Eigenvalues and Canonical Forms

$p(T)=T^2(T-c)$ with c real (degree-3 real polynomial). So all eigenvalues are real. The geometric multiplicity of 0 may be 1 (Jordan block), so (1), (4) fail; A need not be nilpotent (c ≠ 0).

(a) fails: for the nilpotent Jordan block on ${\mathbb{R}}^3,f=X^3,g=X^2$, and $g(T)=T^2\ne 0.(b)$ If 0 has geometric multiplicity ≥ 2 in dimension 3, the 0-Jordan blocks all have size 1, so the minimal polynomial is X(X − c) (or X if T = 0), which divides g(X) = X(X − c); hence g(T) = 0.

Spectral theorem: A = QDQᵀ with real D. Aᵏ = I forces each real eigenvalue to satisfy ${\lambda }^k=1$, so $\lambda =\pm 1$ and $A^2=I;A^k=0$ forces all $\lambda =0$ so A = 0.

A = vvᵀ with v = (1, 2, …, n) has rank 1; its only non-zero eigenvalue is $v^{T}v=\sum i^2=n(n+1)(2n+1)/6$.

Minimal polynomial divides $x^3-x=x(x-1)(x+1)$, distinct linear factors ⇒ diagonalisable over $\mathbb{R}$.

$A^2=u(v^{T}u)v^T=0.A\ne 0$ and nilpotent ⇒ not diagonalisable.

Minimal polynomial has a repeated factor ⇒ not diagonalisable. Eigenvalue 1: single-size blocks, two of them ⇒ rank(A − I) = 4 − 2 = 2. Eigenvalue 2: one block of size 2 ⇒ geometric multiplicity 1 ⇒ rank(A − 2I) = 3.

m | $x^3-1=(x-1)(x^2+x+1)$: distinct roots over $\mathbb{C}\Rightarrow$ diagonalisable over $\mathbb{C}.A\ne I$ forces the complex pair $\omega ,\bar{\omega }$ to appear, so not diagonalisable over $\mathbb{R}$.

(c) fails: nilpotent 4×4 with Jordan blocks {2,2} vs {2,1,1} share $\chi =x^4,m=x^2.(d)$ BA = A⁻${}^1($AB)A.

A satisfies $x^3-x=x(x-1)(x+1)$, so the minimal polynomial divides a product of distinct linear factors over $\mathbb{R}$ — hence A is diagonalisable over $\mathbb{R}$, giving (1). Every eigenvalue lies in {0, 1, −1}, so the trace is a sum of three such numbers and is an integer, giving (3). (2) fails for A = 0. (4) fails: conjugating a diagonal matrix diag(1,0,0) by any invertible non-orthogonal matrix still satisfies $A^3=A$ but need not be symmetric.

The eigenvalues are the roots 0, 1, −1: distinct, so A is diagonalisable; the trace is their sum, 0, and the determinant their product, 0 — in particular A is singular.

Rearrange: $A^2-A=I$, so A(A − I) = I. Hence A is invertible with inverse A − I — Cayley–Hamilton-style manipulation without ever computing entries.

(a) minimal poly divides x(x−1). (b) diagonalisable nilpotent ⇒ zero. (c) standard. (d) false: zero matrix.

A is a rotation by $\theta \notin$ {$0,\pi$}; eigenvalues $e^{\pm i\theta }$ are non-real and distinct.

(b): eigenvalues ±i, not even real — not diagonalisable over $\mathbb{R}(e.g$. the 90° rotation).

(b) fails: A = B = a non-zero nilpotent matrix commutes with itself.

(1) Distinct eigenvalues give n independent eigenvectors. (2) is the exact algebraic criterion for diagonalisability. (3) The spectral theorem diagonalises every normal matrix unitarily. (4) is no condition at all over $\mathbb{C}$ — every characteristic polynomial splits there, yet the Jordan block [[0,1],[0,0]] is not diagonalisable.

dim W = number of Jordan blocks. Equal numbers of blocks do not fix the largest block (minimal polynomial): blocks {2,2} vs {3,1} in dimension 4. (1) similar ⇒ equal kernel dimension. (3) W = V ⇒ T = 0. (4) nilpotent ⇒ characteristic polynomial $x^n$ always.

(1) Nilpotency forces every eigenvalue to satisfy ${\lambda }^3=0.(2)$ The largest Jordan block has size equal to the nilpotency index, here 3; in dimension 4 the only possible structure is 3 + 1. (3) fails: $a{J}_3$ block contributes rank 2 and $a{J}_1$ block rank 0, so rank N = 2, not 3. (4) A nonzero nilpotent matrix is never diagonalisable — its diagonal form would have to be the zero matrix.

The invariant factors come from the Smith normal form of xI − A over $\mathbb{Q}[x]$, so they are field-independent and no splitting is needed.