# Bases for spaces of solutions

I have these spaces of solutions and im supposed to find an orthogonal basis for them. I can do this,if i have the regular basis for them. And i have no idea how to compute them:

a) $$\begin{cases} 2x+y-z=0 \\ y+z=0\end{cases}$$

For this the rank is obviously 2. Trying to solve the system I get: $$\begin{cases} z=-y \\ 2x = -2y \end{cases}$$ But I'm not sure where to go from here.

b) $$x-y+z = 0$$

Rank is 1 obviously. Solving the system gets me nowhere useful from what I can tell. If the rank is 1, shouldn't a valid basis for the space be $$[1;0;0]$$?

• For a) just take arbitrary $y$, for example $y = 1$. For b), this vector doesn't belong to our subspace. Just take $y = 1, z = 0$ and $y = 0, z = 1$. – Jakobian May 21 at 21:52
• Given that the rank is 2 in a), shouldn't the space be defined with 2 variables? It being defined by only y would make it 1-dimentional if i am correct. – user569685 May 21 at 21:54
• No, rank is the dimension of the image of a matrix. How much dimensional the subspace being generated by an equation $Ax = 0$ is the same as asking what is the dimension of the kernel of $A$. If $r$ is rank and $n$ the dimension of the space we are working in, then that is $n-r$ – Jakobian May 22 at 13:54

## 2 Answers

In the first case, the single vector $$(-1,1,-1)$$ forms an orthogonal basis.

In the second case a basis is $$\left\{\, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \,\right\}$$ and you can orthogonalize it with Gram-Schmidt.

Why is it a basis? The equation is $$x=y-z$$, so it has two free variables. You get a basis by giving them suitable values, typically $$y=1$$ and $$z=0$$ for the first vector and then $$y=0$$ and $$z=1$$ for the second vector. This ensures linear independence.

I will be writing all vectors as tuples rather than column vectors simply for ease.

In the first case, after row reduction you get

$$1x+0y-1z=0$$
$$0x+1y+1z=0$$

The $$1x$$ in the first row and the $$1y$$ in the second row are pivots, and thus can be given in terms of non-pivots, in this case just $$z$$: $$x=z$$ and $$y=-z$$. This means $$(x, y, z) = (z, -z, z)$$. Factor out $$z$$ and you get that all solutions are of the form $$z*(1,-1,1)$$. So $$(1,-1,1)$$ is a basis.

In the second case, since you only have one equation, it's already row reduced, and $$y$$ and $$z$$ are your two non-pivot variables. This means that you can take the two dimensional space that you get just varying $$y$$ and $$z$$, and any basis of that space will translate to a basis of the solution space. For instance, if you take $$(y,z) = (1,0)$$ and $$(y,z)= (0,1)$$, the first one gives you that $$x = y-z = 1-0=1$$, so the full vector is $$(1,1,0)$$. The second one gives you $$x = -1$$, or $$(-1,0,1)$$. This gives you the basis $$\{(1,1,0), (-1,0,1)\}$$.

So:

Row reduce, and then for each non-pivot variable, set it to one and the other non-pivots to zero, and solve for the pivot variables. This gives you a vector for each non-pivot variable, and the set of all such vectors is a basis.