Let $T$ be a linear operator on the finite-dimensional space $V$. Let $\lambda_l,…,\lambda_k$ be the distinct characteristic values of $T$ and let $W_i$ be the space of characteristic vectors associated with the characteristic value $\lambda_i$. If $W=W_1+…+W_k$, then $\text{dim}(W)= \text{dim}(W_1)+…+ \text{dim}(W_k)$.
Approach (1): We use mathematical induction. $\forall j\in J_k$, $P(j)$: $ \text{dim}(W_1+…+W_j)= \text{dim}(W_1)+…+ \text{dim}(W_j)$. Base case: $j=2$. Since $W_i=N_{T-\lambda_i I}$, we have $W_i$ is subspace of $V$. By theorem 6 section 2.3, $\mathrm{dim}(W_1)+ \mathrm{dim}(W_2)= \mathrm{dim}(W_1\cap W_2)+ \mathrm{dim}(W_1+W_2)$. It’s easy to check, $\mathrm{dim}(W_1\cap W_2)=0$$\iff$$W_1\cap W_2=\{0_V\}$. Let $x\in W_1\cap W_2$. Then $T(x)=\lambda_1\cdot x=\lambda_2\cdot x$. By distributive law, $(\lambda_1 -\lambda_2)\cdot x=0_V$. Which implies $\lambda_1-\lambda_2=0_F$ or $x=0_V$. Since $\lambda_1\neq \lambda_2$, we have $x=0_V$. Thus $W_1\cap W_2=\{0_V\}$. So $\mathrm{dim}(W_1\cap W_2)=0$. Hence $\mathrm{dim}(W_1)+ \mathrm{dim}(W_2)= \mathrm{dim}(W_1+W_2)$. Inductive step: Let $S_j=\sum_{i=1}^jW_i$. Suppose $\text{dim}(S_j)= \text{dim}(W_1)+…+ \text{dim}(W_j)$, for some $2\leq j\lt k$. Since $S_j=\text{span}(\bigcup_{i=1}^jW_i)$, we have $S_j$ is subspace of $V$. By theorem 6 section 2.3, $\text{dim}(S_j)+\text{dim}(W_{j+1})= \text{dim}(S_j \cap W_{j+1})+ \text{dim}(S_j+W_{j+1})$. So $\text{dim}(W_1)+…+ \text{dim}(W_j)+ \text{dim}(W_{j+1})= \text{dim}(S_j \cap W_{j+1})+ \text{dim}(S_{j+1})$. We need to show $S_j\cap W_{j+1}=\{0_V\}$. Let $x\in S_j\cap W_{j+1}$. Then $x\in S_j$ and $x\in W_{j+1}$. So $x=x_1+…+x_j$ for some $x_i\in W_i$. Since $T$ is linear map, we have $T(x)=T(x_1+…+x_j)=T(x_1)+…+T(x_j)=\lambda_1\cdot x_1+…+\lambda_j \cdot x_j=\lambda_{j+1}\cdot x$. So $\lambda_1 \cdot x_1+…+\lambda_j \cdot x_j= \lambda_{j+1} \cdot x_1+…+\lambda_{j+1}\cdot x_j$. By distributive law, $(\lambda_1-\lambda_{j+1})\cdot x_1+…+(\lambda_j-\lambda_{j+1})\cdot x_j=0_V$. We claim $x_i=0_V$, $\forall 1\leq i\leq j$. Let $A=\{i\in J_j|\ x_i=0_V\}$ and $B=\{i\in J_j|\ x_i\neq 0_V\}$. Assume towards contradiction, $\exists p\in J_j$ such that $x_p\neq 0_V$, i.e. $p\in B$. Then $\sum_{i=1}^j(\lambda_i-\lambda_{j+1})\cdot x_i=\sum_{i\in B} (\lambda_i-\lambda_{j+1})\cdot x_i=0_V$. $\forall i\in B$, $x_i$ is eigenvector of $\lambda_i$. It’s a standard result, if $\lambda_1,…,\lambda_m$ are distinct eigenvalue of $T$ and $v_1,…,v_m$ are corresponding eigenvector, then $\{v_1,…,v_m\}$ is linearly independent. So $\{x_i|\ i\in B\}$ is independent. Which implies $\lambda_i-\lambda_{j+1}=0_F$, $\forall i\in B$. So $\lambda_p=\lambda_{j+1}$. Thus we reach contradiction. Hence $x_i=0_V$, $\forall 1\leq i\leq j$. So $x=x_1+…+x_j=0_V$. Thus $S_j\cap W_{j+1}=\{0_V\}$$\iff$$\text{dim}(S_j\cap W_{j+1})=0$. Hence $\text{dim}(W_1)+…+ \text{dim}(W_j)+ \text{dim}(W_{j+1})=\text{dim}(S_{j+1})$. By principle of mathematical induction, $P(j)$ holds $\forall j\in J_k$. Is my proof correct?
Potential Approach (2): Since $\text{dim}(V)=n\in \Bbb{N}$, we have $\text{dim}(W_i)=r_i\leq n$. Let $B_i=\{\alpha_{i1},…,\alpha_{ir_i}\}$ be basis of $W_i$, $\forall i\in J_k$. It’s easy to check, $W_i\cap W_j=\{0_V\}$, if $i\neq j$. So $B_i\cap B_j=\emptyset$, if $i\neq j$. So $|\bigcup_{i=1}^k B_i|=|B_1|+…+|B_k|=r_1+…+r_k$. We claim $\bigcup_{i=1}^k B_i$ is basis of $W=W_1+…+W_k$. We show $\text{span}(\bigcup_{i=1}^k B_i)=W$. Let $x\in W$. Then $\exists x_i\in W_i$ such that $x=x_1+…+x_k$. Since $\text{span}(B_i)=W_i$, we have $x_i=\sum_{j=1}^{r_i}b_{ij}\cdot \alpha_{ij}$, $\forall i\in J_k$. So $x= \sum_{j=1}^{r_1}b_{1j}\cdot \alpha_{1j}+…+ \sum_{j=1}^{r_k}b_{kj}\cdot \alpha_{kj}\in \text{span}(\bigcup_{i=1}^kB_i)$. Thus $\text{span}(\bigcup_{i=1}^kB_i)=W$. We show $\bigcup_{i=1}^k B_i$ is linearly independent. If $\sum_{j=1}^{r_1}c_{1j}\cdot \alpha_{1j}+…+\sum_{j=1}^{r_k}c_{kj}\cdot \alpha_{kj}=0_V$, for some $c_{ij}\in F$. Let $x_i=\sum_{j=1}^{r_i}c_{ij}\cdot \alpha_{ij}\in \text{span}(B_i)=W_i$. Then $x_1+…+x_k=0_V$. We claim $x_i=0_V$, $\forall i\in J_k$. Let $A=\{i\in J_k|\ x_i=0_V\}$ and $B=\{i\in J_k|\ x_i\neq 0_V\}$. So $\sum_{i=1}^kx_i=\sum_{i\in B}x_i=0_V$. Since $T$ is linear map, $T(\sum_{i\in B}x_i)=\sum_{i\in B}T(x_i)=\sum_{i\in B}\lambda_i \cdot x_i=0_V$. $\forall i\in B$, $x_i$ is eigenvector of $\lambda_i$. lemma: if $\lambda_1,…,\lambda_m$ are distinct eigenvalue of $T$ and $v_1,…,v_m$ are corresponding eigenvector, then $\{v_1,…,v_m\}$ is linearly independent. So $\{x_i|i\in B\}$ is independent. Which implies $\lambda_i=0_F$, $\forall i\in B$. Since $\lambda_i \neq \lambda_j$, if $i\neq j$, we have $|B|=0$ or $1$. If $|B|=0$, then we’re done. If $|B|=1$, then $\exists p\in J_k$ such that $x_p\neq 0_V$. How to progress from here?
My approach (2) is partial proof. I didn’t know to use polynomial notion. Hoffman’s proof: If $\sum_{j=1}^{r_1}c_{1j}\cdot \alpha_{1j}+…+\sum_{j=1}^{r_k}c_{kj}\cdot \alpha_{kj}=0_V$, for some $c_{ij}\in F$. Let $f\in \Bbb{F}[x]$. Since $f(T)\in L(V,V)$, we have $0_V=[f(T)](\sum_{j=1}^{r_1}c_{1j}\cdot \alpha_{1j}+…+\sum_{j=1}^{r_k}c_{kj}\cdot \alpha_{kj})=\sum_{j=1}^{r_1}c_{1j}\cdot [f(T)](\alpha_{1j})+…+ \sum_{j=1}^{r_k}c_{kj}\cdot [f(T)](\alpha_{kj})$. By lemma 2 section 6.2, $[f(T)](\alpha_{ij})=f(\lambda_i)\cdot \alpha_{ij}$. So $\sum_{j=1}^{r_1}c_{1j}\cdot [f(T)](\alpha_{1j})+…+ \sum_{j=1}^{r_k}c_{kj}\cdot [f(T)](\alpha_{kj})= \sum_{j=1}^{r_1}c_{1j}\cdot f(\lambda_1)\cdot \alpha_{1j}+…+ \sum_{j=1}^{r_k}c_{kj}\cdot f(\lambda_k)\cdot \alpha_{kj}=0_V$, $\forall f\in F[x]$. Define $f_i=\prod_{j\in J_k-\{i\}}(x-\lambda_j)$, $\forall i\in J_k$. Then $\sum_{j=1}^{r_1}c_{1j}\cdot f_i(\lambda_1)\cdot \alpha_{1j}+…+ \sum_{j=1}^{r_k}c_{kj}\cdot f_i(\lambda_k)\cdot \alpha_{kj}=\sum_{j=1}^{r_i}c_{ij}\cdot f_i(\lambda_i)\cdot \alpha_{ij}=0_V$. Since $B_i$ is independent, we have $c_{ij}\cdot f_i(\lambda_i)=0_F$, $\forall 1\leq j\leq r_i$. Which implies $c_{ij}=0_F$ or $f_i(\lambda_i)=0_F$. Since $f_i(\lambda_i)\neq 0_F$, we have $c_{ij}=0_F$, $\forall 1\leq j\leq r_i$. Since $i$ was arbitrary, $c_{ij}=0_F$, $\forall i\in J_k$. Hence $\bigcup_{i=1}^kB_i$ is independent.
Edit: We can make potential approach (2) work. Claim, $x_1+…+x_k=0_V$$\implies$$x_i=0_V$, $\forall i\in J_k$ is same as $W_1\oplus \dotsb \oplus W_k$. Here is proof of $W_1\oplus \dotsb \oplus W_k$ using mathematical induction. Below (in comment) my proof of $\oplus_{i=1}^k W_i$ is incorrect & stupid.
Best Answer
Suppose $$v_1+\dots+v_k=0$$ for $v_j\in W_j$. Because eigenvectors corresponding to distinct eigenvalues are linearly independent, we have that $v_1=\cdots =v_k=0$.
Thus, $W_1+\dots+W_k$ is a direct sum.