Question

Find a formula in terms of k for the entries of A^k, where A is the diagonalizable matrix below, and the eigenvalues of A are −1 and 2.
A = -10 18
-6 11

Answer 1

P = 3 2
2 1
P^-1 = -1 2
2 -3
Then what?

Answer 2

Now you have the eigenvalues, so depending on how you set up P and P inverse you'll have this:

\[A=P^{-1}DP\]
So any power of A, such as \(A^2\) will be
\[A^2=P^{-1}DPP^{-1}DP=P^{-1}D^2P\]
Notice on the inside there we have P and its inverse, so we just have the identity! \(PP^{-1}=I\)

This will work for larger powers just as well,
\[A^k=P^{-1}D^k P\]

What are the entries of D? They're the eigenvalues of A, check this to make sure, and then you can see what happens when you raise a diagonal matrix to a power.

Answer 3

so D = -1 0
0 -1
or is it
D = 2 0
0 2

Answer 4

Neither, it will be either diag(-1, 2) or diag(2, -1) depending on which eigenvalue is which, really you're writing the eigenvalue equation in this new way:

\[A v_1 = \lambda_1 v_1\]\[A v_2 = \lambda_2v_2\]
Combine them so that these column vectors fill a 2x2 matrix: (work out the matrix multiplication and try to understand as best you can, tell me where/if you get stuck:)
\[A\begin{bmatrix}
v_1 & v_2
\end{bmatrix} = \begin{bmatrix}
v_1 & v_2
\end{bmatrix}\begin{pmatrix}
\lambda_1 &0 \\0
&\lambda_2
\end{pmatrix}\]

This matrix of eigenvectors are our P or \(P^{-1}\) matrix (depending on what you call P) and can be inverted:

\[A\begin{bmatrix}
v_1 & v_2
\end{bmatrix} = \begin{bmatrix}
v_1 & v_2
\end{bmatrix}\begin{pmatrix}
\lambda_1 &0 \\0
&\lambda_2
\end{pmatrix}\]
Multiply on the right by the inverse,
\[A\begin{bmatrix}
v_1 & v_2
\end{bmatrix}\begin{bmatrix}
v_1 & v_2
\end{bmatrix}^{-1} = \begin{bmatrix}
v_1 & v_2
\end{bmatrix}\begin{pmatrix}
\lambda_1 &0 \\0
&\lambda_2
\end{pmatrix}\begin{bmatrix}
v_1 & v_2
\end{bmatrix}^{-1}\]
\[A= \begin{bmatrix}
v_1 & v_2
\end{bmatrix}\begin{pmatrix}
\lambda_1 &0 \\0
&\lambda_2
\end{pmatrix}\begin{bmatrix}
v_1 & v_2
\end{bmatrix}^{-1}\]

Answer 5

So, basically in order to find D, you have to follow this formula each time
D =P^-1AP

Answer 6

.

Answer 7

Yeah, basically, but it's really just lumping all the eigenvalue/eigenvector formulas of the form:

\[Ax_1 = \lambda_1x_1\]\[Ax_2 = \lambda_2x_2\]\[\cdots\]

into just one matrix equation where the matrix P has all the eigenvectors and the diagonal has all the eigenvalues
\[AP=PD\]