6
$\begingroup$

We can generalize matrix inverses from non-singular square matrices to rectangular matrices in general, for example, the well-known Moore–Penrose pseudoinverse. I am wondering how this can be done for eigenvalues and eigenvectors.

Though $\det(|A-\lambda I|)=0$ cannot be used any more when $A$ is not square, there is nothing that prevents one to consider $Av=\lambda v$ for non-zero vector $v$ except the possibility of having an inconsistent linear system.

Please give your comments and provide some references if there are some.

Many thanks.

Edit

I know SVD. But it does not seem to be the one I wanted. For SVD of real matrix $A$, $A=UDV^T$ where $U, V$ are orthogonal matrices and $D$ is diagonal (with possibly zeros in the diagonal). We only have $AV_{*k}=\sigma_{k}U_{*k}$, $V_{*k}$ is the $k^\text{th}$ column of $V$. Since $V_{*k}$ and $U_{*k}$ are in general different, it does not resemble $Av=\lambda v$ for non-zero vector $v$ in the definition of eigenvectors. Also, even if we can have $A^TAV_{*k}=\lambda V_{*k}$, but this is for the (square) matrix $A^TA$, rather than $A$ itself.

$\endgroup$
2
  • 5
    $\begingroup$ But isn't the problem that when you write $Av$, this is a vector of different dimension than $v$ itself? Eigenvectors are kind like fixed points. If you think of a (non-singular) linear map as really acting on projective space, the eigenvectors are just the fixed points of the map. But fixed points are all about mapping a space to itself. If you have an $m \times n$ matrix... it is a linear map from one space to a completely different space. With that said, I'm sure there is something one can say about some kind of analogous ideas, but I do not know about them. $\endgroup$
    – Bill Karr
    Apr 19, 2011 at 5:20
  • 2
    $\begingroup$ As a concrete example, if you multiply a 4-by-3 matrix and a 3-vector, you get a 4-vector. I thus don't see how you can make sense of $\mathbf A\mathbf v=\lambda \mathbf v$ if $\mathbf A$ is non-square... $\endgroup$ Apr 19, 2011 at 8:42

2 Answers 2

6
$\begingroup$

Check out singular values.

$\endgroup$
3
  • $\begingroup$ Ah, I forgot about SVD. $\endgroup$
    – Bill Karr
    Apr 19, 2011 at 5:29
  • 1
    $\begingroup$ As a matter of fact, the pseudoinverse can be cleanly expressed in terms of the SVD factors... $\endgroup$ Apr 19, 2011 at 5:41
  • $\begingroup$ @Bill, please see my updates. $\endgroup$
    – Qiang Li
    Apr 19, 2011 at 6:01
2
$\begingroup$

Rectangular matrices do not have eigenvectors/eigvalues. Instead they have singular values & singular vectors. (Square matrices have both - eigenvalues and singular values).

However, there is a simple relationship between eigenvalues and the singular values.

(Singular-values of matrix $A$) == sqroot (eigValues of $A^TA$). $A^TA$ has have eigenvalues as it will be a square matrix irrespective of the dimensions of A.

This can be seen below:

$ A = UDV^T ; A^T = VDU^T $

(Here, D is a diagonal matrix with singular values along its diagonal)

$ A^T*A = VDU^T * UDV^T $

As U and V are orthogonal matrices, the following relation holds: $ U^TU = I and V^TV = I $

=> $ A^TA = VD^2V^T $

=> $ A^TA * (V^T)^{-1} = VD^2V^T * (V^T)^{-1} $

=> $ A^TA(V) = VD^2 $

Here, the diagonal values of $D^2$ are the eigenvalues of $A^TA$, which is the square of the singular values of matrix $A$.

Additional Source: http://www-users.math.umn.edu/~olver/num_/lnv.pdf

$\endgroup$
1
  • $\begingroup$ Use \implies for $\implies$ $\endgroup$ Jun 1, 2018 at 1:57

You must log in to answer this question.

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