# Orthogonal complement not resulting ok

I have the subspaces:

$$S = \langle(2,1,-1), (-1,2,0)\rangle, \qquad T = \{ X + Y + 2Z =0; X - Y - Z = 0\}.$$

I got that $$T = \langle(1, 3,-2)\rangle$$.

All vectors are linearly independent, so $$S + T = \mathbb{R}^3.$$ Then I tried to calculate the orthogonal complement for both.

It is my understanding that the orthogonal complement for $$S$$ should give a vector that is a multiple of $$(1,3,-2)$$ (generates the same subspace as $$T$$) and the orthogonal complement for $$T$$ should give vectors that are linearly dependent from the ones in $$S$$ (generates the same subspace).

However, complement of $$S = t(2,1,3)$$ from my calculations, and complement of $$T = t(-3,1,0) + s(2,0,1)$$.

What I did was resolving the system

$$2x + y - z = 0 \\ - x + 2y = 0$$

For $$S$$, and for $$T$$:

$$x + 3y - 2z = 0 .$$

Am I doing something wrong?

Edit: Maybe I'm understanding wrong a part of the theory, but I read that $$S$$ and its orthogonal complement will generate $$\mathbb{R}^n$$. The orthogonal complement for T is indeed resulting in $$0$$ for the scalar product. But if $$\dim(S) + \dim(S^\bot) = 3$$, and $$\dim(S) + \dim(T) = 3$$, shouldn't $$T$$ and $$S^\bot$$ be the same subspace?

• $S^\perp=\{(2,1,5)\}$ and $T^\perp=\{(1,1,2)\}$. Orthogonal complements need not be unique. – Yadati Kiran Nov 22 '18 at 13:45

There is absolutely no reason why a vector orthogonal to $$S$$ should be an element of $$T$$. This would be true if $$T$$ were actually the orthogonal complement of $$S$$, but it is not. Clearly $$(1,3,-2)$$ is not orthogonal to $$S$$.
You seem to believe that a complement of $$S$$ is unique. It is not the case (take the line generated by any vector outside $$S$$).
PS. Note that your calculations for the complement of $$S$$ are certainly wrong, since the vector you obtain is not orthogonal to the generators of $$S$$...
• Your example shows that the answer to your edit is NO. Once again, $T$ is NOT the orthogonal complement of $S$ because it is NOT orthogonal to $S$. Ypu have infinitely many possible complements for $S$. – GreginGre Nov 22 '18 at 13:46