Definition: Let $V$ be finite dimensional vector space over $F$ and $T\in L(V,V)$. We say $T$ is diagonalizable if $\exists B=\{\alpha_1,…,\alpha_n\}$ basis of $V$ such that $\alpha_i$ is eigenvector of $T$, $\forall i\in J_n$.
Suppose that $T$ is a diagonalizable linear operator. Let $c_1,…,c_k$ be the distinct characteristic values of $T$. Then there is an ordered basis $B$ in which $T$ is represented by a diagonal matrix which has for its diagonal entries the scalars $c_i$, each repeated a certain number of times. If $c_i$ is repeated $d_i$ times, then (we may arrange that) the matrix has the block form $$\label{eq}\tag{1} [T]_B= \begin{bmatrix}
c_1I_1 & 0 & \cdots & 0 \\
0 & c_2I_2 & \cdots & 0 \\
\vdots & \vdots & & \vdots \\
0 & 0 &\cdots &c_kI_k
\end{bmatrix}$$ where $I_j$ is the $d_j \times d_j$ identity matrix. From that matrix we see two things. (1) The characteristic polynomial for $T$ is the product of (possibly repeated) linear factors: $f=(x-c_1)^{d_1}\cdots (x-c_k)^{d_k}$. (2) $d_i$, the number of times which $c_i$ is repeated as root of $f$, is equal to the dimension of the space of characteristic vectors associated with the characteristic value $c_i$. That is because the nullity of a diagonal matrix is equal to the number of zeros which it has on its main diagonal, and the matrix $[T-c_i I]_B$ has $d_i$ zeros on its main diagonal.
Que(1): I don’t really understand, $T$ is diagonalizable $\Rightarrow$ Characteristic polynomial for $T$ is $f=(x-c_1)^{d_1}\cdots (x-c_k)^{d_k}$. “Let $c_1,…,c_k$ be the distinct characteristic values of $T$“ I assume it means all possible eigenvalues of $T$, i.e. $\nexists c\in F$ such that $c$ is eigenvalue of $T$ and $c_i\neq c$, $\forall i\in J_k$. “Then there is an ordered basis $B$ in which $T$ is represented by a diagonal matrix” let $B_i=\{\alpha_{i,1},…,\alpha_{i,d_i}\}$ such that $\alpha_{i,j}$ is eigenvector corresponding to $c_i$, $\forall j\in J_{d_i}$. Then $B=\bigcup_{i=1}^kB_i$ is ordered basis of $V$ such that $[T]_B$ is \eqref{eq}. What if $\exists B’=\{\beta_1,…,\beta_n\}$ basis of $V$ such that $\beta_i$ is eigenvector corresponding to $c_j$, $\forall i\in J_n$ and $j\lt k$? Then for former case $f(c_k)=0$ and for later case $f(c_k)\neq 0$. Which is absurd.
Que(2): Since I skipped beginning part of Hoffman’s book, I can’t fill in details. I don’t know to find dimension of null space without dimension of range space info.
Best Answer
Firstly, we assume $\mathscr{dim}~V=n$.
From the definition of Hoffman’s Linear Algebra, $T$ is diagonalizable if there is a basis for $V$ each vector of which is a characteristic vector of $T$.
The author has supposed $c_1,c_2,\cdots,c_k$ are all the distinct characteristic values of $T$.
So there exists $({\alpha_1,\alpha_2,\cdots,\alpha_n})$ which are basis.
Here is a step the author omits that we can change the positions of $\alpha_p$ and $\alpha_q$ if necessary such that $1\le i\le d_1$, $\alpha_i$ is the eigenvector of $c_1$ , $d_{1}+1\le j\le d_2+d_1$, $\alpha_j$ is the eigenvector of $c_2$ and so on.
Thus
$$T({\alpha_1,\alpha_2,\cdots,\alpha_n})=({\alpha_1,\alpha_2,\cdots,\alpha_n})\left( \begin{matrix} c_1I_1& & & \\ & c_2I_2& & \\ & & \ddots& \\ & & & c_kI_k\\ \end{matrix} \right).$$
We use the following property:
If the matrix of $T$ under a basis of $V$ has the form $$\left( \begin{matrix} A_1& *\\ 0& A_2\\ \end{matrix} \right) ,$$we define the characteristic polynomial for $T$ is $f(x)$, the characteristic polynomial for $A_1$ is $f_1(x)$ and the characteristic polynomial for $A_2$ is $f_2(x)$, then $f(x)=f_1(x)f_2(x)$.
So we get the characteristic polynomial for $T$ is $f(x)=(x-c_1)^{d_{1}}\cdots(x-c_k)^{d_k}$.