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)

  • $\begingroup$ 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 $sLw$ $\endgroup$ 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.

  • $\begingroup$ corresponding eigenvalue $\lambda+\theta$ $\endgroup$ May 4 '20 at 19:59
  • 1
    $\begingroup$ Of course, thanks, edited. $\endgroup$
    – Jan
    May 4 '20 at 20:04

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