Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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
up vote 10 down vote accepted

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.

share|cite|improve this answer
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 be honest the "common false belief" linked to is not that a sum of subspaces is direct whenever the sum of every pair of them is direct; anyone doing linear algebra should realise that this is much to weak a hypothesis. The false belief is that a complicated formula holds between dimensions of intersections of subspaces, which tries to mimic the principle of inclusion/exclusion. It happens that this false belief would imply that a sum of pairwise independent subspaces is direct, but those that hold the false belief probably do not realise that (and would cease to believe when they do). – Marc van Leeuwen Oct 21 '14 at 6:09

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.

share|cite|improve this answer

For the record, here is one way to prove that any finite sum of eigenspaces for distinct eigenvalues is a direct sum. (One may in fact drop the "finite", since an infinite direct sum of subspaces is direct if and only if all sums of a finite subset of those subspaces are direct.)

Using induction on the number $n$ of eigenvalues involved (the cases $n\leq1$ are trivial), one may suppose that the sum of $V_{\lambda_1},\ldots,V_{\lambda_{n-1}}$ is already known to be direct. Then it suffices to show that the sum of $V_{\lambda_1}\oplus\cdots\oplus V_{\lambda_{n-1}}$ and $V_{\lambda_n}$ is direct (in this form one can conclude directness of a large sum of subspaces by considering only two subspaces at a time!). For this we may show that $(V_{\lambda_1}\oplus\cdots\oplus V_{\lambda_{n-1}})\cap V_{\lambda_n} = \{0\}$. But this amounts to showing that $\lambda_n$ is not an eigenvalue of the restriction of our linear operator $\alpha$ to the direct sum $V_{\lambda_1}\oplus\cdots\oplus V_{\lambda_{n-1}}$; this is obvious since on a basis of eigenvectors this restriction has a diagonal matrix with all diagonal entries among $\{\lambda_1,\ldots,\lambda_{n-1}\}$. Or equivalently, $\alpha-\lambda_nI$ acts as an invertible operator on each of the summands $V_{\lambda_i}$ (with $i<n$), and therefore also on their sum (combine the inverses).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.