3
$\begingroup$

I want to solve the matrix equation for $X$

$$aX^{3} + bX^{2} = I,$$ where $a,b \in \mathbb{R}$ and $X \in \mathbb{R}^{n\times n}$.

My thoughts:

If $a = 0$ or $b = 0$, the solution is easy.

If $a, b \neq 0$ I have tried to find matrices $U$ and $V$ such that $X^3 =UD_3V$ and $X^2 =UD_2V$, where $D_2$ and $D_3$ are diagonal matrices, but I did not have success until now. Maybe this is not possible.

Do you have any idea?

Thanks in advance!

$\endgroup$

2 Answers 2

2
$\begingroup$

Use Cayley-Hamilton theorem to find the eigenvalues that $X$ need to have. If $\lambda_1,\lambda_2,\lambda_3$ are the three eigenvalues then $diag(\lambda_1,\lambda_2,\lambda_3)$ is an immediate solution. But there can be many other solutions for different choices of eigenvectors corresponding to these eigenvalues.

$\endgroup$
5
  • $\begingroup$ Note that the real polynomial $ax^3 + bx^2 - 1$ will have at least one real root (since it is of odd degree for $a\neq 0$), and that $X$ might have only one of the roots as eigenvalues (in order to satisfy the matrix equation). If there are a pair of conjugate complex roots, we have to be a little tricky in order to get real matrix $X$ to have those eigenvalues. $\endgroup$
    – hardmath
    Jan 1, 2015 at 21:24
  • $\begingroup$ Thank you very much for the idea. :) $\endgroup$
    – Alex Silva
    Jan 1, 2015 at 21:39
  • $\begingroup$ Yes @hardmath, it may be difficult to get real matrix if the some of the eignvalues are complex. I stated the general idea. $\endgroup$ Jan 2, 2015 at 6:34
  • $\begingroup$ Hmm, in the OP, $X$ is an $n\times n$ matrix and $n$ does not necessarily equal to $3$. $\endgroup$
    – user1551
    Jan 17, 2015 at 9:04
  • $\begingroup$ Yes, @user1551, there can be many other solutions, but the ones Cayley-Hamilton theorem offers have to have $n=3$. $\endgroup$ Jan 17, 2015 at 10:33
1
$\begingroup$

Although this is a classic problem, I see no precise idea (except the Hardmath's comment). Let $f(x)=ax^3+bx^2-1$ where $a\not= 0$ and assume that $X$ is a real matrix. We decompose $f$ in product of real irreducible polynomials (with multiplicity). There are $3$ cases:

Case 1. $f(x)=a(x-u)(x-v)(x-w)$. Then $X\sim diag(u I_p,vI_q,wI_{n-p-q})$.

Case 2. $f(x)=a(x-u)^2(x-v)$. Then $X\sim diag(uI_p,U_1,\cdots,U_q,vI_{n-p-2q})$ where $U_i=\begin{pmatrix}u&1\\0&u\end{pmatrix}$

Case 3. $f(x)=a(x-u)(x^2+vx+w)$. Then $X\sim diag(uI_p,V_1,\cdots,V_{(n-p)/2})$ where $V_i=\begin{pmatrix}0&-w\\1&-v\end{pmatrix}$.

$\endgroup$

You must log in to answer this question.

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