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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 $[1;0;0]$?

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  • $\begingroup$ 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 $y = 0, z = 1$. $\endgroup$ – Jakobian May 21 at 21:52
  • $\begingroup$ 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. $\endgroup$ – user569685 May 21 at 21:54
  • $\begingroup$ 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 $n-r$ $\endgroup$ – Jakobian May 22 at 13:54
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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.

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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.

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