Diagonalizable, trigonalizable কী?

Example 3.8 (Diagonalization put to work)

$A = I + J = \left(\begin{smallmatrix}2 & 1 & 1\\ 1 & 2 & 1\\ 1 & 1 & 2\end{smallmatrix}\right)$, with $J$ the all-ones matrix: from $\operatorname{Sp}(J) = \{3, 0\}$ (Example 2.19), $\operatorname{Sp}(A) = \{4, 1\}$, with eigenspaces $\R(1,1,1)$ and the plane $\{x + y + z = 0\}$: dimensions $1 + 2 = 3$, diagonalizable (Theorem 3.6 (2)). Powers without any change-of-basis matrix: with $\Pi = J/3$ the projector onto $\R(1,1,1)$,

(Check $k = 1$: $\frac{4-1}3 J + I = A$.) The closing insight: when the eigenspaces are visible, spectral projectors compute powers faster than $PDP^{-1}$ ever will — and the formula displays the dynamics: $A^k$ grows like $4^k$ along $(1,1,1)$ and stays put on the orthogonal plane.

Example 3.10 (Trigonalizing by hand)

$B = \begin{pmatrix}3 & -1\\ 1 & 1\end{pmatrix}$: $\chi_B = X^2 - 4X + 4 = (X - 2)^2$, and $\ker(B - 2I) = \ker\left(\begin{smallmatrix}1 & -1\\ 1 & -1\end{smallmatrix}\right)$ is the line spanned by $e_1' = (1, 1)$: one eigenvalue, a one-dimensional eigenspace — not diagonalizable, but trigonalizable (Theorem 3.9). Complete the basis with $e_2' = (1, 0)$ and compute:

so in the basis $(e_1', e_2')$ the matrix is $T = \left(\begin{smallmatrix}2 & 1\\ 0 & 2\end{smallmatrix}\right)$. The closing insight: the diagonal of $T$ was forced (both entries must be the double eigenvalue $2$); only the corner entry depended on the choice of $e_2'$, and rescaling $e_2'$ can make it any nonzero value — the resistant “$1$” is the shadow of the nilpotent part that Dunford will isolate.

Example 3.4 (Same $\chi$, different geometry)

The matrices

share the characteristic polynomial $(X - 2)^2$, the trace, the determinant, the spectrum — yet are not similar: the first has $E_2$ of dimension $2$ (geometric multiplicity $2$), the second of dimension $1$. The characteristic polynomial sees only algebraic multiplicities; the eigenspace dimensions are the finer invariant, and the minimal polynomial arbitrates ($X - 2$ versus $(X - 2)^2$). Moral for all diagonalizability discussions: $\chi$ shortlists the candidates, but kernels cast the votes.