How can I find the basis of a subspace in $R^n$?

Question

Linear algebra test tomorrow. How can I find the basis of a vectorial subspace of $R^n$? The exam question that made me ask this was:

Find the basis of the subspace of $R^3$ generated by $S=\{(-1, 2, 5), (3, 0, 3), (5, 1, 8)\}$

Answer 1

That's a set of 3 vectors so any space spanned by has dimension less than or equal to 3. It will have dimension 3 if and only if all three vectors are independent. It that case the three given vectors are a basis. It will have dimension 2 if and only if two of the vectors are independent but the third is a linear combination of those two vectors. In that case the two independent vectors form a basis. It will have dimension three if only all vectors are multiples of one. In that case any one of the three is a basis by itself.

Answer 2

Write the vectors as rows of a matrix and perform row reduction. The nonzero rows will form a basis of the subspace generated by the original nows.

In your example, the reduced matrix is $$ \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{pmatrix} $$ The two nonzero rows are linearly independent because in a linear combination $a_1 r_1 + a_2 r_2 =0$, we must have $a_1=0$ because the first coordinate of $r_2$ is zero. Likewise, $a_2=0$ because the second coordinate of $r_1$ is zero.

Answer 3

A basis of a subspace is a set of vectors that both spans the entire subspace and is linearly independent.

Given a set of vectors $S=\{(-1,2,5),(3,0,3),(5,1,8)\}\subset\mathbb{R}^3$

Lets check which vectors are linearly independent and spans S. We want to find $x,y,z$ such that :

$x(-1,2,5)+y(3,0,3)+z(5,1,8)=0$, by doing so, we get a system of equations on $x,y,z$.

As a matrix, ths system of equation would look like this :

$ \begin{gather} \begin{bmatrix} -1&3&5\\2&0&1\\5&3&8 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix}=0 \end{gather} $

We can simplify using row reduction without changing the span of the three vectors, and we get :

$ \begin{bmatrix} 1&0&0.5\\0&1&\frac {11}{6} \\0&0&0 \end{bmatrix} $

Which translates into:

$x+0.5z=0$

$y+\frac {11} {6}z=0$

$0=0$

since $z$ is arbitrary (We have no constraints on $z$), we can pick $z=1$ and get :

$\frac {1}{2}(-1,2,5)+\frac {11}{6}(3,0,3)=(5,1,8)$

So the third vector is a linear combination of the first two, meaning the span of the two first vectors is the same span of S. In addition, the first two vectors are linearly independent. So $E=\{(-1,2,5),(3,0,3)\}$ is a basis of S.

A more simple approach would be to arrange the given vectors as rows in a matrix, and find the reduced row form. Since the span does not change, we can take the non-zero rows as our basis vectors.