0
$\begingroup$

Does there exist a non-diagonalizable 3x3 matrix that has precisely 1 real eigenvalue and a multiplicity of 1? When it comes to multiplicity I'm trying to find a matrix that would give me something like $(\lambda-1)^3$ as the eigenvalue. This factors down to $\lambda^3 - 3\lambda^2+3\lambda-1$ so you could say the multiplicity is 3 but you can also say that it only has 1 real root. So could I use this to find a non-diagonalizable 3x3 matrix with only 1 eigenvalue. So would such a matrix exist?

$\endgroup$
  • 1
    $\begingroup$ For one real eigenvalue with algebraic multiplicity 1, you'd have a characteristic polynomial of $(\lambda-a)(\lambda^2+b\lambda +c)$ where the quadratic $(\lambda^2+b\lambda +c)$ is irreducible (i.e. it has no real roots). Example: $(\lambda -1)(\lambda^2+\lambda+1)$. $\endgroup$ – user137731 Nov 29 '16 at 23:43
  • $\begingroup$ non-daigonazable? you are adding additional criteria. $\endgroup$ – Doug M Nov 29 '16 at 23:47
  • $\begingroup$ Would it make a difference? $\endgroup$ – david mah Nov 29 '16 at 23:48
  • $\begingroup$ I have given you an example of a daigonalizable matrix with one real eigenvalue. $\endgroup$ – Doug M Nov 29 '16 at 23:48
  • $\begingroup$ I thought that a matrix can't be diagonalizable if the number of eigenvalues does not equal the dimension of the matrix. $\endgroup$ – david mah Nov 29 '16 at 23:49
0
$\begingroup$

Assuming the matrix to be real, one real eigenvalue of multiplicity one leaves the only possibility for other two to be nonreal and complex conjugate. Thus all three eigenvalues are different, and the matrix must be diagonalizable.

If the matrix can be complex then it is possible to find a non-diagonalizable matrix with the only real eigenvalue of multiplicity one, for example $$ \begin{bmatrix} 1 & 0 & 0\\ 0 & i & 1\\ 0 & 0 & i \end{bmatrix} $$

$\endgroup$
  • $\begingroup$ I forgot to add an important detail. It also has to be non-diagonalizable. $\endgroup$ – david mah Nov 30 '16 at 1:50
  • $\begingroup$ @davidmah Sorry for being unclear. I was talking about non-diagonalizable matrices too (edited). $\endgroup$ – A.Γ. Nov 30 '16 at 2:07
  • $\begingroup$ So then it's not possible without complex numbers? $\endgroup$ – david mah Nov 30 '16 at 2:08
  • $\begingroup$ @davidmah For real matrices it is not possible (however, the diagonalization is going to be complex). $\endgroup$ – A.Γ. Nov 30 '16 at 2:10
3
$\begingroup$

Sure

\begin{bmatrix} 0&-1&0\\1&0&0\\0&0&1\end{bmatrix}

has one real eigenvalue of multiplicity 1.

$\endgroup$
0
$\begingroup$

For example

$$\begin{pmatrix}x&1&0\\0&x&1\\0&0&x\end{pmatrix}$$

has one unique eigenvalue $\;x\;$ of algebraic multiplicity $\;3\;$ and geometric multiplicity $\;1\;$ (if this is what you meant) , for any $\;x\in\Bbb R\;$ .

$\endgroup$

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.