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 10 months ago
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.
 10 months ago
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.

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

jcaruso
 10 months ago
Best ResponseYou've already chosen the best response.0how do i determine that

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

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

jtvatsim
 10 months ago
Best ResponseYou've already chosen the best response.3ok. 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?

jtvatsim
 10 months ago
Best ResponseYou've already chosen the best response.3excellent! that is right.

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

jtvatsim
 10 months ago
Best ResponseYou've already chosen the best response.3Well, 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?

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

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

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

jtvatsim
 10 months ago
Best ResponseYou've already chosen the best response.3OK, well I personally wouldn't actually calculate a number to the 10th power, just leave it with an exponent. I'll type it below.

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

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

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

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

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

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

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

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

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

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

helder_edwin
 10 months ago
Best ResponseYou've already chosen the best response.0therefore \[\large A^{10}x=(S^{1}\Lambda S)^{10}x=(S^{1}\Lambda^{10}S)x \]
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