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 $\mathbb{R}^4$?
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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