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Let $X = \mathbb{R}^2$. Find the annihilator $M^{\perp}$ if $M$ is:

(a) $\{x\}$, where $x = (x_1,x_2) \not=0$.

(b) a linearly independent set $\{x_1,x_2\} \subseteq X$.

Here is what I tried so far. In both parts, I am stuck, not sure how to understand this fundamentally.

For part (a): \begin{align} M^{\perp}&=\{y \in X :y \perp M\} & \text{by definition of } M^\perp \\ &= \{y \in \mathbb{R}^2 : y \perp \{x\}\}\\ &= \{y \in \mathbb{R}^2 : \langle y,x \rangle=0\} \\ &= \{y \in \mathbb{R}^2 : y=c(-x_2,x_1)\} \end{align} and for part (b): \begin{align} M^{\perp}&=\{y \in X :y \perp M\} & \text{by definition of } M^\perp \\ &= \{y \in \mathbb{R}^2 : y \perp \mathbb{R}^2\}\\ &= \{y \in \mathbb{R}^2 : \langle y,\underbrace {\underset{\begin{subarray}{c} \text{linearly} \\ \text{independent} \end{subarray}}{\{x_1,x_2\}}}\rangle=0\} \\ &= \{y \in \mathbb{R}^2 : y=(0,0)\} \\ &= \{(0,0)\} \end{align}

Cookie
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    You might consider changing your notation since you are using $x$ as a fixed point in $\Bbb{R}^2$ and $x$ as the general symbol for an arbitrary point in $\Bbb{R}^2$. For example, in part a, your last equality has $\langle x, x \rangle = 0$, but the $x$ in the left and right arguments of the inner product aren't the same element. – Tom Oct 05 '14 at 21:19
  • Have in mind what you are trying to get geometrically. In case $a)$ $M$ consists of a line through the origin. So $M^{\perp}$ is the line through the origin perpendicular to line $M.$ To part $b),$ two linearly independent vectors generate all the plane. So, $M$ is the whole plane and $M^{\perp}$ only contains one vector. – mfl Oct 05 '14 at 21:20
  • @mfl oop just realized your comment basically coincides with my answer! Sorry! – DanZimm Oct 05 '14 at 21:36
  • @DanZimm There is no problem. Sometimes it also happens to me. – mfl Oct 05 '14 at 21:38
  • @mfl I made my corrections to part b. Since the set of two linearly independent vectors generate $\mathbb{R}^2$ as you pointed out, which means there is only one vector in $M^\perp$, which is $(0,0)$. Is that a good answer? – Cookie Oct 05 '14 at 22:16
  • Yes, it is the correct answer. – mfl Oct 05 '14 at 22:19

1 Answers1

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Hint: Given a vector $v \in \mathbb{R}^2$ how can we find an orthogonal vector, i.e. if we draw this vector $v$ what must we do to find an orthogonal vector (i.e. what angle do we have to rotate through).

Hint:$^2$ Gram-Schmidt.

Hint:$^3$ If we have $n$ linearly independent vectors in $\mathbb{R}^n$ what can we say about the span of those vectors?

DanZimm
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  • Okay, for the first hint, we would rotate that vector 90 degrees counterclockwise or clockwise. For the second hint, is using the Gram-Schmidt process necessary here? It isn't introduced in my book (Functional Analysis by Kreyszig) until the next section, after the section containing this exercise I posted. – Cookie Oct 05 '14 at 21:37
  • It's not "Graham Schmidt", but Gram-Schmidt, with Gram as in Jørgen Pedersen Gram. – Daniel Fischer Oct 05 '14 at 21:39
  • @DanielFischer right, sorry – DanZimm Oct 05 '14 at 21:40
  • @legâteauaufromage exactly, try and think about how one could write a vector that's been rotated 90 degrees. Gram-Schmidt isn't necessary but it does give an alternate way of understanding how to rotate this vector. – DanZimm Oct 05 '14 at 21:41
  • If I have $(x_1,x_2) \in \mathbb{R}^2$, and I want to rotate this 90 degrees counterclockwise for example, would I end up with $(-x_2,x_1)$? Same thing with 90 degrees clockwise: $(x_2,-x_1)$? – Cookie Oct 05 '14 at 21:44
  • @legâteauaufromage yep I would agree with this. Now consider any scalar times $(x_2, -x_1)$, do all of these vectors satisfy the orthogonality condition to $(x_1, x_2)$? Are these all the possible vectors that do so? (I would try using double inclusion to prove this). – DanZimm Oct 05 '14 at 21:51
  • Yes, the scalar times could be all real numbers. So $c(x_2,-x_1)$ and $c(-x_2,x_1)$ (for all $c \in \mathbb{R}$) both work, because taking the dot product with $x=(x_1,x_2)$ for each orthogonal vector will result in $0$, and the constant $c$ will be absorbed with it. – Cookie Oct 05 '14 at 21:55