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Let $v_1,v_2,v_3,p,q \in \mathbb{R}_2[x]$. We have that $p=a+bx+cx^2, q = a'+b'x+c'x^2$. The scalar product in $\mathbb{R}_2[x]$ is defined as $\left \langle p,q \right \rangle=aa'+2bb'+cc'$

$v_1= \begin{pmatrix} 2\\ 0\\ 0 \end{pmatrix}, v_2=\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}, v_3=\begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix}$

Calculate $\left \| v_1 \right \|, \left \langle v_1,v_2 \right \rangle, \left \| v_1+v_2 \right \|$

Could you please tell me if I do it correct?

$\left \| v_1 \right \|= \sqrt{2^2+0^2+0^2}= \sqrt{4}=2$

$\left \langle v_1,v_2 \right \rangle= 2 \cdot 0+2 \cdot 0 \cdot 1 +0 \cdot 1 =0$

$\left \| v_1+v_2 \right \|= \sqrt{2^2+0^2+0^2+0^2+1^2+1^2}= \sqrt{6}$

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    $\begingroup$ The first one is correct numerically, but only because the $2^{nd}$ component of the vector is $0\,$. To make that more obvious, it might be better written as $\left \| v_1 \right \|= \sqrt{2^2+\color{red}{2} \cdot 0^2+0^2}=\cdots\,$. Second one is fine. $\endgroup$
    – dxiv
    Mar 16, 2017 at 23:05
  • $\begingroup$ @dxiv What's the general formula for $\left \| v_1 \right \|$? $\endgroup$
    – tenepolis
    Mar 16, 2017 at 23:12
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    $\begingroup$ Given that you said nothing to the contrary, the norm would be assumed to be defined in terms of the scalar product $||v||=\sqrt{\langle v,v \rangle}\,$, and the posted answers show how that works. $\endgroup$
    – dxiv
    Mar 16, 2017 at 23:15
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    $\begingroup$ @dxiv Thank you, now I got it ! :) $\endgroup$
    – tenepolis
    Mar 16, 2017 at 23:25

2 Answers 2

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$\|v_1 + v_2\| = (\langle v_1+v_2,v_1+v_2\rangle)^{\frac 12}$

Two ways you can go by.

$v_1+v_2 = \begin{pmatrix} 2\\1\\1\end{pmatrix}$

$\|(2,1,1)\| = \sqrt {2^2 + 2\cdot 1^2 + 1^2} = \sqrt 7$

or

$\langle v_1+v_2,v_1+v_2\rangle = \langle v_1,v_1\rangle + 2\langle v_1,v_2 \rangle + \langle v_2,v_2\rangle = 4+0+3$

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Seems fine besides the last one though the answers coincide.

$$\left \| v_1+v_2 \right \|= \sqrt{\langle \begin{bmatrix} 2 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 1\end{bmatrix}\rangle}=\sqrt{2^2+2(1)(1)+1^2}$$

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  • $\begingroup$ Could I also do this instead? $\left \| v_1+v_2 \right \|= \left \| v_1 \right \|+\left \| v_2 \right \|= \sqrt{2^2+0^2+0^2}+\sqrt{0^2+1^2+1^2}$ $\endgroup$
    – tenepolis
    Mar 16, 2017 at 22:46
  • $\begingroup$ Nope. $$\sqrt{6} \neq 2 + \sqrt{2}$$ $\endgroup$
    – Siong Thye Goh
    Mar 16, 2017 at 22:48
  • $\begingroup$ Oh okay.. Can you please give me a link that gives me these formula? Or some topic name. Thank you for answer man! $\endgroup$
    – tenepolis
    Mar 16, 2017 at 22:49
  • $\begingroup$ I made a mistake just now due to definition of your inner product. Correcting now. $\endgroup$
    – Siong Thye Goh
    Mar 16, 2017 at 22:53
  • $\begingroup$ $\|v_1 + v_2\| \ne \|v_1\| +\|v_2\|$ $\endgroup$
    – Doug M
    Mar 16, 2017 at 22:57

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