Linear Combinations of Eigenvectors

Working with eigenvectors can greatly simplify computations, since matrix multiplication with an eigenvector is just scalar multiplication. As we will see in the following theorem, there are also computational advantages to working with linear combinations of eigenvectors.

Theorem: Let AA be an n$$nn$$n matrix and let v$1,$v$2,...,$vmv$1,$v$2,...,$vm be eigenvectors of AA corresponding to the eigenvalues $1,$$2,...$m$1,$$2,...$m$,$ respectively. If xx is a linear combination of these eigenvectors, say

x$=$c$1$v$1+$c$2$v$2++$cmvm$,$x$=$c$1$v$1+$c$2$v$2++$cmvm$,$

then for any non$-$negative integerkk$($or any integer ifAAis invertible$),$

Akx$=$c$1$k$1$v$1+$c$2$k$2$v$2++$cmkmvmAkx$=$c$1$$1$kv$1+$c$2$$2$kv$2++$cmmkvm

Let's illustrate how to use this theorem through an example.

Example: Let A$=102122100$A$=[111020220].$

Recall from Question $9$ of the learning activity "Eigenvalues, Eigenvectors, and Eigenspaces: Part $1"$ that the eigenvalues of AA are $=1$$=1$ and$$ $=2$$=2$ and the corresponding eigenspaces are

E$1=$Span$$u$=102$andE$2=$Span$$v$=110,$w$=101$E$1=$Span$($u$=[102])$andE$2=$Span$($v$=[110],$w$=[101])$

Note: For any integer kk$,$since u in E$1$u in E$1,$ we have Aku$=(1)$ku$=$uAku$=(1)$ku$=$u and since v$,$w in E$2$v$,$w in E$2,$ we have$$Akv$=(2)$kvAkv$=(2)$kv and$$Akw$=(2)$kwAkw$=(2)$kw$.$

$($a$)$ Compute A$32413$A$3[2413].$

We first note that $2413[2413]$ can be written as a linear combination of the eigenvectors uu$,$ vv$,$ and ww$.$ Indeed,$$

$2413=51024110+3101[2413]=5[102]4[110]+3[101]$

Thus,$$A$32413$A$3[2413]==$A$351024110+3101$A$3(5[102]4[110]+3[101])==$A$35102$A$34110+$A$33101$A$3(5[102])$A$3(4[110])+$A$3(3[101])==5$A$31024$A$3110+3$A$31015$A$3[102]4$A$3[110]+3$A$3[101]==5(5()31024()3[102]4()3110+3()3[110]+3()3101)3[101],$since u in E$1$u in E$1$ and v$,$w in E$2$v$,$w in E$2==102[102]++110[110]101[101]$

$$

$==$

$$

$$

$$

$$

$($b$)$ Compute A$5126$A$5[126].$

We first note that $126[126]$ can be written as a linear combination of the eigenvectors uu$,$ vv$,$ and ww$.$ Indeed,$$

$126=[126]=102[102]110+[110]+101[101]$

$$

$$

Thus, A$5126=$A$5[126]=$

$$

$$