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Say that we have a square matrix $M$, and that a non-zero vector $v$ can be an eigenvector of $M$ if $Mv = kv$ for some real number $k$. This real number, $k$, can be called the eigenvalue of $v$ with respect to $M$.

(a) Let v be an eigenvector of matrix $M$, with eigenvalue $k$. What is a simple expression for $M^nv$, and where $n$ is a positive integer?

(b)Set $A = \begin{pmatrix} 3 & -2 & 3 \\ 1 & 2 & 1 \\ 1 & 3 & 0 \end{pmatrix}.$

Using this information, show that $v_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad v_2 = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}, \quad v_3 = \begin{pmatrix} 11 \\ 1 \\ -14 \end{pmatrix}$ are $A$'s eigenvectors, and for each of these, find what their corresponding eigenvalue is.

(c) Show that any 3D vector $v$ can be expressed as a linear combination of $v_1$, $v_2$, and $v_3$.

(d) From the previous three parts above, demonstrate how $A^nv$ can be determined for any positive integer $n$ and any 3D vector $v$. Specifically, find the value of $A^{10} \begin{pmatrix} 12 \\ 2 \\ -13 \end{pmatrix}.$


Since it seems like each part refers to the one above/previous to it, I am starting with what I think I know about part a), and then try to tackle the rest of the parts one at a time.

For part a): So if "$v$ is an eigenvector of matrix $M$, with eigenvalue $k$, then $Mv=kv$ from what is stated in the original problem statement. I'm not sure what this would mean for $M^{n}v$, however. I am puzzled.

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2 Answers 2

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Note that, since $Mv=kv$, then $$ M^2v=M(Mv)=M(kv)=kMv=k^2v. $$ Try to take it from here.

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  • $\begingroup$ I believe I know how to solve part b): Plug in the set value for A and each of the $v_1, v_2, v_3$ values separately into the $Mv=kv$ equation. From there, I can find $k$. I have no clue what to do for parts c) and d), though. I would appreciate any hints you may have. $\endgroup$
    – Grace
    Jul 23, 2015 at 22:30
  • $\begingroup$ For part c), you should have been taught some way to determine that a set of vectors is linearly independent, and what a basis is. If you don't know that, I would venture to say that you are way over your head with this exercise. $\endgroup$ Jul 23, 2015 at 22:33
  • $\begingroup$ yes, I have learned that before. $\endgroup$
    – Grace
    Jul 23, 2015 at 22:34
  • $\begingroup$ For part d), one you write $v=c_1v_1+c_2v_2+c_3v_3$, you have $Av=c_1\lambda_1v_1+c_2\lambda_2v_2+c_3\lambda_3v_3$, and so $$A^nv=Av=c_1\lambda_1^nv_1+c_2\lambda_2^nv_2+c_3\lambda_3^nv_3.$$ $\endgroup$ Jul 23, 2015 at 22:34
  • $\begingroup$ Thanks, I think I can solve them both now. :) $\endgroup$
    – Grace
    Jul 23, 2015 at 22:36
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For part (a), you know that $$Mv = kv$$ since $v$ is an eigenvector of $M$ with a corresponding eigenvalue of $k$.

Try applying $M$ to both sides of this equation. The properties of the matrix $M$ will help you transform and simplify the result.

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  • $\begingroup$ Thank you--I have solved part a) and think I know how to solve part b). I'm completely clueless on parts c) and d)--if you have any hints, that would be great. $\endgroup$
    – Grace
    Jul 23, 2015 at 22:31

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