Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If I have a equation in the form

$${\lambda ^2}{I_N} + \lambda {M_1} + {M_2} = {0_N}$$

where ${I_N}$ is the identity matrix of order $N$, $M_1$ and $M_2$ are matrices of ($N\times N$) order and $\lambda \in \mathbb C$ belongs to the complex numbers .

What are the mathematical tools or the mathematical framework to solve this kind of equations?

share|cite|improve this question
Looking at the equation for one coefficient gives you two possible values for $\lambda$. You can test both of them to see if they solve the equation. – Quimey Jan 8 '13 at 14:18
up vote 2 down vote accepted

For the equation to have a solution, your matrices $M_1$ and $M_2$ must necessarily commute. This would only be one requirement. To see this, consider the case for diagonalizable $M_2$ so that $ PM_2P^{-1} = D$ for some invertible $P$ and diagonal $D$.

\begin{align} P\left(\lambda^2 I_N + \lambda M_1 + M_2 = 0_N\right)P^{-1} \\ \lambda^2 I_N + \lambda PM_1P^{-1} + D = 0_N \end{align}

Here we can see that $M_1$ must not only have the same spectrum as $M_2$ (since otherwise the non-zero elements in the off-diagonal would not be canceled in the sum), but it must have appropriate eigenvalues such that the single $\lambda$ simultaneously solves for each diagonal term.

tl:dr: solve for $\lambda_1$ and $\lambda_2$ at any desired coordinate. Check if either one works globally. If not, then there is no solution.

share|cite|improve this answer
I've comproved that this eq does not have solution, this because M1 and M2 do not commute and they do not share the same spectra. Thanks a lot. – Daniel.B. Jan 10 '13 at 10:17
+1 and thanks for noting me that gap there. – Babak S. Jan 16 '13 at 15:56
@Babak Thanks, and let me know when to delete my other comments on your question, I will look up and refresh my memory on the array syntax in the meantime – adam W Jan 16 '13 at 16:04
You noted me a very important thing and I want you to accept my thanks. Let them be there. Your comments contain some useful points I didn't know them. ;-) – Babak S. Jan 16 '13 at 16:08

If the equality is to be true, we need to check that for $ \{M_1\}_{ij} $ and $ \{M_2\}_{ij} $ we have $$ \lambda^2 \delta_{ij} + \lambda \{M_1\}_{ij} + \{M_2\}_{ij} = 0 $$

where $\delta_{ij}$ is the Kronecker delta and $ 1 \leq i,j \leq n $.

There's probably a better way of investigating it - I'm having a think about that now. Certainly, from the $i \neq j$ case, we are left with a single variabled equation, so at most we have one $\lambda $ as a solution.

share|cite|improve this answer
The determinant does not distribute over the sum. – Quimey Jan 8 '13 at 15:11
Indeed - I guess I thought that the determinant was a ring homomorphism. – Andrew D Jan 8 '13 at 15:15

Indeed, by adam W's answer, $M_1$ is without loss of generality in Jordan canonical form, so that $M_1 = \bigoplus_{k=1}^M (\mu_k I_{n_k} + N_{n_k})$ for $\mu_k$ the eigenvalues (counted with the relevant notion of multiplicity) and $N_{n_k}$ the appropriate nilpotent matrices. Then $$\lambda^2 I_N + \lambda M_1 = \bigoplus_{k=1}^M ((\lambda^2 + \lambda\mu_k) I_{n_k} + \lambda N_{n_k}),$$ so that $M_2$ must necessarily have the analogous block diagonal form $$M_2 = \bigoplus_{k=1}^M (\alpha_k I_{n_k} + \beta N_{n_k})$$ for some constants $\alpha_k$ and $\beta$. Hence, when the dust settles, you're left with the system of quadratic equations $$\lambda^2 + \mu_k \lambda - \alpha_k =0$$ together with the additional equation $\lambda = \beta$ whenever $M_1$ (and hence also $M_2$) is not diagonal. I think this should be correct?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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