W is generated by {(1,-2,0,3), (1,-1,-1,4), (1,0,-2,5)}

I have found out that the third vector is linearly dependent on the previous two. Thus the basis of the solution space is W = {(1,-2,0,3), (1,-1,-1,4)}

Now I haven't been able to figure out what to do next for two days. I have Googled similar problems, run through different texts and read some previous answers but couldn't understand the concept behind the methodology employed to solve this problem.

If someone can give me the basic algorithm itself on how to make an approach to such problems, I will be grateful.

Thank you for your time.

  • $\begingroup$ What is the question? How does one find a basis of a given vector space? $\endgroup$ Jan 20, 2016 at 13:38
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    – Frentos
    Jan 20, 2016 at 13:44
  • $\begingroup$ @Travis Hi. The primary question is what I've put in the title. I"m sorry for ambiguity posed. $\endgroup$
    – Vibhu
    Jan 25, 2016 at 17:21

2 Answers 2


I assume that you want to find two linearly independent equations

$$ a_{11}x + a_{12}y + a_{13}z + a_{14}w = 0, \\ a_{21}x + a_{22}y + a_{23}z + a_{24}w = 0 $$

whose solution space is the vector space $W$. To do that, write a general linear equation

$$ ax + by + cz + dw = 0 $$

and plug in for $(x,y,z,w)$ a basis of $W$ yielding the equations

$$ a - 2b + 3d = 0, \\ a - b - c + 4w = 0. $$

Solving the equations for $(a,b,c,d)$ will give you a two-dimensional subspace of solutions and choosing any two linearly independent vectors $(a_{11}, a_{12}, a_{13}, a_{14})$ and $(a_{21}, a_{22}, a_{23}, a_{24})$ from the solution space will give you two linearly independent equations whose solution space is the vector space $W$.

There is an important high level concept called duality that can explain why this works. Solving a system of linear equations using Gauss eliminations allows you to pass from a subspace $W \subseteq \mathbb{F}^n$ described as a solution space of linear equations

$$ W = \{ v \in \mathbb{F}^n \, | \, \varphi_1(v) = \ldots = \varphi_k(v) = 0 \} $$

(here $\varphi \colon \mathbb{F}^n \rightarrow \mathbb{F}$ are linear functionals, not necessarily independent) to a description of $W$ of the form

$$ W = \mathrm{span} \{ v_1, \ldots, v_l \} $$

where $v_1, \ldots, v_l$ are vectors in $\mathbb{F}^n$ that form a basis for $W$. By treating the vectors $v_i$ as functionals on $(\mathbb{F}^n)^{*}$, one can see that the same procedure (under some identifications) allows you to pass from a subspace $W$ described as

$$ W = \mathrm{span} \{ u_1, \ldots, u_k \} $$

where the $u_i$ are (not necessarily linearly independent) vectors in $\mathbb{F}^n$ to a description of $W$ of the form

$$ W = \{ v \in \mathbb{F}^n \, | \, \varphi_1(v) = \ldots = \varphi_l(v) = 0 \} $$

where the $\varphi_i$ are linearly independent functionals (interpreted as equations) whose common solution space is precisely $W$.

  • $\begingroup$ Oh thank you so much for this. :-) $\endgroup$
    – Vibhu
    Jan 25, 2016 at 18:34

$W$ is the set of all linear combinations of the basis vectors, i.e. all $c_1\begin{bmatrix}1\\-2\\0\\3\end{bmatrix} +c_2\begin{bmatrix}1\\-1\\-1\\4\end{bmatrix}$ for $c_1,c_2\in\mathbb{R}$. We can ask: when can $c_1\begin{bmatrix}1\\-2\\0\\3\end{bmatrix} +c_2\begin{bmatrix}1\\-1\\-1\\4\end{bmatrix}$ give me a particular vector $\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}$? What rules do $x_1,x_2,x_3,x_4$ have to follow to make it part of $W$?

Setting that up as $\begin{bmatrix}1&1\\-2&-1\\0&-1\\3&4\end{bmatrix}\begin{bmatrix}c_1\\c_2\end{bmatrix}=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}$ and solving we get: $\left[\begin{array}{cc|c}1&1&x_1\\-2&-1&x_2\\0&-1&x_3\\3&4&x_4\end{array}\right]$ $\rightarrow$ $\begin{array}{l}R2=R2+2R1\\R4=R4-3R1\end{array}$ $\rightarrow$ $\left[\begin{array}{cc|c}1&1&x_1\\0&1&2x_1+x_2\\0&-1&x_3\\0&1&x_4-3x_1\end{array}\right]$ $\rightarrow$ $\begin{array}{l}R3=R3+R2\\R4=R4-R2\end{array}$ $\rightarrow$ $\left[\begin{array}{cc|c}1&1&x_1\\0&1&2x_1+x_2\\0&0&2x_1+x_2+x_3\\0&0&x_4-5x_1-x_2\end{array}\right]$.

For consistency (for this to be a solution) we need RHS=$0$ for each row where LHS is all zeros. This gives us two constraints:

$2x_1+x_2+x_3=0$ and $x_4-5x_1-x_2=0$

Those are the homogenous equations you're looking for.

Sometimes when you've got fewer rows or more columns you'll get a reduced matrix without any $\left[0~0~|~\text{blah}\right]$ rows. In this case, the system is always consistent and the solution space is the whole space. You can think of the whole space as matching the rather odd equation $0x_1+0x_2+\cdots+0x_n=0$, which works for any $x_1,x_2,\cdots,x_n$.

  • $\begingroup$ Oh thank you so much! I'd tried one more method which makes use of the rref of the coefficient matrix. I am getting the same result. This method is more straight forward. :-) $\endgroup$
    – Vibhu
    Jan 25, 2016 at 19:21

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