Find a basis for the subspace \begin{align} V=\{ (x_1,x_2,x_3,x_4)^T\in \mathbb{R}^4 | x_1-3x_2+5x_3 -6x_4=0\} \end{align} where $V\subset \mathbb{R}^4$

The basis ends up being spanned by the vectors of $S=\{(3,1,0,0)^T,(-5,0,1,1)^T,(6,0,0,1)^T\}$

My question:

Don't I need 4 linearly independent vectors for a basis of $\mathbb{R}^4$?

  • $\begingroup$ 4 Vectors would span all $\mathbb{R}^4$ but since you restrict yourself to the hypersurface $x_1-3x_2+5x_3-6x_4=0$ you are left with 3 vectors. $\endgroup$ – k1next Nov 13 '14 at 8:20
  • $\begingroup$ Yes, $\mathbb{R}^4$ needs a set of 4 independent vectors to form its basis. However, since you are considering $V$ (i.e, $\mathbb{R}^4$'s subspace), so the number of vector in its basis is $\le 4$, depending on $V$'s dimension. And in this case, $V$'s dimension is 3. $\endgroup$ – user49685 Nov 13 '14 at 8:24
  • $\begingroup$ Note that $V\neq \mathbb{R}^4$. Note that its all solution to $x_1-3x_2+5x_3 -6x_4=0$ which only has 3 free parameters which implies it has dimension 3 thus since set $S$ has 3 linear independent vectors of $\mathbb{R}^{4}$ and are in $V$ this must be a basis for $V$ $\endgroup$ – Kamster Nov 13 '14 at 8:27
  • $\begingroup$ If you add $(1,-3,5,-6)^{T}$ to your basis, then you'll get a full basis of $\mathcal{R}^{4}$. Your basis elements are, by definition, orthogonal to this vector. $\endgroup$ – DisintegratingByParts Nov 13 '14 at 10:11

Let's take a look at a $3D$ example.


We can ignore the colors (it's just some image I found in a quick websearch). The plane could be described by the following set \begin{align} V=\{ (x_1,x_2,x_3)^T\in \mathbb{R}^3 | ax_1+bx_2+cx_3=0\} \end{align} Where $a,b,c$ depend on the plane. Clearly, the dimension of the space $V$ is only $2$ and not $3$, since all elements of $V$ lie on the (colored) plane. Thus it is possible to give a basis of $V$ by two elements $v_1,v_2\in \mathbb{R}^3$, such that $V=span\{v_1,v_2\}$.



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