Proof of a matrix

Question

Suppose $ M= \begin{pmatrix} 5 & -1 \\ 4 & 1 \end{pmatrix} $. Prove that $\forall n\in \mathbb N $ the following equation holds :

$M^n = 3^{n-1} \begin{pmatrix} 2n+3 & -n\\ 4n & 3- 2n \end{pmatrix}$

I've already proven this as follows:

$\underline{n=0:}$ is trivial just as ${n=1}$ which is given.

$\underline{I.H.:}$ (n=k) Suppose the above statement is true $\forall n \in \mathbb N. $

$\underline{n=k+1:}$

$\begin{align}M^{k+1}& =^{I.H.} 3^{k-1}\begin{pmatrix}2k+3 & -k\\ 4k&3-2k \end{pmatrix} \cdot 3 \begin{pmatrix}5 & -1 \\ 4 & 1 \end{pmatrix}\\ &= 3^{k-1} \begin{pmatrix} 6k+15 & -3k-3\\ 12k+12 & -6k+3 \end{pmatrix}\\ & = 3^k \begin{pmatrix} 2k+5 & -k-1\\ 4k+4 & -2k+1 \end{pmatrix}\\ & = 3^n \begin{pmatrix} 2n+3 & -n\\ 4n & -2n+3 \end{pmatrix} \end{align} $

My question now is how do I determine a matrix $P^2= M$?

What I've tried is the following:

$\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix} \cdot \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix} = \begin{pmatrix} 5 & -1\\ 4 & 1\end{pmatrix} $

Now I got stuck on this system of equations. I tried using other variables such as x, y, z and u to make it clearer for me but nevertheless I still didn't solve this problem. Would someone help me with this problem please? And maybe check my proof above?

Answer 1

Your proof is somewhat confused. First of all the induction hypothesis in not that the statement is true for all $n$, nothing good can come from such an assumption. Instead you assume that it's true for a specific $n$ and show that it's true for the next integer $n+1$.

Then the next question about finding $P$ such that $P^2=M$ the identity that you've shown is a clue here. Insert $n=1/2$ into the formula. Of course your proof has not shown it to work for non-integers, but let's give it a try:

$$\left(3^{1/2-1}\begin{pmatrix} 2 (1/2) + 3 & -(1/2) \\ 4(1/2) & 3 -2(1/2) \\ \end{pmatrix}\right)^2 \\= {1\over3}\begin{pmatrix} 4 & -1 \\ 2 & 2 \end{pmatrix}^2\\ = \begin{pmatrix} 5 & -1 \\ 4 & 1\end{pmatrix} = M$$

Well it worked out - and that's all you need for an answer (no reason to justify a "lucky" guess).

Answer 2

The problem that I see (beyond the typos) is that the inductive step is never related to $M^{k+1}$. What is done is that you've shown how to make the necessary simplification, but not quite how to get there (how is what you write related to $M$)?

For the inductive hypothesis:

Assume that the statement is true when $n=k$. In other words: $$ M^k=3^{k-1}\begin{bmatrix}2k+3&-k\\4k&3-2k\end{bmatrix}. $$

Multiplying both sides by $M$ gives \begin{align} M^{k+1}&=3^{k-1}\begin{bmatrix}2k+3&-k\\4k&3-2k\end{bmatrix}\begin{bmatrix}5&-1\\4&1\end{bmatrix}\\ &=3^{k-1}\begin{bmatrix}6k+15&-3k-3\\12k+12&3-6k\end{bmatrix}\\ &=3^k\begin{bmatrix}2k+5&-k-1\\4k+4&1-2k\end{bmatrix}\\ &=3^{(k+1)-1}\begin{bmatrix}2(k+1)+3&-(k+1)\\4(k+1)&3-2(k+1)\end{bmatrix} \end{align} which is of the desired form. The work above starts somewhere in the middle of the final computation, but doesn't relate the computation to $M^{k+1}$.

Answer 3

For the question about $P^2=M$, one way to solve it would be to use the Jordan form:

$$ M=\begin{bmatrix}1&.5\\2&0\end{bmatrix}\begin{bmatrix}3&1\\0&3\end{bmatrix}\begin{bmatrix}0&.5\\2&-1\end{bmatrix} $$

Then, it is enough to find a matrix whose square is $\begin{bmatrix}3&1\\0&3\end{bmatrix}$, this can be done by hand to find $$\begin{bmatrix}\sqrt{3}&\frac{1}{2\sqrt{3}}\\0&\sqrt{3}\end{bmatrix}$$

Then $$ P=\begin{bmatrix}1&.5\\2&0\end{bmatrix}\begin{bmatrix}\sqrt{3}&\frac{1}{2\sqrt{3}}\\0&\sqrt{3}\end{bmatrix}\begin{bmatrix}0&.5\\2&-1\end{bmatrix} $$