## jcaruso one year 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.

• This Question is Open
1. jcaruso

2. jtvatsim

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

3. jcaruso

how do i determine that

4. jtvatsim

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

5. jcaruso

not quite im very confused with this topic

6. 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?

7. jcaruso

ok i see that

8. jcaruso

so X=2v1+3v2

9. jtvatsim

excellent! that is right.

10. jtvatsim

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

11. jcaruso

yea

12. 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?

13. jcaruso

ok i get that

14. jtvatsim

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

15. jcaruso

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

16. jtvatsim

precisely!

17. jcaruso

i just dont know how to get the final answer

18. 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.

19. jtvatsim

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

20. jtvatsim

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

21. jtvatsim

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

22. jtvatsim

but however you want to write it :)

23. jcaruso

ok i tired that final answer and i got it incorrect

24. jtvatsim

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

25. jtvatsim

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

26. jcaruso

i got it correct thanks so much for your help

27. jtvatsim

no problem, good luck on eigenvectors! :)

28. 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 $\large S=\begin{pmatrix} 1 & 1\\ -1 & 1 \end{pmatrix}\qquad\text{and}\qquad \Lambda=\begin{pmatrix} 1/2 & 0\\ 0 & 2 \end{pmatrix}$

29. helder_edwin

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