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OpenStudy (decentnabeel):
OpenStudy (fwizbang):
Z is a column vector.
\[Z=\left(\begin{matrix}3 \\ 5\end{matrix}\right)\]
The line is a stray mark on he paper not a fraction. Try multiplying
Z by A once and compare the answer to Z. What happens?
OpenStudy (decentnabeel):
A ^n multiple by Z
OpenStudy (decentnabeel):
so how to find n
OpenStudy (fwizbang):
I know, humor me....
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OpenStudy (fwizbang):
If YOU multiply AZ, what do you get?
OpenStudy (fwizbang):
Oops, make that if you multiply AZ, what do you get?
Ok, now I see what's happening. We need to write Z as a linear combination of X and Y(which are eigenvectors of A.) So, find an a and a b such that Z=aX+bY
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OpenStudy (decentnabeel):
yes
OpenStudy (fwizbang):
Then if we now the eigenvalues, we can write the answer for a general integer n. If\[AX=\lambda _{x}X\] then \[A^nX=(\lambda _{x})^n X\]
OpenStudy (fwizbang):
Same idea for Y, so do yu know the eigenvalues for X and Y?
OpenStudy (decentnabeel):
no
OpenStudy (fwizbang):
Ok. Multiply X by A and compare the answer to X. Same for Y.
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OpenStudy (fwizbang):
It looks like you already did this on your sheet....
OpenStudy (decentnabeel):
Multiply X by A
\[\left(\begin{matrix}2 \\ -2\end{matrix}\right)\]