How can I find the formula for M^n if M is a diagonalizable matrix?

[UrbanXrisis]

M = \left(\begin{array}{cc}4&-1 \\ 2&7 \end{array}\right)

I need to show M^n as a formula of entries where n>0:

so say M = \left(\begin{array}{cc}4&-1 \\ 2&7 \end{array}\right) is diagonalizable: M = SDS^{-1}

then M^2=S^2 D^2 S^{-2}

and... M^3=S^3 D^3 S^{-3}

I can see from this that D^n = \left(\begin{array}{cc}\alpha^n&0 \\ 0&\beta^2 \end{array}\right) assuming \alpha= \lambda_1 and that \beta= \lambda_2

\alpha= 5 and \beta= 6

D^n = \left(\begin{array}{cc}5^n&0 \\ 0&6^n \end{array}\right)

so to find S...

\left(\begin{array}{cc}4-\alpha&-1 \\ 2&7-\beta \end{array}\right)
\left(\begin{array}{cc}-1&-1 \\ 2&2 \end{array}\right)

v_1= \left(\begin{array}{c}1\\ -1 \end{array}\right)

\left(\begin{array}{cc}2&1 \\ 2&1 \end{array}\right)

v_2= \left(\begin{array}{c}1\\ -2 \end{array}\right)

S = \left(\begin{array}{cc}1&1 \\-1&-2 \end{array}\right)

det(S)=-1

S^{-1} = \left(\begin{array}{cc}-1&-1 \\1&2 \end{array}\right)

this is where I am stuck, i don't know how to get S^n or S^-n

any ideas?

 

[0rthodontist]

then M^2=S^2 D^2 S^{-2}

and... M^3=S^3 D^3 S^{-3}

Not true.
M = SDS^-1.
MM = SDS^-1SDS^-1
How does this simplify?

 

[UrbanXrisis]

AKG helped me out and I realized that M^n=S D^n S^{-1}

what i don't understand now is now a diagonal matrix can be multiplied with other matricies to get a matrix with non-zero numbers?

D^n = \left(\begin{array}{cc}5^n&0 \\ 0&6^n \end{array}\right)

so D multiplied by some other matrix will always be \left(\begin{array}{cc}a&0 \\ 0&b \end{array}\right)

so how does this produce the original matrix M = \left(\begin{array}{cc}4&-1 \\ 2&7 \end{array}\right)?

 

[AKG]

Multiply D with the matrix full of 1's. It has no zeroes. Anyways, we want to know how to find an invertible matrix S such that M = SDS^-1. You were on the right track to finding S in that other thread.

 

[HallsofIvy]

AKG helped me out and I realized that M^n=S D^n S^{-1}

what i don't understand now is now a diagonal matrix can be multiplied with other matricies to get a matrix with non-zero numbers?

D^n = \left(\begin{array}{cc}5^n&0 \\ 0&6^n \end{array}\right)

so D multiplied by some other matrix will always be \left(\begin{array}{cc}a&0 \\ 0&b \end{array}\right)

No, that's not true! Did you actually try it?

so how does this produce the original matrix M = \left(\begin{array}{cc}4&-1 \\ 2&7 \end{array}\right)?

How did you get D? Wasn't it D= SMS^-1? So then M= S^-1DS.

You saw, of course, that the numbers on the diagonal of D are the eigenvalues of M. A standard way of finding S is to use the corresponding eigenvectors as columns of S.

 

[UrbanXrisis]

Lets give it a try:

M = \left(\begin{array}{cc}4&-1 \\ 2&7 \end{array}\right)

M^n=S D^n S^{-1}

Find the eigenvalues:
\left(\begin{array}{cc}4-\lambda&-1 \\ 2&7-\lambda \end{array}\right)

\lambda_1=5, \lambda_2=6

This mean:

D = \left(\begin{array}{cc}5&0 \\ 0&6 \end{array}\right)

D^n = \left(\begin{array}{cc}5^n&0 \\ 0&6^n \end{array}\right)

To find the eigenvectors:
\left(\begin{array}{cc}4-\lambda_1&-1 \\ 2&7-\lambda _1 \end{array}\right)

\left(\begin{array}{cc}4-5&-1 \\ 2&7-5 \end{array}\right)

\left(\begin{array}{cc}-1&-1 \\ 2&2 \end{array}\right)

Solving the matrix:
\left(\begin{array}{ccc}-1&-1&0\\ 2&2 &0 \end{array}\right)

v_1=\left(\begin{array}{c}1\\-1 \end{array}\right)

\left(\begin{array}{cc}4-\lambda_2&-1 \\ 2&7-\lambda_2 \end{array}\right)

\left(\begin{array}{cc}4-6&-1 \\ 2&7-6 \end{array}\right)

\left(\begin{array}{cc}-2&-1 \\ 2&1 \end{array}\right)

Solving the matrix:
\left(\begin{array}{ccc}-2&-1&0\\ 2&1 &0 \end{array}\right)

v_2=\left(\begin{array}{c}1\\-2 \end{array}\right)

The eigenspace is then:
S=\left(\begin{array}{cc}1&1 \\ -1&-2 \end{array}\right)

det\left(\begin{array}{cc}1&1 \\ -1&-2 \end{array}\right)=-1

S^{-1}=\left(\begin{array}{cc}-1&-1 \\ 1&2 \end{array}\right)

So this means that:
M= \left(\begin{array}{cc}1&1 \\ -1&-2 \end{array}\right)\left(\begin{array}{cc}5&0 \\ 0&6 \end{array}\right) \left(\begin{array}{cc}-1&-1 \\ 1&2 \end{array}\right)

M^n= \left(\begin{array}{cc}1&1 \\ -1&-2 \end{array}\right)\left(\begin{array}{cc}5^n&0 \\ 0&6^n \end{array}\right) \left(\begin{array}{cc}-1&-1 \\ 1&2 \end{array}\right)

When I solve this out.. I get:

M^n= \left(\begin{array}{cc}-5^n+6^n&-5^n+2(6^n) \\ 5^n-2(6^n)&5^n-4(6^n) \end{array}\right)

When I sub in n=1, I do not get back the original matrix... could someone help?

 

[AKG]

Solving the matrix:
\left(\begin{array}{cc}-1&-1&0\\ 2&2 &0 \end{array}\right)

You don't solve a matrix, you solve an equation, and the equation you want to solve is:

\left(\begin{array}{cc}-1&-1\\ 2&2 \end{array}\right)v_1 = \left (\begin{array}{c}0\\0\end{array}\right)

for v₁, which coincidentally you have:

v_1=\left(\begin{array}{c}1\\-1 \end{array}\right)

Next

Solving the matrix:
\left(\begin{array}{cc}-2&-1&0\\ 2&1 &0 \end{array}\right)
v_2=\left(\begin{array}{c}1\\-2 \end{array}\right)

Again, you're solving the equation:

\left(\begin{array}{cc}-2&-1\\ 2&1 \end{array}\right)v_2 = \left(\begin{array}{c}0\\0\end{array}\right )

which you have already done.

The eigenspace is then:
S=\left(\begin{array}{cc}1&-1 \\ 2&-2 \end{array}\right)

Just as a matrix is not something you solve, a matrix is not an eigenspace either. Moreover, it's not clear as to how you got this matrix. You're supposed to let S be the matrix whose rows (or columns, I can't remember) are the eigenvectors you found. You had vectors (1 -1)^T and (1 -2)^T, so I don't know where you're getting this S from.

det\left(\begin{array}{cc}-1&-1 \\ 2&2 \end{array}\right)=-1

First of all, this matrix you're finding the determinat of is not the S you wrote down just a few lines above, and moreover it still isn't a matrix whose rows/columns are eigenvectors. It's yet another unexplained matrix. Thirdly, the determinant of that matrix there isn't -1, it's 0. Just look at the columns, they're the exact same, so the columns are clearly linearly dependent, so the determinant is 0. Of course, a very simply computation: (-1)(2) - (-1)(2) = 0 shows this as well. The original matrix for S that you had also has determinant 0, since its second row is just 2x the first, so they rows are dependent, hence determinant is 0. Again, a computation shows this: 1(-2) - (2)(-1) = 0.

S^{-1}=\left(\begin{array}{cc}-1&1 \\ -2&2 \end{array}\right)

No matter which of the two matrices you're talking about, this is wrong since neither of those matrices are invertible (both have det 0). And I have no idea how you came up with this answer. I mean, if you did the thing you learned from high school: Take S, switch the things in positions a and d, replace c and d with their negatives, then divide the whole thing by det(S) [which you thought to be -1], you still wouldn't get what you have above, no matter which of the two matrices were taken to be S.

 

[AKG]

Okay, it looks as though you've edited your post. Now everything's right up to the point where you find S^-1 (well, you still don't solve matrices or call them eigenspaces, but anyways...). It looks like all you've done is taken S and divided by det(S) [which is indeed -1].

 

[UrbanXrisis]

sorry... still not so good with LaTeX.

 

[UrbanXrisis]

thanks for catching that mistake and then help AKG, I've got it figured out!