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

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?

• The subspace that has only the 0 vector is the trivial subspace it is easy to verify that it satisfies all the axioms of vector spaces. With the other one think about proper subspaces and their direct sum. Commented Sep 25, 2014 at 2:47

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 $\mathbb{R}^2$.

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.

• By this I intend to say that since the sum of these vectors do not belong to the set, it is violating the principal rule of a subspace. Hence it does not form a subspace. Commented Jun 24, 2018 at 11:08
• Welcome to MSE. It is in your best interest that you use MathJax. Commented Jun 24, 2018 at 11:22
• Thank you. I will look into it. Commented Jun 24, 2018 at 12:44