# Diagonalizable matrix

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

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 dont know how to get S^n or S^-n

any ideas?

Last edited:

0rthodontist
UrbanXrisis said:
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?

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

what i dont 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
Homework Helper
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
Homework Helper
UrbanXrisis said:
AKG helped me out and I realized that $$M^n=S D^n S^{-1}$$

what i dont 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.

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?

Last edited:
AKG
Homework Helper
UrbanXrisis said:
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:

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

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 dependant, 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