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!
On
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 $V_i=\begin{pmatrix}0&-w\\1&-v\end{pmatrix}$.
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.