Question

A is a 2 × 2 matrix with eigenvectors
v1=[1;-1] v2=[1;1] corresponding to eigenvalues
λ1 = (1/2) and λ2 = 2,respectively, and
X =[5;1]. Find A^10X.

Answer 1

please help im lost

Answer 2

hint: try to write X as a linear combination of v1 and v2

Answer 3

how do i determine that

Answer 4

ok, first of all have you learned what a linear combination is?

Answer 5

not quite im very confused with this topic

Answer 6

ok. A linear combination is just the addition and scalar multiplication of two vectors. Take for example \[2\left(\begin{matrix}1 \\ 0\end{matrix}\right) + -3\left(\begin{matrix}0 \\ 1\end{matrix}\right)\] this is a linear combination of (1 0) and (0 1) because we are multiplying by scalars and adding vectors... does that make sense?

Answer 7

ok i see that

Answer 8

so X=2v1+3v2

Answer 9

excellent! that is right.

Answer 10

OK, so now you are probably thinking... now what? :)

Answer 11

yea

Answer 12

Well, let's write A^10 X using our new expression for X as a linear combination. We get \[A^{10}X = A^{10}(2v_1 + 3v_2) = A^{10}(2v_1) + A^{10}(3v_2)\] by distributing. Do you see that?

Answer 13

ok i get that

Answer 14

Cool! Now, we get \[2A^{10}v_1 + 3A^{10}v_2\] since scalars can be moved around.

Answer 15

so then we can get 2(\[\lambda ^{10}\]v1+3(\[\lambda _{2^{10}}\]v2

Answer 16

precisely!

Answer 17

i just dont know how to get the final answer

Answer 18

OK, well I personally wouldn't actually calculate a number to the 10th power, just leave it with an exponent. I'll type it below.

Answer 19

Maybe something like this \[\left(\begin{matrix}2^{-9} + 3 \times 2^{10} \\ -2^{-9} + 3 \times 2^{10}\end{matrix}\right)\]

Answer 20

After I've just added to the two vector together after multiplying by the scalars.

Answer 21

That 2^(-9) came from \[2(\frac{1}{2})^{10} = 2 \times 2^{-10} = 2^{-9}\]

Answer 22

but however you want to write it :)

Answer 23

ok i tired that final answer and i got it incorrect

Answer 24

could you take a screenshot maybe of what the input looks like?

Answer 25

or else you could just input the decimal answer... maybe the computer would like that better...

Answer 26

i got it correct thanks so much for your help

Answer 27

no problem, good luck on eigenvectors! :)

Answer 28

since u have the eigenvectors and eigenvalues of the matrix A then A can be "factored" as
\[\large A=S^{-1}\Lambda S \]
where
\[\large S=\begin{pmatrix}
1 & 1\\ -1 & 1
\end{pmatrix}\qquad\text{and}\qquad
\Lambda=\begin{pmatrix}
1/2 & 0\\ 0 & 2
\end{pmatrix} \]

Answer 29

therefore
\[\large A^{10}x=(S^{-1}\Lambda S)^{10}x=(S^{-1}\Lambda^{10}S)x \]