Proving the $0$ Vector and the Set of Eigenvectors of a Linear Map are a Subspace

Question

Given the map $T\in L(V)$ where $L(V)$ is the set of all linear maps from $V$ to $V$. I'm wondering whether it can be proven that the set of {the $0$ vector and all the eigenvectors of $T$} can be shown to be a subspace of $V$. I think it can, but I can't seem to prove that the set is closed under addition and multiplication (making it a vector space). Is there some example that doesn't fit and this, disproves this? Or is there a really simple way to prove it that I'm missing?

Answer 1

In general this is only a subspace if $T$ has a single eigenvalue. For example, the transformation $T$ of $\mathbb{R}^2$ given by the diagonal matrix $$\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$$ has eigenvalues $1$ and $-1$, with respective eigenspaces $$\text{span}\left\{\left(\begin{array}{c} 1 \\ 0 \end{array}\right)\right\} \qquad \text{and} \qquad \text{span}\left\{\left(\begin{array}{c} 0 \\ 1 \end{array}\right)\right\}.$$

So, the space $$\{0\} \cup \{\mathbf{x} : \mathbf{x} \text{ is an eigenvalue of $T$}\}$$ is the union of two distinct lines, which is not closed under addition, and hence is not a subspace of $\mathbb{R}^2$.

Answer 2

In general this statement is false.

Considering a vector space V, and the linear transformation as E, the set S is :

S = {v: Ev = kv where k is an eigenvalue of the linear transformation}

Now considering two eigenvalues k1 and k2 and corresponding eigenvectors v1 and v2, the equations are

Ev1 = k1v1

Ev2 = k2v2

By adding these two equations we get

E(v1+v2) = k1v1+k2v2

These two are linearly independent and cannot be written in the form of a multiple of v1+v2. Hence the misconception is false that the sum of eigenvectors is and eigenvector.