Which of the following matrices are similar over \(\mathbb R\)? \[ (a)\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}, \qquad (b)\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}, \qquad (c)\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix}, \] \[ (d)\begin{pmatrix}1&0&0\\1&1&0\\0&1&1\end{pmatrix}, \qquad (e)\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}, \qquad (f)\begin{pmatrix}0&1&1\\0&1&0\\1&0&0\end{pmatrix}. \]
Solution
Denote the six matrices by \(A_a,A_b,A_c,A_d,A_e,A_f\).
1. The characteristic polynomials are\[ \chi_{A_a} = \chi_{A_c} = \chi_{A_d} = \chi_{A_e} = (t-1)^3 \]and\[ \chi_{A_b} = \chi_{A_f} = (t-1)^2(t+1). \]
The matrices \(A_a,A_c,A_d,A_e\) are triangular with all diagonal entries equal to \(1\), giving the first identity.
For \(A_b\), \[ A_b \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}z\\y\\x\end{pmatrix}, \] so it fixes the two-dimensional subspace \(x=z\) and acts by \(-1\) on the line spanned by \((1,0,-1)^T\). Hence its characteristic polynomial is \((t-1)^2(t+1)\).
Finally, \[ \det(tI-A_f) = \det \begin{pmatrix} t&-1&-1\\ 0&t-1&0\\ -1&0&t \end{pmatrix} = (t-1)(t^2-1) = (t-1)^2(t+1). \]
2. The minimal polynomials of \(A_a,A_c,A_d,A_e\) are respectively\[ t-1, \qquad (t-1)^2, \qquad (t-1)^3, \qquad (t-1)^2. \]
For \(A_a=I\), the minimal polynomial is \(t-1\).
Set \(N_c=A_c-I\) and \(N_e=A_e-I\). Each is nonzero and has exactly one nonzero entry, so \[ N_c^2=0=N_e^2. \] Thus both minimal polynomials are \((t-1)^2\).
For \[ N_d=A_d-I = \begin{pmatrix} 0&0&0\\ 1&0&0\\ 0&1&0 \end{pmatrix}, \] one has \[ N_d^2 = \begin{pmatrix} 0&0&0\\ 0&0&0\\ 1&0&0 \end{pmatrix} \neq0, \qquad N_d^3=0. \] Hence the minimal polynomial of \(A_d\) is \((t-1)^3\).
3. Among \(A_a,A_c,A_d,A_e\), precisely \(A_c\) and \(A_e\) aresimilar.
Similar matrices have the same minimal polynomial, so step 2 excludes every pair except \(A_c,A_e\).
For both \(A_c\) and \(A_e\), the nilpotent matrix \(A-I\) is nonzero, has rank \(1\), and has square \(0\). Therefore each has Jordan form \[ \begin{pmatrix} 1&1&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}. \] Thus \(A_c\) and \(A_e\) are similar over \(\RR\).
4. The matrix \(A_b\) is diagonalizable over \(\RR\).
Step 1 already exhibited a two-dimensional eigenspace for eigenvalue \(1\) and a one-dimensional eigenspace for eigenvalue \(-1\). These eigenspaces have total dimension \(3\), so \(A_b\) is diagonalizable.
5. The matrix \(A_f\) is not diagonalizable.
The equation \[ (A_f-I) \begin{pmatrix}x\\y\\z\end{pmatrix} = 0 \] is \[ \begin{pmatrix} -x+y+z\\ 0\\ x-z \end{pmatrix} = 0. \] Thus \(z=x\) and \(y=0\), so \[ \ker(A_f-I) = \operatorname{span} \left\{ \begin{pmatrix}1\\0\\1\end{pmatrix} \right\} \] has dimension \(1\). But step 1 shows that the algebraic multiplicity of eigenvalue \(1\) is \(2\). Therefore \(A_f\) is not diagonalizable.
6. The matrices \(A_b\) and \(A_f\) are not similar.
Diagonalizability is invariant under similarity. Step 4 shows that \(A_b\) is diagonalizable, while step 5 shows that \(A_f\) is not.
7. The complete similarity classification is\[ \boxed{(c)\sim(e),\quad\text{and no other distinct pair is similar}.} \]
Step 1 separates the four matrices \(A_a,A_c,A_d,A_e\) from \(A_b,A_f\) by characteristic polynomial. Step 3 classifies the first group, and step 6 separates the two matrices in the second group.
8. Q.E.D.
Step 7 answers the classification question completely.