Review : Eigenvalues & Eigenvectors — Question 8

Question 8 – Solution

Strategy. Follow the eigenvectors as well as the roots. Distinct roots can merge into one direction at the transition.

Step 1: Classify the roots. The polynomial is $\lambda^2-2\lambda+\tau$. For $\tau<1$, set $d=\sqrt{1-\tau}>0$; the roots are $1\pm d$. At $\tau=1$ the root is $1$ twice. For $\tau>1$, set $b=\sqrt{\tau-1}>0$; the roots are $1\pm ib$. Their sum is $2=\operatorname{tr}A_\tau$ and product is $\tau=\det A_\tau$ in all cases.

Step 2: Find the eigenspaces and field restrictions. The first row of $A_\tau x=\lambda x$ forces $x_2=\lambda x_1$. The second row is then exactly the characteristic equation. Thus every root has complex eigenspace spanned by $(1,\lambda)^T$; for real roots, the real eigenspace is the real span of the same vector. At $\tau=1$ this is only one-dimensional, despite algebraic multiplicity $2$. Consequently $$\boxed{\begin{array}{c|cc} &\mathbb R\text{-diagonalizable}&\mathbb C\text{-diagonalizable}\\\hline \tau<1&\text{yes}&\text{yes}\\ \tau=1&\text{no}&\text{no}\\ \tau>1&\text{no}&\text{yes} \end{array}}$$ The real failure above $1$ comes from nonreal eigenvalues, not a shortage of complex eigenvectors.

Step 3: Construct a real representation for the complex pair. For $\lambda=1+ib$, write $(1,1+ib)^T=p+iq$ with $p=(1,1)^T$, $q=(0,b)^T$. Equating parts gives $A_\tau p=p-bq$ and $A_\tau q=bp+q$. Since $\det(p\ q)=b\ne 0$, this is a real basis, with representation $$\boxed{(p\ q)^{-1}A_\tau(p\ q)=\begin{pmatrix}1&b\\-b&1\end{pmatrix}.}$$ It is a real block, not a real diagonalization.

Step 4: Inspect the collapsing eigenvector basis. Below $1$, use $P=\begin{pmatrix}1&1\\1+d&1-d\end{pmatrix}$. Its determinant is $-2d$, and $$P^{-1}=\frac 1{2d}\begin{pmatrix}d-1&1\\1+d&-1\end{pmatrix}.$$ In particular, the second column is unbounded as $d\to 0^+$. Both columns tend to $(1,1)^T$ as $d\to 0^+$. The limiting basis is singular and cannot diagonalize the limiting matrix. Diagonalizability need not persist when distinct eigendirections coalesce.