Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
jcaruso
Group Title
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.
 4 months ago
 4 months ago
jcaruso Group Title
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.
 4 months ago
 4 months ago

This Question is Open

jcaruso Group TitleBest ResponseYou've already chosen the best response.0
please help im lost
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
hint: try to write X as a linear combination of v1 and v2
 4 months ago

jcaruso Group TitleBest ResponseYou've already chosen the best response.0
how do i determine that
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
ok, first of all have you learned what a linear combination is?
 4 months ago

jcaruso Group TitleBest ResponseYou've already chosen the best response.0
not quite im very confused with this topic
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
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?
 4 months ago

jcaruso Group TitleBest ResponseYou've already chosen the best response.0
ok i see that
 4 months ago

jcaruso Group TitleBest ResponseYou've already chosen the best response.0
so X=2v1+3v2
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
excellent! that is right.
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
OK, so now you are probably thinking... now what? :)
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
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?
 4 months ago

jcaruso Group TitleBest ResponseYou've already chosen the best response.0
ok i get that
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
Cool! Now, we get \[2A^{10}v_1 + 3A^{10}v_2\] since scalars can be moved around.
 4 months ago

jcaruso Group TitleBest ResponseYou've already chosen the best response.0
so then we can get 2(\[\lambda ^{10}\]v1+3(\[\lambda _{2^{10}}\]v2
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
precisely!
 4 months ago

jcaruso Group TitleBest ResponseYou've already chosen the best response.0
i just dont know how to get the final answer
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
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.
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
Maybe something like this \[\left(\begin{matrix}2^{9} + 3 \times 2^{10} \\ 2^{9} + 3 \times 2^{10}\end{matrix}\right)\]
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
After I've just added to the two vector together after multiplying by the scalars.
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
That 2^(9) came from \[2(\frac{1}{2})^{10} = 2 \times 2^{10} = 2^{9}\]
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
but however you want to write it :)
 4 months ago

jcaruso Group TitleBest ResponseYou've already chosen the best response.0
ok i tired that final answer and i got it incorrect
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
could you take a screenshot maybe of what the input looks like?
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
or else you could just input the decimal answer... maybe the computer would like that better...
 4 months ago

jcaruso Group TitleBest ResponseYou've already chosen the best response.0
i got it correct thanks so much for your help
 4 months ago

jtvatsim Group TitleBest ResponseYou've already chosen the best response.3
no problem, good luck on eigenvectors! :)
 4 months ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.0
since u have the eigenvectors and eigenvalues of the matrix A then A can be "factored" as \[\large A=S^{1}\Lambda S \] where \[\large S=\begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix}\qquad\text{and}\qquad \Lambda=\begin{pmatrix} 1/2 & 0\\ 0 & 2 \end{pmatrix} \]
 4 months ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.0
therefore \[\large A^{10}x=(S^{1}\Lambda S)^{10}x=(S^{1}\Lambda^{10}S)x \]
 4 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.