find power of matrix using matrix diagonal

Question

there is matrix $ B=\left(\begin{array}{ccc}1 & -1 \\ 1 & 3\\ \end{array}\right)$ ,

for vector $ v_1=\left(\begin{array}{ccc}1 \\ -1\\ \end{array}\right), B v_1=2v_1$ holds.
using $v_1$ and $v_2=\left(\begin{array}{ccc}a \\ b\\ \end{array}\right) $ which is a linear independent vector of magnitude 1, we can express $B.v_2$ as linear combination of $v_1$ & $v_2$ such as $B.v_2=\alpha v_1+2v_2$.
a.) find $v_2$ & $\alpha$

b.) $P=\left(\begin{array}{ccc}1 & a \\ -1 & b\\ \end{array}\right)$, then we can express $BP=PC$, find matrix C, and find $B^n$

first try to find $v_2$ & $\alpha$, $(B-2E)v_2=v_1$ $ B=[\left(\begin{array}{ccc}1 & -1 \\ 1 & 3\\ \end{array}\right)$ - $ \left(\begin{array}{ccc}2 & 0 \\ 0 & 2\\ \end{array}\right) ] .v_2 =\left(\begin{array}{ccc}1 \\ -1\\ \end{array}\right)$
got $v_2=\left(\begin{array}{ccc}-b-1 \\ b \end{array}\right)$ = $\left(\begin{array}{ccc}-1 \\ 1\\ \end{array}\right)s + \left(\begin{array}{ccc}-1 \\ 0\\ \end{array}\right)t$, since $v_2$ is vector with magnitude 1 , i chose $ \left(\begin{array}{ccc}-1 \\ 0\\ \end{array}\right)$ with $\alpha =1$, is my answer right?
then find matrix $C$ , i know matrix $C$ is matrix diagonal, $P^{-1}BP=C$, so with $P=\left(\begin{array}{ccc}1 & -1\\ -1 & 0\\ \end{array}\right)$ , after some calculation $ C=\left(\begin{array}{ccc}2 & 0 \\ 1 & 2\\ \end{array}\right)$

But i wasnt so sure to find $B^n$ ?

Answer 1

Credit: José Carlos Santos for pointing out that $C= \begin{bmatrix} 2 & 1 \\ 0 & 2\end{bmatrix}$

Guide:

$$P^{-1}BP=C$$

$$P^{-1}B^nP=C^n$$

$$C=\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$$

$$C^2 = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}=\begin{bmatrix} 2^2 & 2(2) \\ 0 & 2^2 \end{bmatrix}$$

$$C^3 = \begin{bmatrix} 8 & 0 \\ 12 & 8 \end{bmatrix}=\begin{bmatrix} 2^3 & 3(2^2) \\ 0 & 2^3 \end{bmatrix}$$

Compute a few more terms, spot a pattern and prove the pattern of $C^n$ perhaps by induction.

Edit:

$\begin{bmatrix} a \\ b\end{bmatrix}= \begin{bmatrix} -1 \\ 1\end{bmatrix} b + \begin{bmatrix} -1 \\ 0\end{bmatrix}$

Note that there is no $t$ at the end.

Since you were solving $$(B-2I)v_2 = v_1$$

You have chosen $\alpha = 1$.

If you want to impose the condition that $v_2$ is of length $1$.

You just have to solve for $b$ in $\left\|\begin{bmatrix} -1 \\ 1\end{bmatrix} b + \begin{bmatrix} -1 \\ 0\end{bmatrix}\right\|^2=1$, no guessing is required.

Answer 2

It looks like this exercise is taking you step-by-step through calculating what’s known as the Jordan normal form of $B$ (in this exercise, the matrix $C$) and then using that to compute $B^n=(PCP^{-1})^n=PC^nP^{-1}$. It’s not too hard to work out the pattern for $C^n$, as illustrated in other answers, but you can save yourself some guesswork by decomposing $C$ into the sum of the diagonal matrix $2I$ and a nilpotent matrix $N$, i.e., $$C=\begin{bmatrix}2&1\\0&2\end{bmatrix} = \begin{bmatrix}2&0\\0&2\end{bmatrix}+\begin{bmatrix}0&1\\0&0\end{bmatrix} = 2I+N.$$ Now, use the Binomial theorem and the fact that $2I$ and $N$ commute to expand $$(2I+N)^n = 2^nI+n2^{n-1}N+\binom n 2 2^{n-2}N^2 + \cdots$$ but $N^2=0$, as you can verify for yourself, therefore $$C^n = 2^nI+n2^{n-1}N = \begin{bmatrix}2^n & n2^{n-1} \\ 0 & 2^n\end{bmatrix}.$$

In fact, you can use a similar decomposition to compute $B^n$ more directly, without finding the matrix $P$ first. We have $$B = PCP^{-1} = P(2I+N)P^{-1} = 2PIP^{-1}+PNP^{-1} = 2I+PNP^{-1}.$$ Now, $$(PNP^{-1})^2 = PNP^{-1}PNP^{-1} = PN^2P^{-1} = 0,$$ so by a similar computation to that above we get $$B^n = 2^nI + n2^{n-1}(B-2I) = \begin{bmatrix}2^n&0\\0&2^n\end{bmatrix}+\begin{bmatrix}-n2^{n-1}&-n2^{n-1}\\n2^{n-1}&n2^{n-1}\end{bmatrix} = \begin{bmatrix}2^n-n2^{n-1} & -n2^{n-1} \\ n2^{n-1} & 2^n+n2^{n-1}\end{bmatrix}.$$ This is a lot less work than working out generalized eigenvectors of $B$ (the matrix $P$ in this exercise) and then doing all of those matrix multiplications. You can use this method to compute powers and exponentials of any $2\times2$ matrix with a repeated eigenvalue that isn’t a multiple of the identity (and the method can be generalized to some extent to larger matrices).