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?

  • $\begingroup$ $S^\perp=\{(2,1,5)\}$ and $T^\perp=\{(1,1,2)\}$. Orthogonal complements need not be unique. $\endgroup$ – 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 $S$...

  • $\begingroup$ Please see my edit $\endgroup$ – JorgeeFG Nov 22 '18 at 13:43
  • $\begingroup$ 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 $S$. $\endgroup$ – GreginGre Nov 22 '18 at 13:46

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