 0's and 1's in there, so how do you know just by these 0's and 1's in there that the two vectors are linearly independent?

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