# Eigenvector of two matrices

Suppose that the vector $$w$$ in $$\mathbb{R}^n$$ is an eigenvector for the $$n \times n$$ matrices $$L$$ and $$M$$, with (possibly different) eigenvalues $$𝜆$$ and $$𝜃$$.

a) Is $$w$$ an eigenvector of $$L + M$$? If so, what is the corresponding eigenvalue? Explain.

b) Is $$w$$ an eigenvector of the matrix $$sL$$, where $$s$$ is a scalar? If so, what is the corresponding eigenvalue? Explain.

c) What do your answers in parts a) and b) suggest about the set of all matrices for which w is an eigenvector? Explain.

I don't know if there is an easier way to do it but I made L and M into two 2x2 matrices.

$$L=\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$$

$$M= \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix}$$

Both of these I get the same eigenvectors $$[1,0]$$ and $$[0,1]$$ with the different eigenvalues of $$L=1, 1$$ and $$M=0,1$$.

If doing it correctly so far:

a) Yes, $$w$$ is an eigenvector of $$L+M$$. The corresponding eigenvalues are $$1$$.

b) Yes, $$w$$ is an eigenvector of $$sL$$ when $$s$$ is a scalar

c) I know this once a and b are correct. (Disregard C)

• Welcome to Mathematics Stack Exchange. Just because it's true for some examples doesn't mean it's always true. If you know $Lw=\lambda w$ and $Mw=\mu w$, using linearity you should know $(L+M)w$ and $sLw$ May 4 '20 at 19:39

You have only looked at one example, but one should do it more generally. Note that

$$(L + M)w = Lw + Mw = \lambda w + \theta w = (\lambda + \theta) w$$

implies that $$w$$ is an eigenvector of $$(L + M)$$ with corresponding eigenvalue $$(\lambda + \theta)$$.

For the second property, we have

$$(sL)w = s(Lw) = s(\lambda w) = (s\lambda)w,$$

hence $$w$$ is an eigenvector of $$sL$$ with corresponding eigenvalue $$s\lambda$$.

For c), you now should recall what one calls a vector space.

• corresponding eigenvalue $\lambda+\theta$ May 4 '20 at 19:59
• Of course, thanks, edited.
– Jan
May 4 '20 at 20:04