Which of the following matrix equations have a real matrix solution \(X\)? (It is not necessary to exhibit solutions.)
\[ X^3=\begin{pmatrix}0&0&0\\1&0&0\\2&3&0\end{pmatrix}, \]
\[ 2X^5+X=\begin{pmatrix}3&5&0\\5&1&9\\0&9&0\end{pmatrix}, \]
\[ X^6+2X^4+10X=\begin{pmatrix}0&-1\\1&0\end{pmatrix}, \]
\[ X^4=\begin{pmatrix}3&4&0\\0&3&0\\0&0&-3\end{pmatrix}. \]
Solution
1. Equation (1) has no real matrix solution.
Let \[ N=\begin{pmatrix} 0&0&0\\ 1&0&0\\ 2&3&0 \end{pmatrix}. \] Suppose that a real \(3\times3\) matrix \(X\) satisfied \(X^3=N\). Then every eigenvalue \(\lambda\) of \(X\) satisfies \(\lambda^3=0\), because every eigenvalue of \(X^3=N\) is \(0\). Hence every eigenvalue of \(X\) is \(0\), so the characteristic polynomial of \(X\) is \(t^3\). By Cayley–Hamilton, \[ X^3=0, \] contrary to \(N\ne0\). Thus no such \(X\) exists.
2. Equation (2) has a real matrix solution.
Let \[ B=\begin{pmatrix} 3&5&0\\ 5&1&9\\ 0&9&0 \end{pmatrix} \] and let \(f(t)=2t^5+t\). The matrix \(B\) is real symmetric, so there are an orthogonal matrix \(Q\) and real numbers \(\lambda_1,\lambda_2,\lambda_3\) such that \[ B=Q\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)Q^T. \] Moreover \[ f'(t)=10t^4+1>0 \] for every real \(t\), and \(f(t)\to\pm\infty\) as \(t\to\pm\infty\). Thus \(f:\RR\to\RR\) is bijective. Choose real \(\mu_i\) with \[ f(\mu_i)=\lambda_i. \] Then \[ X=Q\operatorname{diag}(\mu_1,\mu_2,\mu_3)Q^T \] is real and satisfies \[ 2X^5+X=B. \]
3. Equation (3) has a real matrix solution.
Identify \(\RR^2\) with \(\CC\), so that the matrix \[ J=\begin{pmatrix}0&-1\\1&0\end{pmatrix} \] represents multiplication by \(i\). Consider the complex polynomial \[ p(z)=z^6+2z^4+10z. \] By the fundamental theorem of algebra, the nonconstant polynomial \[ p(z)-i \] has a complex root \(z_0=a+bi\). Let \[ X=\begin{pmatrix}a&-b\\b&a\end{pmatrix}, \] the real matrix representing multiplication by \(z_0\) on \(\CC\cong\RR^2\). The map \(a+bi\mapsto\begin{pmatrix}a&-b\\b&a\end{pmatrix}\) is a ring homomorphism \(\CC\to M_2(\RR)\), so \(p(X)\) represents multiplication by \(p(z_0)=i\). Therefore \[ X^6+2X^4+10X=J. \]
4. Equation (4) has no real matrix solution.
Let \[ C=\begin{pmatrix} 3&4&0\\ 0&3&0\\ 0&0&-3 \end{pmatrix}. \] Suppose \(X^4=C\). Since \(X\) commutes with every power of itself, \[ XC=X^5=CX. \] Hence \(X\) preserves every eigenspace of \(C\). In particular, the \((-3)\)-eigenspace \[ E_{-3}=\ker(C+3I)=\operatorname{span}(e_3) \] is \(X\)-invariant. Because this space is one-dimensional over \(\RR\), the restriction of \(X\) to it is multiplication by some real number \(r\). Restricting \(X^4=C\) to \(E_{-3}\) gives \[ r^4=-3, \] which is impossible over \(\RR\).
5. The equations with a real matrix solution are exactly\[ \boxed{\text{equations (2) and (3).}} \]
Steps 2 and 3 give solutions of equations (2) and (3), and steps 1 and 4 rule out equations (1) and (4).
6. Q.E.D.
Step 5 answers the question.