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This question already has an answer here:

I just noticed that for any two Cartesian vectors their dot product is precisely the only non-zero eigenvalue (if such exists) of the following matrix:

$$\vec{a}=(a_1,a_2,a_3,\dots)$$

$$\vec{b}=(b_1,b_2,b_3,\dots)$$

$$A_{mn}=a_m b_n$$

For example:

$$\vec{a}=(a_1,a_2)$$

$$\vec{b}=(b_1,b_2)$$

$$A= \left[ \begin{matrix} a_1 b_1 & a_1 b_2 \\ a_2 b_1 & a_2 b_2 \end{matrix} \right]$$

$$\left| \begin{matrix} a_1 b_1-x & a_2 b_2 \\ a_2 b_1 & a_2 b_2-x \end{matrix} \right|=x^2-(a_1b_1+a_2b_2)x=0$$

$$x_1=0,~~~x_2=a_1b_1+a_2b_2=\vec{a} \cdot \vec{b}$$

The same works for vectors of any dimension. We obtain the following characteristic polynomial of $A$:

$$x^d-(a_1b_1+a_2b_2+\dots+a_db_d)x^{d-1}=0$$

Of course for two perpendicular vectors their dot product will be zero, so the characteristic polynomial will just be:

$$x^d=0$$

Is there some deeper meaning behind all this?

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marked as duplicate by 6005, Community Jul 19 '16 at 0:00

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    $\begingroup$ The trace of a matrix is the sum of its eigenvalues. Here the trace is just the dot product. $\endgroup$ – smcc Jul 18 '16 at 23:38
  • $\begingroup$ @smcc, yes, I know. I'm more interested in the fact that there is only one non-zero eigenvalue $\endgroup$ – Yuriy S Jul 18 '16 at 23:39
  • $\begingroup$ But the way you have constructed it don't all the other terms cancel out because of the symmetry of the matrix. $\endgroup$ – smcc Jul 18 '16 at 23:40
  • $\begingroup$ @smcc, I see it for $2 \times 2$ matrix, but for higher dimesions? Not obvious to me $\endgroup$ – Yuriy S Jul 18 '16 at 23:41
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    $\begingroup$ See: math.stackexchange.com/questions/55165/… $\endgroup$ – user258700 Jul 18 '16 at 23:47