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