## 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. 5 months ago 5 months ago

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1. jcaruso Group Title

2. jtvatsim Group Title

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

3. jcaruso Group Title

how do i determine that

4. jtvatsim Group Title

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

5. jcaruso Group Title

not quite im very confused with this topic

6. jtvatsim Group Title

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?

7. jcaruso Group Title

ok i see that

8. jcaruso Group Title

so X=2v1+3v2

9. jtvatsim Group Title

excellent! that is right.

10. jtvatsim Group Title

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

11. jcaruso Group Title

yea

12. jtvatsim Group Title

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?

13. jcaruso Group Title

ok i get that

14. jtvatsim Group Title

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

15. jcaruso Group Title

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

16. jtvatsim Group Title

precisely!

17. jcaruso Group Title

i just dont know how to get the final answer

18. jtvatsim Group Title

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.

19. jtvatsim Group Title

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

20. jtvatsim Group Title

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

21. jtvatsim Group Title

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

22. jtvatsim Group Title

but however you want to write it :)

23. jcaruso Group Title

ok i tired that final answer and i got it incorrect

24. jtvatsim Group Title

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

25. jtvatsim Group Title

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

26. jcaruso Group Title

i got it correct thanks so much for your help

27. jtvatsim Group Title

no problem, good luck on eigenvectors! :)

28. helder_edwin Group Title

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

29. helder_edwin Group Title

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