General formula for multiplying a matrix by a power

sktsasus

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"Let M = \begin{pmatrix}2&2\\ -4&8\end{pmatrix}


Find formulas for the entries of M^n, where n is a positive integer."

So I think this problem involves multiplying a matrix by itself continuously. But I do not think I have observed a link between the older entries and the newer entries each time I have multiplied this matrix by itself. It just seems to be random changes in the numbers (unless I am doing something wrong). Am I supposed to create a sequence out of each of the entries and then derive a general formula?

Also, I have managed to find the eigenvalues (4 and 6) and eigenvectors, but I am not too sure if they are relevant here. If they are relevant, how exactly would I apply them here?


Any assistance would be highly appreciated!
 
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First, you MAY wish to simplify your life. You have common factors all about.

Second, about eigenvalues. We have the result, If \(\displaystyle \lambda\) is an eigenvalue of Matrix \(\displaystyle A\), then for s, a Natural Number, \(\displaystyle \lambda^{s}\) is an eigenvalue of \(\displaystyle A^{s}\). What does that do for us?
 
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First, simplify your life. You have common factors all about. Please factor that out.

Thank you for your reply! But I am sorry I do not understand what you are referring to when you say common factors. There does not seem to be anything I have written above that seems to require factoring. I have not written an algebraic equation or something along those lines.

I apologize for my ignorance, but could you please clarify what you are referring to.

Edit: Nevermind. I think you are referring to the numbers in the matrix.

Thank you!
 
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First, you MAY wish to simplify your life. You have common factors all about.

Second, about eigenvalues. We have the result, If \(\displaystyle \lambda\) is an eigenvalue of Matrix \(\displaystyle A\), then for s, a Natural Number, \(\displaystyle \lambda^{s}\) is an eigenvalue of \(\displaystyle A^{s}\). What does that do for us?

I see. So would that mean the eigenvalues of the new M^n matrix will be 4^n and 6^n? If so, do I apply this formula to the entries of the initial matrix?

Thank you!
 
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One very easy way to get a formula for the nth power of a matrix is to use diagonalization. Are you familiar with that? The first step is to find the eigenvectors of your matrix. For now we'll call them \(\displaystyle \begin{pmatrix}x_1\\ y_1\end{pmatrix}\) and \(\displaystyle \begin{pmatrix}x_2\\ y_2\end{pmatrix}\). Next, define a new matrix P by:

\(\displaystyle P=\begin{pmatrix}x_1&x_2\\ y_1&y_2\end{pmatrix}\)

Define another new matrix B by:

\(\displaystyle B = P^{-1}AP\)

This matrix will be a diagonal matrix, meaning it's of the form:

\(\displaystyle B=\begin{pmatrix}\alpha &0\\ 0&\beta \end{pmatrix}\)

Which has very nice properties. In particular, with regard to nth powers, we have:

\(\displaystyle B^n=\begin{pmatrix}\alpha^n &0\\ 0&\beta^n \end{pmatrix}\)

Finally, since we have \(\displaystyle B = P^{-1}AP\), we can left-multiply by P and get:

\(\displaystyle PB = PP^{-1}AP \implies PB = AP\)

Then we can right-multiply by P-1 to get:

\(\displaystyle PBP^{-1} = APP^{-1} \implies A = PBP^{-1}\)

And hence:

\(\displaystyle A^n = PB^nP^{-1}\)

See how far that gets you.
 
One very easy way to get a formula for the nth power of a matrix is to use diagonalization. Are you familiar with that? The first step is to find the eigenvectors of your matrix. For now we'll call them \(\displaystyle \begin{pmatrix}x_1\\ y_1\end{pmatrix}\) and \(\displaystyle \begin{pmatrix}x_2\\ y_2\end{pmatrix}\). Next, define a new matrix P by:

\(\displaystyle P=\begin{pmatrix}x_1&x_2\\ y_1&y_2\end{pmatrix}\)

Define another new matrix B by:

\(\displaystyle B = P^{-1}AP\)

This matrix will be a diagonal matrix, meaning it's of the form:

\(\displaystyle B=\begin{pmatrix}\alpha &0\\ 0&\beta \end{pmatrix}\)

Which has very nice properties. In particular, with regard to nth powers, we have:

\(\displaystyle B^n=\begin{pmatrix}\alpha^n &0\\ 0&\beta^n \end{pmatrix}\)

Finally, since we have \(\displaystyle B = P^{-1}AP\), we can left-multiply by P and get:

\(\displaystyle PB = PP^{-1}AP \implies PB = AP\)

Then we can right-multiply by P-1 to get:

\(\displaystyle PBP^{-1} = APP^{-1} \implies A = PBP^{-1}\)

And hence:

\(\displaystyle A^n = PB^nP^{-1}\)

See how far that gets you.

Thank you very much for your helpful reply!

I did what you said and got the matrix half correct. Two of my entries still seem to be incorrect. Here is what I did:

P = \begin{pmatrix}\frac{1}{2}&1\\ \:\:1&1\end{pmatrix} (I combined both eigenvectors)

P^-1 = \begin{pmatrix}-2&2\\ \:\:2&-1\end{pmatrix}

B = \begin{pmatrix}6&0\\ \:0&4\end{pmatrix}

B^n = \begin{pmatrix}6^n&0\\ \:0&4^n\end{pmatrix}

A^n = \begin{pmatrix}-2^n\cdot \:3^n+2^{2n+1}&-3^n\cdot \:2^{n+1}+2^{2n+1}\\ 2^n\cdot \:3^n-4^n&3^n\cdot \:2^{n+1}-4^n\end{pmatrix}

My top left and bottom right entries seem to be correct but my top right and bottom left entries still appear to be wrong. I have doubled checked and can't seem to find anything.

Would you happen to know why?

Thank you!
 
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My top left and bottom right entries seem to be correct but my top right and bottom left entries still appear to be wrong. I have doubled checked and can't seem to find anything.

Would you happen to know why?

Thank you!

Your process looks right to me, but you got one of the eigenvectors wrong. The eigenvector corresponding to \(\displaystyle \lambda_2 = 4\) is indeed \(\displaystyle \begin{pmatrix}1\\ 1\end{pmatrix}\), but the eigenvector for \(\displaystyle \lambda_1 = 6\) is wrong. Recall that to find the eigenvector, we need to solve \(\displaystyle (A-6I)x=0\) for x. This can be done by forming an augmented matrix and row reducing it. What did you get for this? Perhaps you made an error in this step?
 
Your process looks right to me, but you got one of the eigenvectors wrong. The eigenvector corresponding to \(\displaystyle \lambda_2 = 4\) is indeed \(\displaystyle \begin{pmatrix}1\\ 1\end{pmatrix}\), but the eigenvector for \(\displaystyle \lambda_1 = 6\) is wrong. Recall that to find the eigenvector, we need to solve \(\displaystyle (A-6I)x=0\) for x. This can be done by forming an augmented matrix and row reducing it. What did you get for this? Perhaps you made an error in this step?

Yes, perhaps I made an error in that step. But I still keep getting the same eigenvector.

This is what I did for the eigenvalue of 6:

\begin{pmatrix}2&2\\ \:-4&8\end{pmatrix} -6\begin{pmatrix}1&0\\ 0&1\end{pmatrix} = \begin{pmatrix}-4&2\\ -4&2\end{pmatrix}

I then reduced that matrix and got \begin{pmatrix}1&-\frac{1}{2}\\ 0&0\end{pmatrix}.

This then leads to an eigenvector of \begin{pmatrix}\frac{1}{2}\\ 1\end{pmatrix}.

So I am not entirely sure where I have went wrong here. Perhaps it was in the reduction but I can't see why.

Note: Sorry it is a little unclear but I multiplied the identity matrix by 6. I subtracted the initial matrix from the 6I.
 
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Ah, okay. I see now. Your row reduction is spot on, but the conclusion is not. Let's call the two components of the eigenvector x and y. The matrix

\(\displaystyle \begin{pmatrix}1&-\frac{1}{2}\\ 0&0\end{pmatrix}\)

gives us the equation \(\displaystyle x-\dfrac{1}{2}y = 0\). Rearranging that gives \(\displaystyle x=\dfrac{1}{2}y\). It's customary when finding eigenvectors to ensure that all of the components are integers and pick the smallest such vector. y cannot be 1, because that would make x = 1/2, so y must be 2. Thus, the eigenvector must be \(\displaystyle \begin{pmatrix}1\\ 2\end{pmatrix}\).
 
Ah, okay. I see now. Your row reduction is spot on, but the conclusion is not. Let's call the two components of the eigenvector x and y. The matrix

\(\displaystyle \begin{pmatrix}1&-\frac{1}{2}\\ 0&0\end{pmatrix}\)

gives us the equation \(\displaystyle x-\dfrac{1}{2}y = 0\). Rearranging that gives \(\displaystyle x=\dfrac{1}{2}y\). It's customary when finding eigenvectors to ensure that all of the components are integers and pick the smallest such vector. y cannot be 1, because that would make x = 1/2, so y must be 2. Thus, the eigenvector must be \(\displaystyle \begin{pmatrix}1\\ 2\end{pmatrix}\).

OK, that makes sense now. Thank you for your help!
 
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