A matrix has $n$ independent eigenvectors $\Rightarrow\Bbb R^n$ is the direct sum of the eigenspaces

Question

Let $A\in M_n(\Bbb R)$. I want to prove that if $A$ has $n$ linearly independent eigenvectors, then $\Bbb R^n$ is the direct sum of the eigenspaces. I want to prove it more directly (i.e., don't through the equivalent proposition that $A$ is similar to a diagonal matrix). In other words, if we pretend not knowing any preknowledge of diagonalization, how can we prove the result "$\Bbb R^n$ is the direct sum of the eigenspaces"? It seems straightforward, but aftering thinking it for a while, it is not so easy as I first expected.

Answer 1

Let $v_1,\ldots,v_n$ be $n$ linearly independent eigenvectors; for each $i\in\{1,2,\ldots,n\}$, let $\lambda_i\in\mathbb R$ be such that $Av_i=\lambda_iv_i$. We can assume without loss of generality that $\lambda_1\leqslant\lambda_2\leqslant\cdots\leqslant\lambda_n$. Suppose that $\lambda_1=\lambda_2=\cdots=\lambda_k$ , for some $k\in\{1,2,\ldots,n\}$. Then the eigenspace $E_{\lambda_1}$ associated with $\lambda_1$ is $\bigoplus_{i=1}^k\mathbb{R}v_i$. And if $\lambda_{k+1}=\lambda_{k+2}=\cdots=\lambda_l$ for some $l\in\{k+1,\cdots,n\}$, then $E_{\lambda_{k+1}}=\bigoplus_{i=k+1}^l\mathbb{R}v_i$. And so on. So, if $k_1,\ldots,k_m\in\{1,2,\ldots,n\}$ are such that $\lambda_{k_1}<\cdots<\lambda_{k_m}$ and that $\{\lambda_{k_1},\ldots,\lambda_{k_m}\}=\{\lambda_1,\ldots,\lambda_n\}$, then$$\mathbb{R}^n=\bigoplus_{j=1}^mE_{\lambda_j}.$$

Answer 2

For $\lambda\in \Bbb R$, let $E_\lambda:=\ker(A-\lambda\operatorname{id})$ denote the corresponding eigenspace.

The special situation we have in the OP is that there are $n$ linearly independent eigenvectors. Therefore, $$ \dim \sum_\lambda E_\lambda\ge n$$ and as all proper subspaces of $\Bbb R^n$ have smaller dimension, we conclude $$ \Bbb R^n=\sum_\lambda E_\lambda.$$ (At this point we use that we are in the finite-dimensional case; all that follows does not rely on that).

What remains is that the sum on the right is in fact direct. But recall that this is always the case: Assume $$ v\in E_{\lambda_1}\cap \sum_{\lambda\ne\lambda_1} E_\lambda$$ and write $v=\sum_{\lambda\ne\lambda_1}v_\lambda$ with $v_\lambda\in E_\lambda$ and with $v_\lambda=0$ for all but finitely many $\lambda$. Then applying the operator $$T:=\prod_{\lambda\ne\lambda_1\atop v_\lambda\ne0}(A-\lambda\operatorname{id})$$ to $v$, we see that $Tv_\lambda=0$ for all $\lambda\ne \lambda_1$, hence $Tv=0$. On the other hand, $v\in E_{\lambda_1}$ implies $Tv=cv$ with $c=\prod(\lambda_1-\lambda)$. Each factor in $c$ is non-zero, hence $c\ne0$. We conclude that $v=0$ and therefore that the sum is direct, i.e., $$\sum_\lambda E_\lambda=\bigoplus_\lambda E_\lambda.$$

Answer 3

Group the eigenvectors by eigenvalues; so we'll have $$ B_1=\{v_{11},v_{12},\dots,v_{1n_1}\} \quad B_2=\{v_{21},v_{22},\dots,v_{2n_2}\} \quad \dots \quad B_k=\{v_{k1},v_{k2},\dots,v_{kn_k}\} $$ where the $v_{ij}$ are eigenvectors relative to $\lambda_i$, with $\lambda_1,\lambda_2,\dots,\lambda_k$ pairwise distinct. By assumption, $n_1+n_2+\dots+n_k=n$.

Observe that $\operatorname{span}(B_i)\subseteq E_A(\lambda_i)$ (the eigenspace relative to $\lambda_i$). In particular, the subspaces $\operatorname{span}(B_i)$ are independent, because so are the eigenspaces. Hence $$ \dim\sum_{i=1}^k \operatorname{span}(B_i)= \sum_{i=1}^k \dim\operatorname{span}(B_i)=\sum_{i=1}^k n_i=n $$ Since $n_i\le\dim E_A(\lambda_i)$, we also have $$ n=\sum_{i=1}^k n_i\le\sum_{i=1}^k\dim E_A(\lambda_i) =\dim\sum_{i=1}^k E_A(\lambda_i)\le n $$ and therefore equality holds. Hence $$ \sum_{i=1}^k E_A(\lambda_i)=\mathbb{R}^n $$