# Finding a basis for a subspace in $\;\Bbb R^4\;$

I know this might be a really simple question to ask but I just don't understand how to obtain the answer to this question. I've tried to understand subspaces (and even the difference between a space and a subspace) but I just can't seem to wrap my head around the concept. Any help with this question would be much appreciated.

• A subspace of a vector space is simply a subset that is also a vector space in its own right. – Travis Sep 27 '14 at 14:22
• A basis for the subspace of $\Bbb{R}^4$ spanned by the set $\{v_1,v_2,v_3,v_4\}$ can be a subset of such set with the linearly independent vectors. Note that $v_1=v_2+v_4$, so that vector is a linear combination of others vector of your basis. – DiegoMath Sep 27 '14 at 14:26

Mechanically, the problem is straightforward. Make your vectors the columns of a matrix. Row-reduce the matrix and locate the pivot columns in the reduced form. The original vectors in those columns constitute a basis for the subspace spanned by the given vectors.

• What do you mean by turning the vectors into columns of a matrix? – Amanda Sep 27 '14 at 14:32
• Construct a matrix having columns as your vectors i.e $\displaystyle A=(v_1 v_2 v_3 v_4)$ – Bhauryal Sep 27 '14 at 14:39

Problem solving: Change this matrix to reduced row echelon form: $$\begin{bmatrix} 2 & 1 & 1 & 1 \\ -1 & 0 & 1 & -1 \\ 2 & 1 & 1 & 1 \\ 1 & 0 & -1 & -1\\ \end{bmatrix}$$

which gives, (you should finally get 3 linearly independent columns).

$$\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\\ \end{bmatrix}$$

So, the basis (B) is $$B=\lbrace v_1, v_2,v_4 \rbrace$$ because 1,2 and 4 solved out to be the linearly independent columns.

Understanding: Imagine $$v_1= \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}, \ v_2= \begin{bmatrix} 2 \\ 1 \\ \end{bmatrix},\ v_3= \begin{bmatrix} 4 \\ -1 \\ \end{bmatrix}$$

These three vectors represent 3 different points in $\mathbb R^2$. We don't know right now whether these 3 are linearly independent or not. But, clearly these are 3 differently looking vectors in $\mathbb R^2$. Also remember, span of 1 vector in $\mathbb R^2$ is a straight line. Span of 2 different (or more accurately, linearly independent) vectors is a plane. Now, the rref of the example matrix will be $$\begin{bmatrix} 1 & 0 & 6 \\ 0 & 1 & -1 \\ \end{bmatrix}$$ which means basis = $\lbrace v_1, v_2 \rbrace$

Because $v_3$ here was the extra information, we left it out while writing our basis. Our plane can be spanned by the linear combination of the 2 basis vectors $\lbrace v_1, v_2 \rbrace$ only.