Let \[ A=\begin{pmatrix} 1&-1&0\\ -1&2&-1\\ 0&-1&1 \end{pmatrix}. \] Determine all real numbers \(a\) for which \[ \lim_{n\to\infty}a^nA^n \] exists and is nonzero as a matrix.
Solution
1. The matrix \(A\) has eigenvalues \(0,1,3\), with respectiveeigenvectors\[ u_0=(1,1,1)^{\mathsf T}, \qquad u_1=(1,0,-1)^{\mathsf T}, \qquad u_3=(1,-2,1)^{\mathsf T}. \]
Direct multiplication gives \[ Au_0=0, \qquad Au_1=u_1, \qquad Au_3=3u_3. \] The three vectors are nonzero and mutually orthogonal, so they form a basis of \(\RR^3\).
2. If \(a^nA^n\) converges as a matrix, then both scalar sequences\(a^n\) and \((3a)^n\) converge.
By step 1, \[ a^nA^n u_1=a^n u_1, \qquad a^nA^n u_3=(3a)^n u_3. \] Convergence of the matrices implies convergence of their values on each fixed vector, hence of the two scalar sequences.
3. If \(a^nA^n\) converges to a nonzero matrix, then\[ a=\frac13. \]
For a real number \(c\), the sequence \(c^n\) has a nonzero finite limit only when \(c=1\). If both \(a^n\) and \((3a)^n\) converged to zero, then step 1 would imply \(a^nA^n\to0\) on a basis, hence as a matrix. Therefore a nonzero matrix limit requires either \[ a=1 \qquad\text{or}\qquad 3a=1. \] The first possibility is impossible because then \((3a)^n=3^n\) does not converge. Hence \(a=1/3\).
4. For \(a=1/3\), the sequence \(a^nA^n\) converges to a nonzeromatrix.
On the basis from step 1, \[ \left(\frac A3\right)^n u_0=0, \qquad \left(\frac A3\right)^n u_1=3^{-n}u_1\longrightarrow0, \qquad \left(\frac A3\right)^n u_3=u_3. \] Thus \((A/3)^n\) converges to the projection onto \(\RR u_3\) along \(\RR u_0\oplus\RR u_1\). This projection is nonzero.
5. The unique answer is\[ \boxed{a=\frac13}. \]
Step 3 proves necessity, and step 4 proves sufficiency.
6. Q.E.D.
Step 5 gives the complete classification.