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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.

user264745
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  • Fun fact: My first draft of this post was “potential approach (1) and can we use induction to prove this theorem?”. After lots of draft (trying to make proof work), I eventually reach desired result.It took me long time and I didn’t give up. I made use of one lemma, which is not in Hoffman’s book. I would really appreciate if you check my approach (1) and give feedback for approach (2). I know it is long. – user264745 Nov 06 '22 at 14:15
  • The spaces $W_\bullet$ are linearly disjoint so $W=\bigoplus_i W_i$ and the dimension-counting follows. – FShrike Nov 06 '22 at 16:25
  • @FShrike What’s definition of linearly disjoint. I assume $W_i$‘s are subspace of $V$ and pairwise disjoint. Which implies $W$ is direct sum of $W_1,…,W_k$. – user264745 Nov 06 '22 at 17:52
  • That's what I mean. If you have $W$ as the direct sum, then you can ignore all the eigenvalue stuff. The dimension of a direct sum (in finite case) is just the sum of dimensinos – FShrike Nov 06 '22 at 18:10
  • @FShrike Your claim is really generic! I don’t think it is even true. Though your idea/thought of showing $W$ is direct sum of $W_1,…,W_k$ is nice, since it’s quite standard fact “dimension of a direct sum (in finite case) is just the sum of dimensions”. I can prove $W=\oplus_{i=1}^kW_i$ using eigenvalue notion. Let $x\in W=W_1+…+W_k$. Then $\exists x_i\in W_i$ such that $x=x_1+…+x_k$. Suppose $\exists y_i\in W_i$ such that $x=y_1+…+y_k$. Since $T$ is linear map, we have $T(x)=T(x_1)+…+T(x_k)$ $ =T(y_1)+…+T(y_k)$ $=\lambda_1x_1+…+\lambda_kx_k$ $= \lambda_1y_1+…+\lambda_ky_k$. – user264745 Nov 06 '22 at 18:42
  • @FShrike By distributive law, $\lambda_1\cdot (x_1-y_1)+…+\lambda_k(x_k-y_k)=0_V$. Let $z_i=x_i-y_i\in W_i$, $\forall i\in J_k$. Using result of lemma, $z_i=0$, $\forall i\in J_k$. Thus $x_i=y_i$. Hence $\exists !\ x_i\in W_i$ such that $x=x_1+…+x_k$. This approach is the most efficient proof of theorem. – user264745 Nov 06 '22 at 18:42
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    @user264745 You can also prove that $W_1+\dots+W_k$ is a direct sum by using the theorem that eigenvectors corresponding to distinct eigenvalues are linearly independent. Then the only sum in $W_1+\dots+W_k$ that equals the additive identity of $V$ is $\underbrace{0+\dots+0}_\text{k times}$. – Seeker Nov 06 '22 at 20:24
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    @Seeker yess.. I’m using “eigenvectors corresponding to distinct eigenvalues are linearly independent” theorem. I’m just calling it lemma. – user264745 Nov 06 '22 at 20:33
  • @Seeker I’m wondering what if $z_i=0$, $\forall i\neq j$ and $z_j\neq 0$. Since $W_i$ is null space of $T-\lambda_i I$, $z_i$ could be zero vector. Conventionally we don’t consider zero vector as eigenvector. But I known $W_1\oplus \dotsb \oplus W_k$ is true. So I must be doing something wrong. – user264745 Nov 22 '22 at 20:18
  • @user264745 What do mean if $z_i=0$? Which part are you referring to? – Seeker Nov 22 '22 at 22:07
  • @Seeker one comment above your comment of “you can……… $0+\dotsb +0$ $k$ times”. Above I tryed to prove $W_1 \oplus \dotsb \oplus W_k$. In last step of proof I think proof don’t work. – user264745 Nov 23 '22 at 07:56

1 Answers1

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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.

Seeker
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  • Thank you for the answer. I have following questions: (1) Are you sure vectors $v_1,\dots,v_k$ also form a basis of $W_1+\dots+W_k$? Because I think that is not correct. (2) Let $v\in W_1+\dots+W_k$. Then $v=x_1+…+x_k$, for some $x_i\in W_i$. You wrote $v=a_1v_1+\dots+a_kv_k$. I assume you think $v_i$ spans $W_i$. That is not true in general. – user264745 Nov 23 '22 at 09:00
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    @user264745 I just realised a much simpler proof. Have a look at that. You were right about your questions. I was wrong about those two questions. – Seeker Nov 23 '22 at 09:10
  • Unfortunately your proof is not correct. Actually you get contradiction. $v_j$ should be non zero vector, $\forall j\in J_k$ to use “eigenvectors corresponding to distinct eigenvalues are linearly independent“. ${v_1,…,v_k}$ is independent and $v_1+…+v_k=0$ $\implies$ $1_F=0_F$. ${v_1,…,v_k}$ is not linearly independent if $v_j=0$, for some $j$. And you didn’t show uniqueness part. – user264745 Nov 23 '22 at 09:21
  • @user264745 Yeah, I should have started with: Suppose there exist non-zero vectors $v_1,\dots,v_k$ such that $v_1+\dots+v_k=0$. Then use the theorem to conclude that $v_j=0$ for all $j$. – Seeker Nov 23 '22 at 09:23
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    I think, after using theorem we conclude $1_F=0_F$, not $v_j=0$ for all $j$. We can’t get $v_j=0$ because in beginning of proof we assume $v_j\neq 0$. – user264745 Nov 23 '22 at 09:40
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    @user264745 You make a good point. I wrote up the proof in a hurry and didn't think too much about it. – Seeker Nov 23 '22 at 09:42
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    My point are not good. I also sometimes write proof in hurry and later realize it contains errors. IMO writing proof (right or wrong) is much better than doing nothing. If you read my potential approach (2), you’ll find out the place I got stuck in that proof is essentially same as last step of $W_1\oplus \dotsb \oplus W_k$. – user264745 Nov 23 '22 at 10:08
  • @user264745 I agree with everything! – Seeker Nov 23 '22 at 10:35
  • We don’t have to speculate no more. Here is proof of $W_1\oplus \dotsb \oplus W_k$. Proof is not that “trivial”. It uses induction approach. My approach is incorrect and can’t reach desired result (I think). – user264745 Nov 23 '22 at 14:39