# Finite sum of eigenspaces (with distinct eigenvalues) is a direct sum

I'm trying to show that a finite sum of eigenspaces (with distinct eigenvalues) is a direct sum.

I have $\alpha : V \to V$. The eigenspaces are $V_{\lambda_i} = \ker(\alpha - \lambda_i id_V )$ for $1 \leq i \leq n$. My attempt at a proof:

$A + B$ is a direct sum iff $A \cap B = \{0\}$. If $v \neq 0 \in V_{\lambda_i} \cap V_{\lambda_j}$ for some $i,j, i \neq j$, then $\alpha(v) = \lambda_i v$ and $\alpha(v) = \lambda_j v$. So $(\lambda_i - \lambda_j)v = 0$, and so $\lambda_i = \lambda_j$. This is a contradiction, so any pair of the eigenspaces have trivial intersection. Therefore $\cap_{i=1}^n V_{\lambda_i} = \{0\}$, and so we have a direct sum.

Is this ok?

Thanks

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Sounds good! Why wouldn't it be? –  David Kohler May 23 '11 at 23:59
Because the proof I have in my lecture notes is longer and messier! Thanks. –  user938272 May 24 '11 at 0:00

No, this is not a full proof. It is not true that, if $V = A+B+C$, and $A \cap B = A \cap C = B \cap C = \{ 0 \}$, then $V = A \oplus B \oplus C$. For example, let $V = \mathbb{C}^2$ and let $A$, $B$ and $C$ be the one dimensional subspaces spanned by $(1,0)$, $(1,1)$ and $(0,1)$.

This does give some good intuition for why the claim is true. If you want to build your way to the full proof, you might try the special case of three eigenspaces and see what you can do.

Amusingly, this is currently the top voted example of a common false belief over at MO.

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Thanks, this has been very helpful. –  user938272 May 24 '11 at 1:00
I so much forget this again and again... thanks for the reminder and the link David! –  David Kohler May 24 '11 at 15:05

To clarify some concepts, I list some relevant definitions from Artin's "Algebra". Suppose $\{W_i\}$ $i=1,\ldots,n$ are vector subspaces of $V$, then their sum is given by $$W_1 + \cdots + W_n := \{v\in V\ |\ v = w_1 + \cdots + w_n \mbox{ ,with } w_i\in W_i\}$$ The subspaces $W_1,\ldots,W_n$ are independent if $$w_1 + \cdots + w_n = 0 \mbox{ ,with } w_i\in W_i \mbox{ implies } w_i = 0 \mbox{ for all i}$$ A subspace $U$ is called a direct sum of $W_1,\ldots,W_n$ if $U = W_1 + \cdots + W_n$ and $W_1,\ldots,W_n$ are independent.

From the above definitions, what we real need to show in this problem is that eigenspaces corresponding to distinct eigenvalues are independent.

Let $v_1,\ldots,v_n$ be eigenvectors with eigenvalues $\lambda_1,\ldots,\lambda_n$, respectively, and $v_1 + \cdots + v_n = 0$.

Now, we use induction to show all $v_i = 0$. If $n = 1$, it's trivial. Otherwise, we have the following two observations. If we multiply both sides of the equation by $\lambda_1$, we get $$\lambda_1 v_1 + \cdots + \lambda_1 v_n = 0$$ If we apply the linear map $\alpha$ to both sides of the equation, we get $$\lambda_1 v_1 + \cdots + \lambda_n v_n = 0$$ Then, we subtract the first equation from the second one $$(\lambda_2 - \lambda_1) v_2 + \cdots + (\lambda_n - \lambda_1) v_n = 0$$ Since $\lambda_1,\ldots,\lambda_n$ are distinct, the coefficients in the equation above are non-zero.

By induction $v_1 = \cdots = v_n = 0$.

Hence, eigenspaces corresponding to distinct eigenvalues are independent.

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