1
$\begingroup$

On pp. 280–281 of the book Solar System Dynamics (https://books.google.ca/books?id=aU6vcy5L8GAC&printsec=frontcover#v=twopage&q&f=true), eigenvalues $ g $ and eigenvectors $ \bar e_{ji} $ of matrix $ \mathbf A = \begin{pmatrix} +0.00203738 & -0.00132987 \\ -0.00328007 & +0.00502513 \end{pmatrix} $ (p. 280) are calculated. The results are $ g_1 = 9.63435 \times 10^{-4} $ and $ g_2 = 6.09908 \times 10^{-3} $ (p. 280, which I also get) and $ \bar e = \begin{pmatrix} -0.777991 & 0.332842 \\ -0.628275 & -1.01657 \end{pmatrix} $ (p. 281), or at least the way I understand the problem (I haven’t done matrices in 30+ years, and never did eigenvalues or eigenvectors before now).

I get wholly different results for $ \bar e $, though… $ \begin{pmatrix} -0.35001 & 1.15833 \\ 1 & 1 \end{pmatrix} $, for example, with https://www.symbolab.com/solver/matrix-eigenvectors-calculator. These are not even scalable to the values given in the book, so it’s not a matter of multiplying them by a constant…

What am I doing wrong? or misunderstanding?

Thanks in advance for any help!

P.S. There are published errata for this book, e.g. the value of $ S_2\ \text{sin}\ \beta_2 $ at the bottom of p. 281 should be $ -0.0375549 $ instead of $ -0.375549 $, but never any mention of any error in $ \bar e $

$\endgroup$
0

2 Answers 2

1
$\begingroup$

Maple says $$\overline{e}=\begin{pmatrix} -0.777991219286310 & 0.311162928367653 \\ -0.628275148890516 & -0.950356581504893 \end{pmatrix}$$ This is not so far from the book's answer you quote. Another possibility is (after scaling) $$\overline{e}=\begin{pmatrix} 1,2383 & -0.3274 \\ 1 & 1 \end{pmatrix}$$ which is close to what you found, if you transpose the matrix (representing vectors in columns and not in line, perhaps..).

$\endgroup$
17
  • $\begingroup$ OK, but how do I calculate it? I don’t want to use “ready-made” solutions like apps or websites. I want to be able to calculate it on a piece of paper. $\endgroup$ Apr 10, 2021 at 19:46
  • 1
    $\begingroup$ @PierrePaquette once you found an eigenvalue, say, $\lambda_1$, you solve $(a_{11}- \lambda_1) x =- a_{12} $ and your eigenvector is $\begin{pmatrix}x\\1\end{pmatrix}$. $\endgroup$
    – user145413
    Apr 10, 2021 at 19:57
  • 1
    $\begingroup$ @PierrePaquette formula 7.42 on page 281 is revealing. The vector with both entries $1/\sqrt 2$ is entirely correct. The vector above it is also a vector of length 1, and is correct. However, nonsense in both other vectors $\endgroup$
    – Will Jagy
    Apr 10, 2021 at 19:59
  • 1
    $\begingroup$ @PierrePaquette the $e_{11}$ and $e_{21}$ values are normalised to unit length, in that the sum of their squares is $1.$ $\endgroup$
    – Will Jagy
    Apr 10, 2021 at 21:22
  • 1
    $\begingroup$ @PierrePaquette if you can scan in the pages I can't see, from the beginning of chapter 7 to 282 and a few pages more, email me those, I can make a better guess as to what is needed. My email is visible at mathoverflow.net/users/3324/will-jagy?tab=profile $\endgroup$
    – Will Jagy
    Apr 11, 2021 at 15:57
0
$\begingroup$

From the book___________________________________

On the right hand page, formulas (7.42) show problems.

$$ $$

enter image description here

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .