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What do we get when calculating the eigenvalue and eigenvector of a $4\times4$ matrix? What do those values actually represent?

ohiliouh
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2 Answers2

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Think of the matrix as a function from $\Bbb R^4$ to $\Bbb R^4$. If there is line through the origin in $\Bbb R^4$ that is taken to itself by the function, then it is an eigenvector and the amount that line is stretched or shrunk is the eigenvalue. It's easier to see this in $\Bbb R^2$. If the function is a rotation of $90^\circ$ then there are no (real) eigenvalues because no line is taken to itself. Alternatively suppose a linear function from $\Bbb R^2$ to $\Bbb R^2$ shrinks the $x$-axis by three and stretches the $y$-axis by four. Then there would be two eigenvalues, $1/3$ and 4. Does that help?

Gregory Grant
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  • If it shrinks by 3 then presumably the eigenvalue is $1/3$. – Ian May 24 '15 at 15:40
  • @Ian Thanks bro, yep! I fixed it. – Gregory Grant May 24 '15 at 15:42
  • Thank you for your answer. Sorry I am really having a hard time understanding it, and such abstract explanation makes it even harder for me. So the eigenvalue actually reflects in a way or another the size of the matrix. The part about the eigenvector is not so clear... – ohiliouh May 24 '15 at 15:46
  • The eigenvalue reflects the behavior of the linear transformation in one direction. If the matrix is $4\times4$ then there could be at most four eigenvalues. I highly recommend learning about eigenvalues in $\Bbb R^2$ first and then generalizing to four dimensions. Did you try to read through the wiki page on this subject? https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors – Gregory Grant May 24 '15 at 15:49
  • Yes I did of course. But as in my case(space engineering) there is no transformation in the short mathematical model I am using this does get me confused. page 4-15: http://www.dept.aoe.vt.edu/~cdhall/courses/aoe4140/attde.pdf

    As you can see there is no transformation of the 4x4 matrix K. Is there another possible interpretation of eigenvalues and eigenvectors?

     – ohiliouh May 24 '15 at 16:49
  • A square matrix always corresponds to a transformation. – Gregory Grant May 24 '15 at 16:57
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If this matrix is diagonalizable, that is if it has 4 distinct but not necessarily unique eigenvalues, then the eigenvalues are the entries in the diagonal matrix and the eigenvectors form the two adjacent matrices so we get $PQP^{-1}$. Even more basically the eigenvectors corresponding to eigenvalues are the vectors that when multiplied by the given matrix return the vector multiplied by the eigenvalues. See the following link and check the forum more thoroughly before posting. Eigenvalues and Eigenvectors

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RhythmInk
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  • I fixed a LaTeX error. I think this is fairly clear, but the "distinct but not necessarily unique" phrasing is confusing. There are a bunch of ways of phrasing this; "a basis consisting of eigenvectors of the matrix exists" is one. Counting geometric multiplicities is another. – Ian May 24 '15 at 15:39
  • I am an engineering student not a mathematician or so, in my opinion your explanation is quite abstract, which makes it hard for me to understand it properly. You told me more or less how to find those values. I was wondering what those values actually told us or expressed about that matrix. ( very simple wrong example: Is it like an average of this matrix? ) – ohiliouh May 24 '15 at 15:43
  • @JaqcuesMartin One nice explanation for engineers and physicists arises through the eigenvalues/eigenvectors of the inertia tensor. Here the eigenvectors correspond to the principal axes of a body, and the eigenvalues are the moments of inertia about each of the principal axes. One gap in this example is that the inertia tensor is always symmetric, so this example can't be used to understand, for instance, the interpretation of complex eigenvalues. – Ian May 24 '15 at 15:48
  • To what aspect of those principal axis does the eigenvector correspond? – ohiliouh May 24 '15 at 16:06