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Consider the matrix $$A = \begin{pmatrix}3 & 5 & 11 \\5 & 9 & 20\\11&20&49\end{pmatrix}$$

I have attached an image of the question here.

Could someone please solve the first question, that is the magnitude of $v_2$ . I am not sure which column the vectors belong to, thus I am unable to solve any of the parts. $$\begin{pmatrix}v_1v_1 &v_1v_2 & v_1v_3\\ v_2v_1 &v_2v_2& v_2v_3\\ v_3v_1 &v_3v_2 &v_3v_3\end{pmatrix}$$

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  • $\begingroup$ Here you will find a MathJax tutorial for future questions. Also I think you should read this. $\endgroup$ Jul 10, 2016 at 18:16
  • $\begingroup$ @AaronMaroja Thanks. I also figured out how to solve the problems. $\endgroup$
    – Irina
    Jul 10, 2016 at 18:19

1 Answer 1

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The problem says that the entries of the matrix are given by the dot product of the vectors in $\mathbb R^4$.

For instance, $$ 3 = a_{11} = \vec v_1 \cdot \vec v_1 \\5 = a_{12} = \vec v_1 \cdot \vec v_2\\11 = a_{13} = \vec v_1 \cdot \vec v_3$$ and so on.

Here, the matrix $A$ has the form $$A = \begin{pmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{pmatrix}$$

Extra: Here are some useful hints:

i) $\|\vec v_2\|^2 = \vec v_2 \cdot \vec v_2$;

ii) $\|\vec v_1 + \vec v_2\| \leq \|\vec v_1\| + \|\vec v_2\|$;

iii) $\cos \theta = \frac{\langle \vec v_2, \vec v_3\rangle}{\|\vec v _2\| \|\vec v_3\|}$, where $\theta$ is the angle between $\vec v_2$ and $\vec v_3$.

Just a few ideas so you can get started.

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  • $\begingroup$ I understand that, but if we for instance asked to compute the magnitude of v2 i would think that we would have to find the length of the first column, but this is not correct. So i am not sure how the dot product between the two vectors ties to just v2 for instance. $\endgroup$
    – Irina
    Jul 10, 2016 at 17:52
  • $\begingroup$ So the problem isn't about notation. You want to solve your question. You should rephrase it then. $\endgroup$ Jul 10, 2016 at 17:53
  • $\begingroup$ I believe the issue is that I forgot that the dot product of a vector with itself gives its length. $\endgroup$
    – Irina
    Jul 10, 2016 at 18:05
  • $\begingroup$ @Irina That's not true. It gives you something related to its length, but not the length itself. $\endgroup$ Jul 10, 2016 at 18:06
  • $\begingroup$ @MatthewDrury my mistake, it is the square of the length, thus in order to find the length, do we square root it, here it would be 3. $\endgroup$
    – Irina
    Jul 10, 2016 at 18:11

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