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I noticed that I don't know how to approach True/False questions in Linear Algebra. Since it's a rather abstract branch of mathematics, I find it hard to find an answer intuitively. For example:

Consider a diagonalizable, linear operator $f$ on a finite dimensional $\mathbb{C}$-vector space $V$. Assume that for every two eigenvectors $v,w$ of $f$ holds that $v+w=0$ or $v+w$ is an eigenvector of $f$, then there exists a $\lambda \in \mathbb{C}$ such that $f = \lambda \cdot \operatorname{id}_V$.

When I first read it, I was confused an didn't know how to start. Do I first assume that the statement is true and try to prove it, or should I look for counterexamples first (=disprove it)?

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We are told that the operator is diagonalizable. That means we can speak about it entirely in terms of eigenvectors and eigenvalues. Also, a diagonalizable operator is of the form $\lambda \cdot \operatorname{Id}_n$ iff all its $n$ eigenvalues are the same. Finally, I always try to consider the contrapositive, because some times that's easier to prove (not always, but often enough that it should be among the first things you try if you're stuck).

Putting these known facts together to rewrite your statement, we get

If a diagonalizable linear operator has at least two distinct eigenvalues, then there are two eigenvectors whose sum is neither zero nor an eigenvector

Phrased like this, the answer almost writes itself. You have at least two distinct eigenvalues $\lambda_1,\lambda_2$. Take one eigenvector for each eigenvalue, add them together, and see that the result is non-zero and not an eigenvector (some care needs to be taken in this step in case one of the $\lambda_i$'s is zero). This shows that the statement is true.

The only general advice I can give is to use what you know to continually rephrase the statement you want to prove or disprove. Eventually you will hopefully reach a statement where you have an idea how to start (or rather, this rephrasing process is a start).

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This is something else I tried:

Suppose $\dim(V)=n$ and $f$ is diagonalizable with eigenvalues $\lambda_1,\dots,\lambda_n$. There exists a basis $\mathcal{B} = \{ v_1,\dots,v_n\}$ of eigenvectors such that $f(v_k)=\lambda_k \cdot v_k$ for $k=1,\dots,n$. The sum of eigenvectors of $f$ is an eigenvector if the corresponding eigenvalues are equal. So pick two arbitrary indexes $i,j\in \{1,\dots,n\}, i \ne j$. We have to show that $\lambda_i=\lambda_j$. Consider the corresponding eigenvectors $v_i$ and $v_j$. We know that $v_i + v_j \ne 0$, because $\mathcal{B}$ is a basis (of eigenvectors that are linear independent). The given statements then says that $v_i+v_j$ is an eigenvector of $f$. So, $f(v_i+v_j)=f(v_i)+f(v_j)=\lambda_iv_i+\lambda_jv_j = \lambda_k(v_i+v_j) $ for a $k\in\{1,\dots,n\}$. This implies $(\lambda_i-\lambda_k)v_i + (\lambda_j - \lambda_k)v_j = 0$, and from the linear independence of $v_i$ and $v_j$ follows $\lambda_i = \lambda_j = \lambda_k$. Since $i$ and $j$ were arbitrary indexes, we have shown that all eigenvalues are equal to $\lambda_k =: \lambda$. So for $l \in \{1,\dots,n\}: f(v_l)=\lambda \cdot v_l $. Now we have to prove that $f(v)=\lambda \cdot \operatorname{id}_V(v)$ for all $v\in V$. Since $\mathcal{B}$ is basis for $V$, we can write an element $v\in V$ as follows $v=a_1 v_1 + \dots + a_n v_n$. Now we're looking for an expression for $f(v)$: $f(v) = f(a_1 v_1 + \dots + a_n v_n) = \lambda(a_1 v_1 + \dots + a_n v_n) = \lambda \cdot v = \lambda \cdot \operatorname{id}_V(v).$ This completes the proof.

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Either way. Perhaps try to prove it and see what the difficulties are, and try to get a feel for why it might be true/false.

Now, if $f(v)=\lambda_1v$ and $f(w)=\lambda_2w$, then $f(v+w)=\lambda_1v+\lambda_2w$. So if $f(v+w)=\lambda_3(v+w)$, then we have $(\lambda_3-\lambda_1)v=(\lambda_2-\lambda_3)w$. Thus $v$ is a multiple of $w$, and we have a contradiction (same eigenvector for different eigenvalues), unless $\lambda_1=\lambda_2=\lambda_3$.

So it appears the statement is true.

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