$k = 1, \lambda_1, x_1, x_1 \neq 0 , x_1$ is linearly independent.

Assume that the statement is true for any $(k-1) $ number of distinct eigenvalue.

Consider:

$a_1x_1 + a_2x_2 + … + a_{ k }x_{ k } = 0, a_i \in F$

Apply $(T- \lambda_a I_v)$ on both side.

$$\begin{align*} (T- \lambda_k I_v)(a_ix_i) i \neq k \br &= a_i (T x_i - \lambda_k x_i) \br &= a_i(\lambda_i x_i - \lambda_k x_i) \br &= a_i(\lambda_i - \lambda_k)x_i \end{align*}$$

since, $\lambda_i - \lambda_k \neq 0$

$a_1(\lambda_1 - \lambda_k)x_1 + a_2(\lambda_2 - \lambda_k)x_2 + … + a_{ k }(\lambda_k - \lambda_k)x_{ k } = 0, a_i \in F$

$\therefore a_i (\lambda_i - \lambda_k) = 0 (i = 1, 2, …, k - 1) \implies a_i = 0, \lambda_i \neq \lambda_k$

using (i) and (ii) we get $a_kx_k = 0 $

$\because x_k $ is an eigenvector. $x_k \neq 0$ $\therefore \set{ x_k }$ is linearly independent $\implies a_k = 0 $

$\lambda_1 \neq \lambda_2 \neq … \neq \lambda_k$ eigenvalues

$x_1, $x_2, …, $x_{ n }$ eigenvectors

$\beta = \set{ x_1, x_2, …, x_{ n }}$ is linearly independent.

Notice $\abs{ \beta } = n = \dim V $

$\therefore \beta$ is a basis of $V $

$\beta $ is a basis of $V$ consisting only eigenvectors of $T $, hence $T $ is diagonalizable.

$0 \neq v \in E_ \lambda \implies \dim E_ \lambda \geq 1 $.

$\dim E_\lambda = p$.

Let $\set{ v_1, v_2, …, v_{ p }}$ be a basis of $E_\lambda \subseteq V $

Extend this basis to a basis $\set{ v_1, v_2, …, v_{ p+1 }, v_{p +1}, v_{p+2}, …, v_{ n }} = \beta$ of $V $.

$[ T ]_{ beta }$ observe that $v_1, _2, …, _{ p } $ are all eigenvectors of with respect to $\lambda$.

$\therefore T(v_1) = \lambda v_1 = \lambda \cdot v_1 + 0 \cdot v_2 + … + 0 \cdot v_n$ $T(v_2) = \lambda v_2 =0 \cdot v_1 + \lambda \cdot v_2 + … + 0 \cdot v_n$ $T(v_p) = \lambda v_p =0 \cdot v_1 + 0 \cdot v_2 + … + \lambda \cdot v_p + … + 0 \cdot v_n$

$T(v_{p + 1}) = \sum_{ i=1 }^{ n } a_{i_{p+1}}v_i$

$T(v_n) = \sum_{ i=1 }^{ n } a_{in} v_i $

$A = [ T ]_{ \beta } = {\Mww{ \lambda I_p }{ B }{ 0 }{ C }}$

$(A - tI_n) = {\Mww{ \lambda I_p - t I_p }{ B }{ 0 }{ C - t I_{n-p}}}$

$\begin{align*} \det(A - tI_n) &= \det (\lambda I_p - t I_p) \det (c - t I_{n - p}) \br &= \det ((\lambda - t) I_p) \det(c- t I_{n-p}) \br &= (\lambda -t)^p \det I_p \det(c - t I_ {n - p}) \br &= (\lambda - t)^p g(t), \text{ where } g(t) = \det(c- I_{n-p}) \end{align*}$

By contradiction, assume that there is at least one $v_i$ say $v_{i_1} \neq 0 , i \in \set{ 1, 2, …, k }$,

i.e. $v_{i_1} $ is an eigenvector contradiction to $\lambda_{i_ 1} $

Let $v_{i_1}, v\{ i_2 }, …, v\{ i_{m}} $ are all non-zero vectors, from $E_{ \lambda_{i _ 1}}, E_ { \lambda { i_ 2 }}, … , E_ { \lambda { i_ m }} $, respectively, and the remaining ones are all zero vectors.

$i_1, i_2, …, i_{ m } \in \set{ 1, 2, …, k } $

$\therefore $ the equation $(i) $ looks like the following,

$v_{i_1} + v_{i_2} + … + v_{i_ m } = 0 $(ii)

Since $v_{i_1} \neq 0, v_{i_2} \neq 0, …, v_{i_n} \neq 0$.

Then all eigenvector for corresponding to distinct eigenvalues, and that is contradiction to (ii) since eigenvectors from different eigenvalues are linearly independent.

Let $\dim E_{ \lambda _ i } = n_i , i - 1, 2, …, k$

Let $S_i \subseteq E_ { \lambda_ i } $

$S_i = \set{ v_{i1}, v _{i2}, …, v_{in_i}} , i = 1, 2, …, k$

want to prove $S_i $ linearly independent.

To that end consider the following equation:

$\sum_{ i=1 }^{ k } \sum_{ j = 1 }^{ n_i } a_{ij}v_{ij} = 0$

Let $w_i = \sum_{ j=1 }^{ n_i } a_{ij}v_{ij}, i = 1, 2 , …, k$

Then $w_i \in E_ { \lambda _ i }$, since $v_{i1}, v_{ i2 }, …, v_{ in_i } \in S_i \subseteq E_{ \lambda_ i }$ and $E_ { \lambda_i } is a subspace$

$\therefore $ we get from (i) $\sum_{ i= 1 }^{ k } w_i = 0, w_i \in E_ {\lambda_i}$.

By previous theorem we then get

$w_i = 0, \forall i = 1, 2, …, k $

i.e. $\sum_{ j=1 }^{ n_i } a_{ij} v_{ij} = 0 \forall i = 1, 2, …, k$ since $S_i = \set{ v_{i_1}, v_{i_2}, …, v_{in_i}} $ is linearly independent from (iii) we get

$a_{ij} = 0 \forall i = 1, 2, …, n and j = 1, 2, …, n_i $

$\therefore \underset{i = 1}{\overset{k} \cup} S_i $ is linearly independent.

$T $ is diagonalizable $\iff \exists$ a basis $\beta $ consisting of eigenvectors of $T $.

$\dim V = n $

let $f(t)$ be the characteristic polynomial of $T $.

$deg(f(t)) = n$o $\lambda_1, \lambda_2, …, \lambda_{ k }$ are the distinct roots of $f(t)=(\lambda_1)^{a_1} + (\lambda_2)^{a_2} + … + (\lambda_k)^{a_k}$

where $a_i = $ algebraic multiplicity of $\lambda_i $

$\therefore deg(f(t)) = n $

$\because a_1, a_2, …, a_{ k } = n (*)$

Part 1 that algebraic multiplicity$(\lambda_i)$ = geometric multiplicity $(\lambda_i) , \forall i = 1, 2, …, k$

$\therefore$ geometric multiplicity $(\lambda_i) = a_i, \forall i = 1, 2, …, k$

Recall that, geometric multiplicity$(\lambda_i) = \dim E_ { \lambda_i } $

$\therefore$ let $\beta_i $ be basis of $E_ { \lambda_i } $ Then $\abs{ \beta_i } = a_i, i = 1, 2, …, k $

Then $\beta_1 \cup \beta_2 \cup … \cup \beta_k $ is a linearly independent set by the previous theorem.

Now, observe that $\beta_i \cap \beta_j = \emptyset $ since $i \neq j $.

$\therefore \abs{ \beta_1 \cup \beta_2 \cup … \cup \beta_k} = \abs{ \beta_1 } + \abs{ \beta_2 } + … + \abs{ \beta_K } = n = \dim V by (*)$

$\therefore \underset{i = 1}{\overset{k} \cup} \beta_i $ is a basis of $V $ consisting of eigenvector of $T $. Hence T is diagonalizable.

Part 2

Assume that $T $ is diagonalizable. Then exists an ordered basis $\beta $ of $V $ s.t. [ T ]_{ \beta } is a diagonalizable matrix.

Let $\beta $ = \set{ v_{11}, v_{12}, …, v_{1n_1}, v_{21}, v_{22}, …, v_{2n_2}, …, v_{k1}, v_{k2}, …, v_{kn_k}}

where , $v_{i1}, v_{i2}, …, v_{in_i}$ are eigenvector corresponding to the eigenvalues $\lambda_i \forall i = 1, 2, …, k$.

$\therefore $ geo.mult.$(\lambda_i) = n_i$ … (*)

$T(v_{11}) = \lambda_1v_{11} $

$\vdots $

$T(v_{1n_1}) = \lambda_1v_{1n_i} $

$\vdots $

(you get the idea…)

$\det([ T ]_{ \beta } - tI_n) = (\lambda_1 -t)^{n_1}(\lambda_2-t)^{n_2}…(\lambda_k -t)^{n_k}$

$\therefore$ alg.mult.$(\lambda_i) = n_i = $ geo.mult.$(\lambda_i)$