Let $A$ be an $n \times n$ matrix over $\mathbb{C}$. Use triangular form to show that the trace of $A$ equals the sum of the eigenvalues, where a multiplicity $m$ eigenvalue is counted $m$ times. Show also that an analogous statement holds for the determinant and the product of the eigenvalues.
Solution.
By Corollary 4.6. and ℂ is an algebraically closed field, $A$ is conjugate to an upper triangular matrix $B$. Let the eigenvalues of $A$ be $x_1,…,x_n$. Then $x_1,…,x_n$ are the diagonal entries of $B$. So $\operatorname{tr}A=\operatorname{tr}B=x_1+⋯+x_n$ and $\det A=\det B=x_1⋯x_n$.
Calculate the minimum and characteristic polynomials of the following matrices $$ A=\pmatrix{ 1 & 1 & 0 \\ -9 & -4 & 1 \\ -3 & 3 & 2 },\pmatrix{ -2 & -3 & -3 \\ -1 & 0 & -1 \\ 0 & 1 & -1 },\pmatrix{ 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 } $$ Solution. [Proposition 5.3 in notes]
- $χ_A(x)=-(-2 + x) (1 + x) (2 + x)$. All roots are simple roots, so $m_A(x)=(-2 + x) (1 + x) (2 + x)$.
- $χ_A(x)=-(-1 + x) (2 + x)^2$. Possible values of $m_A(x)$ are $(-1 + x) (2 + x)$ or $(-1 + x) (2 + x)^2$.
  $$(-I+A)(2+A)=\pmatrix{3&0&9\\1&0&3\\-1&0&-3}$$ So $m_A(x)=(-1 + x) (2 + x)^2$.
- $A$ is already in Jordan normal form, it has 1 block of size 2 and 1 block of size 1. So $χ_A(x)=(2-x)^3$ and $m_A(x)=(x-2)^2$.
(a) Find two $2 \times 2$ matrices over $\mathbb{R}$ which have the same characteristic polynomial but which are not conjugate.
(b) Find two $3 \times 3$ matrices over $\mathbb{R}$ which have the same minimal polynomial but which are not conjugate.
(c) Find two $4 \times 4$ matrices over $\mathbb{R}$ which have the same minimal polynomial and the same characteristic polynomial, but which are not conjugate.
(d) What is the smallest value of $n$ for which there are two non-conjugate $n \times n$ matrices which have the same minimum and characteristic polynomials and eigenspaces of equal dimension?
Solution.
- Consider the matrices $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $\begin{pmatrix}0&0\\0&0\end{pmatrix}$
  Each has charateristic polynomial $x^2$, but they are not conjugate. (The only matrix conjugate to the zero matrix is itself.)
- $A=\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$ and $B=\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$ have the same minimal polynomial $x^2-1$, but $\det A=-1≠1=\det B$, so they are not conjugate.
  Also see here
- $A=\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$ and $B=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$ have the same characteristic polynomial $x^4$ and the same minimal polynomial $x^2$.
  $\operatorname{rank}A=2≠1=\operatorname{rank}B$, so they are not conjugate.
- See this post
  Let $M$ be a matrix with coefficients in ℂ. Then $M$ is conjugate to a block-diagonal matrix where each block is a Jordan block. These blocks, up to permutation, uniquely determine the conjugacy classes. Moreover, let $λ$ be an eigenvalue of $M$. Then
  - the number of Jordan blocks corresponding to $λ$ is the dimension of the eigenspace;
    The columns of change-of-basis matrix corresponding to first column of each Jordan block are eigenvectors.
  - the size of the largest block corresponding to $λ$ is the multiplicity of $λ$ in the minimal polynomial;
  - the sum of the sizes of the blocks corresponding to $λ$ is the multiplicity of $λ$ in the characteristic polynomial.
  The problem is equivalent to find the smallest integer $N$ such that there are an integer $k$, and two different $k$-tuples of integers $(n_1,\cdots,n_k)$ and $(m_1,\cdots,m_k)$ such that:
  - $n_1\geq \cdots \geq n_k \geq 1$ and $m_1\geq \cdots \geq m_k \geq 1$;
  - $\sum^k_{i=1} n_i = \sum^k_{i=1} m_i = N$;
  - $n_1 = m_1$;
  If $k=2$, since $n_1+n_2=m_1+m_2$ and $n_1=m_1$, we have $n_2=m_2$, so they are same tuple. So $k≥3$.
  Since $n_1+n_2+n_3=m_1+m_2+m_3$ and $n_1=m_1$, we have $n_2+n_3=m_2+m_3$, the smallest pair is 3+1=2+2, so $n_1≥3$ and smallest tuples are (3,3,1) and (3,2,2), so $n≥7$.
  An example:
  $\left(\begin{array}{ccc:ccc:c} 0&1&0&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&0&0&0&0&0\\\hdashline 0&0&0&0&1&0&0\\ 0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0\\\hdashline 0&0&0&0&0&0&0 \end{array}\right)$ and $\left(\begin{array}{ccc:cc:cc} 0&1&0&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&0&0&0&0&0\\\hdashline 0&0&0&0&1&0&0\\ 0&0&0&0&0&0&0\\\hdashline 0&0&0&0&0&0&1\\ 0&0&0&0&0&0&0 \end{array}\right).$
Let $T: V → V$ be a nilpotent linear map. Show that we have strict inclusions $$ \{0\}<\ker T<\ker\left(T^2\right)<… $$
MA20216: Algebra 2A 4.3.1 Powers of operators and Fitting’s Lemma
Fitting lemma
Proof.
Since $T$ is nilpotent, the minimal polynomial of $T$ is $T^m$ for some $m∈ℕ$, then $\ker(T^k)=V$ iff $k≥m$.
If $v\in \ker T^k$ then $T^k(v)=0$ so that $T^{k+1}(v)=T(T^k(v))=T(0)=0$. Thus $v\in \ker T^{k+1}$. Therefore $\ker T^k≤\ker T^{k+1}$.
Now suppose that $\ker T^k=\ker T^{k+1}$ and induct to prove that $\ker T^k=\ker T^{k+n}$, for $n∈ℕ$.
We already have the $n=1$ case by assumption, so suppose $\ker T^k=\ker T^{k+n}$, for some $n$ and let $v\in\ker T^{k+n+1}$.
Then $0=T^{k+n+1}(v)=T^{k+1}(T^n(v))$, so that $T^n(v)\in \ker T^{k+1}=\ker T^k$. Thus $T^{n+k}(v)=0$ and $v\in \ker T^{n+k}=\ker T^k$ by the induction hypothesis.
Induction now tells us that $\ker T^k=\ker T^{k+n},∀n∈ℕ$. Then $\ker T^k=V$. So $k≥m$ by minimality of $m$.
Alternate Proof: In notes we proved \[\dim(\ker(T^{i+1}))-\dim(\ker(T^i))≤\dim(\ker(T^i))-\dim(\ker(T^{i-1}))\]
Decide whether or not the matrix $$B=\pmatrix{1 & 6 \\3 & 5}$$ can be diagonalised over the field
(i) ℝ
(ii) ℂ
(iii) ℚ
(iv) a field of characteristic 2 (i.e. $1+1=0$)
(v) the field 𝔽₇ of integers modulo 7.
Solution.
$χ_B(x)=-13 - 6 x + x^2$. By Proposition 5.3, $m_B(x)=-13-6x+x^2=(3+\sqrt{22}-x)(3-\sqrt{22}-x)$.
(i)(ii) $B$ has 2 distinct eigenvalues, so is diagonalisable.
(iii) $-13-6x+x^2$ is irreducible, so $m_B(x)$ doesn't split, so $B$ is not even triangularisable.
(iv) $B=\pmatrix{1&0\\1&1}$, so $B-I=\pmatrix{0&0\\1&0}$. Eigenspace $⟨(0,1)⟩$ is of dimension 1, so $B$ is not diagonalisable.
(v) $χ_B(x)=15+8x+x^2⇒B$ has 2 distinct eigenvalues $-3,-5$, so is diagonalisable.
WolframAlpha
Wolfram Language code: Factor[-13 - 6 x + x^2, Modulus -> 7]
No finite field is algebraically closed
Is the matrix $$ B=\left(\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{array}\right) $$ diagonalisable over $ℂ,ℝ$ or $𝔽_3$ ?
Solution.
$χ_B(x)=-1-x^3$. By Proposition 5.3, $m_B(x)=1+x^3=(1+x)\big((-1)^{1/3} - x\big)\big((-1)^{2/3} - x\big)$.
Over ℂ, $B$ has 3 distinct eigenvalues, so it is diagonalisable.
Over ℝ, $1-x+x^2$ is irreducible, so $m_B(x)$ doesn't split, so $B$ is not even triangularisable.
Over $𝔽_3$, $m_B(x)=(1+x)^3$, so $B$ is triangularisable but not diagonalisable. $$P=\begin{pmatrix}1&0&0\\0&1&0\\1&0&1\end{pmatrix}\qquad P^{-1}BP=\begin{pmatrix}-1&0&-1\\0&-1&0\\0&0&-1\end{pmatrix}$$

Some matrices with real entries which are not diagonalizable over ℝ are diagonalizable over the complex numbers ℂ. For instance $\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)$ has $λ^2+1$ as characteristic polynomial. This polynomial doesn’t factor over the reals, but over ℂ it does. Its roots are $λ=±i$. Interpreting the matrix as a linear transformation $ℂ^2→ℂ^2$, it has eigenvalues $i$ and $-i$ and linearly independent eigenvectors $(1,-i),(-i,1)$. So we can diagonalize $A$:$$A=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right)=\left(\begin{array}{cc}1 & -i \\ -i & 1\end{array}\right)\left(\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right)\left(\begin{array}{cc}\frac12 & \frac i2 \\ \frac i2 & \frac12\end{array}\right)$$ But there exist real matrices which aren’t diagonalizable even if complex eigenvectors and eigenvalues are allowed. For example $\pmatrix{0&1\\0&0}$. In fact, the characteristic polynomial is $λ^2$ and it has only one double root $λ=0$. However the eigenspace corresponding to the 0 eigenvalue has dimension 1.\[B\pmatrix{v_1\\v_2}=\pmatrix{0\\0}⇔v_2=0\] and thus the eigenspace is $\ker(B)=\operatorname{span}_ℂ\{(1,0)^{\sf T}\}$, with only one dimension.
PlanetMath

Let $A∈M_n,A^k = 0,A^{k-1}≠0$. Prove that $k≤\operatorname{rank}(A)+1$. aops
Assume $A$ is in Jordan form, so $A=S_{m_1}⊕S_{m_2}⋯⊕S_{m_t}$, where $S_{m_i}$ is a Jordan block of size $m_i$ with zeros along diagonal, and $m_1 + m_2 +⋯+ m_t = n$. Each $S_{m_i}$ contributes $m_i - 1$ to the rank, so $\operatorname{rank}(A) = (m_1 - 1) + (m_2 - 1) + ⋯ + (m_t - 1)$. Now since the index of nilpotency is $k$, we know that at least one $m_i = k$, so $\operatorname{rank}(A)≥k-1$. Equality holds when $t=1$, i.e., when $n=k$.