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In $\mathbb{R}^3 $ with $ \underline{N}= \begin{pmatrix} 1\\ 2\\ 1 \end{pmatrix} $
and the subspace $V={\underline{N}^\bot}$.

(a1) Are the following vectors linearly independent?

(a2)Define the dimension of V.
(a3) Show that the following vectors are vectors of/in V.

(a1)
$\underline{x}_1= \begin{pmatrix} 0\\ 1/2\\ -1 \end{pmatrix}$ , $\underline{x}_2= \begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix}$ , $\underline{x}_3= \begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix}$

$\begin{vmatrix} 0 & 1 & 1\\ 1/2 & 0 & -1 \\ -1 & -1 & 1 \end{vmatrix}=-1/2+1-1/2=0$
$\Rightarrow$ these vectors are linearly dependent.



(a2)
$\begin{vmatrix} 0 & 1\\ 1/2 & 0 \end{vmatrix}=-1/2 \ne 0$

$\begin{vmatrix} 1 & 1\\ 0 & -1 \end{vmatrix}=-1\ne 0$
$\Rightarrow Rg(M)=Rg \begin{pmatrix} 0 & 1 & 1\\ 1/2 & 0 & -1 \\ -1 & -1 & 1 \end{pmatrix}=2$
so that the Dimension of V is 2.

Is (a1) and (a2) done in this way correct?
For (a3)I don't know how to show that, can I use the parametric definition of a plane:
$\underline{N}=\underline{x}_1+r\underline{x}_2+s\underline{x_3}$
and if yes, does the order of the vectors matter?and how?

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    $\begingroup$ Dimension of $V$ equals 2 is correct. $\endgroup$ – Mhenni Benghorbal Apr 2 '13 at 13:41
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    $\begingroup$ A vector $\underline{v}$ is in $V$ if it is orthogonal to $\underline{N}$. Do you know how to check that 2 vectors are orthogonal? $\endgroup$ – Djaian Apr 2 '13 at 13:42
  • $\begingroup$ I think what you did for a2 is calculate the dimension of the vector space spanned by the 3 vectors $x_1$, $x_2$ and $x_3$. This happens to be 2, which is also dimension of $V$. But in my opinion you didn't answer a2 correctly. $\endgroup$ – Djaian Apr 2 '13 at 13:45
  • $\begingroup$ For a2 I calculate the subdeterminants of the Matrix M,because the det(M)=0 <=> rang(M)=< 3, and then I thought if the subdeterminants aren't equal to 0 the rang(M) has to be 2,so that the dim of the vectors building M is also 2. $\endgroup$ – Phil Apr 2 '13 at 13:51
  • $\begingroup$ @Phil but why would your matrix $M$ have any link to vector space $V$? See my answer below for more details. $\endgroup$ – Djaian Apr 2 '13 at 13:55
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Yes, (a1) and (a2) are correct, well, if I understand well, provided only that (a3) is true.

However, we can also argue by simpler arguments: For (a1), note that $\underline x_3=\underline x_2-2\cdot\underline x_1$, so these are linearly dependent. However, $\underline x_1$ and $\underline x_2$ are not parallel (then one would be a scalar times the other), so their rank must be at least $2$, so it is $2$.

For (a3), we need only to check that $\underline x_1$ and $\underline x_2$ are orthogonal to $\underline N$ (i.e. that they are in ${\underline N}^\perp$), then by the above equality,. it will also follow for $\underline x_3$. For this, nothing but the dot product is needed to be computed, and has to be zero: $$\underline N\cdot\underline x_1=1\cdot 0+2\cdot 1/2+1\cdot(-1)=1-1=0\,,$$ and verify similarly that $\underline N\cdot\underline x_2=0$. Since, in 3d, the orthogonal of a nonzero vector has dimension $2$, we have that $\underline x_1,\underline x_2$ forms a basis of $\underline N^\perp$ (actually, any two vectors of $\underline x_1,\underline x_2,\underline x_3$ do so).

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For (a3), to prove the vectors are in $V$, take the dot product of these vectors with $N$ and the dot product should be zero.

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a1 is correct. You calculated the determinant of the matrix containing the 3 vectors and since it is 0 you concluded correctly that they are linearly dependant. Another way to see it is to see that $2x_1 + (-1)x_2 + x_3 = 0$, which gives you a non-trivial linear combination of the 3 vectors giving 0, thus saying they are dependant.

Like I said in my comment I think you have the correct answer to a2 but with an incorrect calculation. Actually you calculated the dimension of the vector space spanned by $x_1, x_2, x_3$. But this might not be the same vector space as $V$ as far as you know.

Probably, the goal of exercise a2 was for you to find that dimension of $V$ was 2 because it is the orthogonal complement of a 1 dimensional vector space in $\mathbb{R}^3$.

[Edit: and here I mean using some argument seen in the course / book. No calculations is needed if you have some theorems on dimensions of orthogonal vector spaces in your book / course.]

Finally, like I said in another of my comments, for a3 you just need to use the definition of orthogonal complement. Other answers show you how to.

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  • $\begingroup$ Ok.I thought that if $\underline{x}_1$,$\underline{x}_2$ and $\underline{x}_3$ would be vectors of V and would be lin. independent, the questions were linked and I could use the rang of the Matrix built by these vectors to define the dimension instead of calculating the dimension by gaussion elimination with these vectors.I have also forgotten to use $V=\underline{N}^\bot$.. $\endgroup$ – Phil Apr 2 '13 at 14:16
  • $\begingroup$ @Phil Maybe you just have a theoretical argument to conclude dimension $V$ is 2 without any calculations. Your book / course might give theorems indicating that the orthogonal complement of a vector has dimension $n-1$ if the original vector space has dimension $n$. $\endgroup$ – Djaian Apr 2 '13 at 14:20

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