11
$\begingroup$

I am confused about the relationship between the invertibility of a matrix and its eigenvalues. What do the eigenvalues of a matrix tell you about whether a matrix is invertible or not? Also, what about the "singular values" of a matrix? If there are good online resources which answer these question I would be grateful for any pointers.

$\endgroup$
  • 1
    $\begingroup$ A matrix is invertible iff its determinant is not zero. The determinant of a triangular matrix equals the product of its diagonal elements. Similar matrices have the same determinant and every matrix is similar to an upper triangular matrix (Jordan form). The diagonal entries of the Jordan form are the eigenvalues of the matrix, with the number of times each one occurs being given by its algebraic multiplicity. $\endgroup$ – Poppy Jan 26 '14 at 13:00
  • 2
    $\begingroup$ So, if $0$ is an eigenvalue, then that matrix would be similar to a matrix whose determinant is 0. Therefore, that matrix would not be invertible, so neither would be ours. $\endgroup$ – Poppy Jan 26 '14 at 13:03
10
$\begingroup$

A square matrix is invertible if and only if it does not have a zero eigenvalue.

The same is true of singular values: a square matrix with a zero singular value is not invertible, and conversely.


The case of a square $n\times n$ matrix is the only one for which it makes sense to ask about invertibility. Its determinant is the product of all the $n$ algebraic eigenvalues (counted as to multiplicity). Since the determinant is nonzero if and only if the matrix is invertible, this is one way to recognize the equivalence of being invertible with not having a zero eigenvalue.

When $A$ is a square matrix, the eigenvalues of $A^TA$ are precisely the squares of the singular values of $A$. So if $A$ has a zero singular value, $A^TA$ has a zero eigenvalue, and conversely. Clearly $A^TA$ is invertible if $A$ is (the determinant of $A^T$ equals the determinant of $A$), and conversely if $A^TA$ is invertible, its determinant $|A^TA| = |A|^2$ is nonzero, and thus $|A|$ is nonzero, and $A$ is invertible.

Something can be said about the non-square case using singular values, but one needs a little care in discerning what it means for $A$ not to have a zero singular value. $A$ and $A^T$ will have equal rank, and at most one of $A^T A$ and $AA^T$ can be invertible in the nonsquare case. In any case one can discern that a non-square matrix is not "invertible" simply from its shape, and singular value computation is unnecessary.

$\endgroup$
  • $\begingroup$ why do you say that the determinant is the product of the eigenvalues? consider $\begin{pmatrix} 0 & -1 \\ 0 & 1 \end{pmatrix}$ over $\mathbb{R}$ which doens't have any eigenvalues but determinant 1. I guess we have to require the underlying field to be algebraically closed. $\endgroup$ – Viktor Glombik Jun 1 at 17:36
  • 1
    $\begingroup$ i'm sorry, my matrix was supposed to be $$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$!! $\endgroup$ – Viktor Glombik Jun 1 at 18:51
  • $\begingroup$ Right, another way to think about it is the determinant is $\pm$ the constant term of the characteristic polynomial, so in any splitting field for that polynomial, the constant term is $\pm$ the product of those roots (eigenvalues), depending on whether there is an even or odd number of them. $\endgroup$ – hardmath Jun 1 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy