anonymous
  • anonymous
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.
Linear Algebra
chestercat
  • chestercat
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chestercat
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anonymous
  • anonymous
please help im lost
jtvatsim
  • jtvatsim
hint: try to write X as a linear combination of v1 and v2
anonymous
  • anonymous
how do i determine that

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jtvatsim
  • jtvatsim
ok, first of all have you learned what a linear combination is?
anonymous
  • anonymous
not quite im very confused with this topic
jtvatsim
  • jtvatsim
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?
anonymous
  • anonymous
ok i see that
anonymous
  • anonymous
so X=2v1+3v2
jtvatsim
  • jtvatsim
excellent! that is right.
jtvatsim
  • jtvatsim
OK, so now you are probably thinking... now what? :)
anonymous
  • anonymous
yea
jtvatsim
  • jtvatsim
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?
anonymous
  • anonymous
ok i get that
jtvatsim
  • jtvatsim
Cool! Now, we get \[2A^{10}v_1 + 3A^{10}v_2\] since scalars can be moved around.
anonymous
  • anonymous
so then we can get 2(\[\lambda ^{10}\]v1+3(\[\lambda _{2^{10}}\]v2
jtvatsim
  • jtvatsim
precisely!
anonymous
  • anonymous
i just dont know how to get the final answer
jtvatsim
  • jtvatsim
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.
jtvatsim
  • jtvatsim
Maybe something like this \[\left(\begin{matrix}2^{-9} + 3 \times 2^{10} \\ -2^{-9} + 3 \times 2^{10}\end{matrix}\right)\]
jtvatsim
  • jtvatsim
After I've just added to the two vector together after multiplying by the scalars.
jtvatsim
  • jtvatsim
That 2^(-9) came from \[2(\frac{1}{2})^{10} = 2 \times 2^{-10} = 2^{-9}\]
jtvatsim
  • jtvatsim
but however you want to write it :)
anonymous
  • anonymous
ok i tired that final answer and i got it incorrect
jtvatsim
  • jtvatsim
could you take a screenshot maybe of what the input looks like?
jtvatsim
  • jtvatsim
or else you could just input the decimal answer... maybe the computer would like that better...
anonymous
  • anonymous
i got it correct thanks so much for your help
jtvatsim
  • jtvatsim
no problem, good luck on eigenvectors! :)
helder_edwin
  • helder_edwin
since u have the eigenvectors and eigenvalues of the matrix A then A can be "factored" as \[\large A=S^{-1}\Lambda S \] where \[\large S=\begin{pmatrix} 1 & 1\\ -1 & 1 \end{pmatrix}\qquad\text{and}\qquad \Lambda=\begin{pmatrix} 1/2 & 0\\ 0 & 2 \end{pmatrix} \]
helder_edwin
  • helder_edwin
therefore \[\large A^{10}x=(S^{-1}\Lambda S)^{10}x=(S^{-1}\Lambda^{10}S)x \]

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