# Union of Bases for Eigenspaces is Linearly Independent

Problem Statement: Suppose $V$ is a vector space and $\Phi \in \mathcal{L}(V)$ has distinct eigenvalues $\lambda_1,...,\lambda_n$, and that each eigenspace $V_{\lambda_j}$ has a basis $B_{\lambda_j}$. Prove that $B = B_{\lambda_1} \cup ... \cup B_{\lambda_n}$ is a linearly independent subset of $V$.

Attempt: My thought is to do this by induction. The base case when $B = B_{\lambda_1}$ seems immediate, but I'm getting stuck on the inductive case.

My inclination is to use the facts that $\text{span}(C) \cap \text{span}(D) = \{0\} \implies C, D$ are linearly independent and that $V_{\lambda_j} \cap V_{\lambda_i} = \{0\}$ for $i \neq j$ to say that $B_{\lambda_1} \cup ... \cup B_{\lambda_n}$ is linearly independent.

However, with the inductive hypothesis that $B^\prime = B_{\lambda_1} \cup ... \cup B = B_{\lambda_{n-1}}$ is a linearly independent subset of $V$, I can't come up with a way to show that $\text{span}(B^\prime) \cap \text{span}(B_{\lambda_{n}}) = \{0\}$.

Is this entire approach flawed or am I missing something?

Let $B_{\lambda_i}$ have $n_i$ elements: $v_{i,1}, v_{i,2}, \dots, v_{i,n_i}$. Let's try your induction; suppose there is some nonzero $u \in \left(B_{\lambda_1}\cup\dots\cup B_{\lambda_n}\right)\cap B_{\lambda_{n+1}}$. Then there is a linear combination

$$\tag{1}\label{eq1}\sum_{i=1}^n\sum_{j=1}^{n_i}c_{i,j}v_{i,j}=u$$

By applying $\Phi$ to both sides we obtain that

$$\tag{2}\label{eq2}\sum_{i=1}^n \sum_{j=1}^{n_i}\big(\lambda_i c_{i,j}\big)v_{i,j}=\lambda_{n+1}u$$

If $\lambda_{n+1}=0$, then we have a linear combination on $B_{\lambda_1}\cup\dots\cup B_{\lambda_n}$ that equals $0$. Moreover, since the $\lambda_i$'s are all distinct, they must all be nonzero for $i<n+1$. It then follows from the induction hypothesis that the $c_{i,j}$ must be all $0$, contradicting the assumption that $u\neq0$.

Now, if $\lambda_{n+1}\neq 0$, then we may write

$$u=\sum_{i=1}^n \sum_{j=1}^{n_i}\left(\frac{\lambda_i}{\lambda_{n+1}} c_{i,j}\right)v_{i,j}.$$

Then, combining equations $\eqref{eq1}$ and $\eqref{eq2}$ we obtain

$$0=u-u=\sum_{i=1}^n \sum_{j=1}^{n_i}\left(\frac{\lambda_i}{\lambda_{n+1}}-1\right)c_{i,j}\,v_{i,j},$$

so that by the induction hypothesis $\left(\frac{\lambda_i}{\lambda_{n+1}}-1\right)c_{i,j}=0$ for all $i,j$. Since the $\lambda$'s are distinct, we have that $\frac{\lambda_i}{\lambda_{n+1}} \neq 1$ so it must be that the $c_{i,j}$ are all $0$, again in contradiction with the assumption that $u\neq0$.

We see that whatever the case, there ca be no nonzero $u\in \left(B_{\lambda_1}\cup\dots\cup B_{\lambda_n}\right)\cap B_{\lambda_{n+1}}$, so the inductive step is complete.

• Great! This answer is just a little simpler for me to understand given my current abilities. – Ben Fogarty Mar 14 '17 at 17:29
• Glad to have helped! – Fimpellizieri Mar 14 '17 at 18:02