3
$\begingroup$

Construct a $2 \times 2$ matrix that is invertible but not diagonalizable.

From this I know that the eigenvalues cannot be equal or greater than, but would having no eigenvalues be a viable option? I don't see anything wrong based on the Diagonalization Theorem.

For the Invertible Matrix Theorem, if there's no eigenvalues then $0$ wouldn't be an eigenvalues because there are no eigenvalues. I'm not sure how having no eigenvalues affects the determinant of a matrix. If anyone could explain that part to me or show me where I went wrong it'd be greatly appreciated.

The Diagonalization Theorem: An $n \times n$ matrix $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors.

The Invertible Matrix Theorem: Let $A$ be an $n \times n$ matrix. Then $A$ is invertible if and only if:

1). The number 0 is not an eigenvalue of $A$

2). The determinant of $A$ is not zero

$\endgroup$
6
  • $\begingroup$ On an algebraically closed field (for example, the complex field), every matrix has an eigenvalue on that field. That is, if you allow complex numbers as matrix entries, then all square matrices will have (complex) eigenvalues. Of course, you can build a real square matrix which has no real eigenvalue, but $\mathbb{R}$ isn't algebraically closed. $\endgroup$
    – AspiringMathematician
    Apr 20, 2017 at 17:05
  • 1
    $\begingroup$ As an example of why @Guilherme's comment requires an algebraically closed field, see this older answer for a counterexample over $\mathbb{R}$: math.stackexchange.com/a/654467/137524 $\endgroup$
    – Semiclassical
    Apr 20, 2017 at 17:06
  • 1
    $\begingroup$ Ah I see, I hadn't considered that possibility. Does the fact of no real eigenvalues but the possibility of complex eigenvalues affect the condition of the matrix being invertible and not diagonalizable? $\endgroup$
    – stumped
    Apr 20, 2017 at 17:11
  • $\begingroup$ On any field which has a Jordan canonical form, then it must have at least one non-zero eigenvalue. Because otherwise all Jordan blocks must be 0 and the matrix be the zero matrix. $\endgroup$
    – mathreadler
    Apr 20, 2017 at 17:21
  • $\begingroup$ The question is, does your definition of "diagonalizable" include the possibility of complex entries? $\endgroup$
    – Robert Israel
    Apr 20, 2017 at 17:22

1 Answer 1

Reset to default
2
$\begingroup$

Hint: a $2\times2$ matrix having a double eigenvalue is diagonalizable if and only if it is diagonal.

Can you find an invertible and not diagonal matrix having a double eigenvalue?

A matrix with no real eigenvalues would be a valid choice, provided you are only working over the reals.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.