# Diagonalize a 2 by 2 Matrix $A$ and Calculate the Power $A^{100}$

## Problem 466

Let
$A=\begin{bmatrix} 1 & 2\\ 4& 3 \end{bmatrix}.$

(a) Find eigenvalues of the matrix $A$.

(b) Find eigenvectors for each eigenvalue of $A$.

(c) Diagonalize the matrix $A$. That is, find an invertible matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

(d) Diagonalize the matrix $A^3-5A^2+3A+I$, where $I$ is the $2\times 2$ identity matrix.

(e) Calculate $A^{100}$. (You do not have to compute $5^{100}$.)

(f) Calculate
$(A^3-5A^2+3A+I)^{100}.$ Let $w=2^{100}$. Express the solution in terms of $w$.

## Solution.

### (a) Find eigenvalues of the matrix $A$.

To find the eigenvalues of $A$, we calculate the characteristic polynomial $p(t)$ as follows.
We have
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
1-t & 2\\
4& 3-t
\end{vmatrix}\\
&=(1-t)(3-t)-8=t^2-4t-5=(t+1)(t-5).
\end{align*}

The eigenvalues of $A$ are roots of its characteristic polynomial $p(t)$.
Hence the eigenvalues of $A$ are $-1$ and $5$.

### (b) Find eigenvectors for each eigenvalue of $A$.

We first determine the eigenvectors of the eigenvalue $-1$ by solving the system $(A+I)\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A+I=\begin{bmatrix}
2 & 2\\
4& 4
\end{bmatrix}
\xrightarrow[\text{then } \frac{1}{2}R_1]{R_2-2R_1}
\begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix}.
\end{align*}
This yields that the eigenvectors corresponding to $-1$ are
$a\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ for any nonzero scalar $a$.

Next, we find the eigenvectors corresponding to the eigenvalue $5$ by solving $(A-5I)\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A-5I=\begin{bmatrix}
-4 & 2\\
4& -2
\end{bmatrix}
\xrightarrow[\text{then } \frac{-1}{4}R_1]{R_2+R_1}
\begin{bmatrix}
1 & -1/2\\
0& 0
\end{bmatrix}.
\end{align*}
It follows that the eigenvectors corresponding to $5$ are
$a\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ for any nonzero scalar $a$.

### (c) Diagonalize the matrix $A$.

From part (a) and part (b), we have seen that $A$ has eigenvalues $-1$ and $5$ with corresponding eigenvectors
$\mathbf{u}=\begin{bmatrix} 1 \\ -1 \end{bmatrix} \text{ and } \mathbf{v}=\begin{bmatrix} 1 \\ 2 \end{bmatrix}.$ (Here we chose the scalars $a$ to be $1$ but you could use any nonzero values for the scalars $a$.)

Let
$S=\begin{bmatrix} \mathbf{u} & \mathbf{v} \end{bmatrix} = \begin{bmatrix} 1 & 1\\ -1& 2 \end{bmatrix}.$ Then the general procedure of the diagonalization yields that the matrix $S$ is invertible and
$S^{-1}AS=D,$ where $D$ is the diagonal matrix given by
$D=\begin{bmatrix} -1 & 0\\ 0& 5 \end{bmatrix}.$

### (d) Diagonalize the matrix $A^3-5A^2+3A+I$.

In part (c), we obtained
$S^{-1}AS=D,$ where
$S=\begin{bmatrix} 1 & 1\\ -1& 2 \end{bmatrix} \text{ and } D=\begin{bmatrix} -1 & 0\\ 0& 5 \end{bmatrix}.$

Note that we have $A=SDS^{-1}$ and
\begin{align*}
A^2&=AA=SDS^{-1}\cdot SDS^{-1}=SD^2S^{-1}\\
A^3&=A^2A=SD^2S^{-1}\cdot SDS^{-1}=SD^3S^{-1}.
\end{align*}

These relations gives
\begin{align*}
A^3-5A^2+3A+I&=SD^3S^{-1}-5SD^2S^{-1}+3SDS^{-1}+I\\
&=S(D^3-5D^2+3D+I)S^{-1}.
\end{align*}

Hence we obtain
\begin{align*}
&S^{-1}(A^3-5A^2+3A+I)S\\
&=D^3-5D^2+3D+I\\
&=\begin{bmatrix}
(-1)^3 & 0\\
0& 5^3
\end{bmatrix}
-5\begin{bmatrix}
(-1)^2 & 0\\
0& 5^2
\end{bmatrix}
+3\begin{bmatrix}
-1 & 0\\
0& 5
\end{bmatrix}
+\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}\6pt] &=\begin{bmatrix} -8 & 0\\ 0& 16 \end{bmatrix}. \end{align*} This completes the diagonalization of the matrix A^3-5A^2+3A+I. ### (e) Calculate A^{100}. In part (d), we have seen that A=SDS^{-1}, A^2=SD^2S^{-1}, A^3=SD^3S^{-1}. Repeating the same argument (or using mathematical induction), we also have \[A^{100}=SD^{100}S^{-1}.

Thus, we have
\begin{align*}
A^{100}&=SD^{100}S^{-1}\\
&=\begin{bmatrix}
1 & 1\\
-1& 2
\end{bmatrix}
\begin{bmatrix}
-1 & 0\\
0& 5
\end{bmatrix}^{100}
\frac{1}{3}\begin{bmatrix}
2 & -1\\
1& 1
\end{bmatrix}\6pt] &=\frac{1}{3}\begin{bmatrix} 1 & 1\\ -1& 2 \end{bmatrix} \begin{bmatrix} (-1)^{100} & 0\\ 0& 5^{100} \end{bmatrix} \begin{bmatrix} 2 & -1\\ 1& 1 \end{bmatrix}\\[6pt] &=\frac{1}{3}\begin{bmatrix} 2+5^{100} & -1+5^{100}\\ -2+2\cdot 5^{100}& 1+2\cdot 5^{100} \end{bmatrix}. \end{align*} ### (f) Calculate (A^3-5A^2+3A+I)^{100}. Let \[B:=A^3-5A^2+3A+I.

In part (d), we obtained
$S^{-1}BS=\begin{bmatrix} -8 & 0\\ 0& 16 \end{bmatrix}.$ Hence we have $B=S\begin{bmatrix} -8 & 0\\ 0& 16 \end{bmatrix} S^{-1}$, and
\begin{align*}
B^{100}&=S\begin{bmatrix}
-8 & 0\\
0& 16
\end{bmatrix}^{100} S^{-1}\6pt] &=S\begin{bmatrix} (-8)^{100} & 0\\ 0& 16^{100} \end{bmatrix} S^{-1}\\[6pt] &=S\begin{bmatrix} 2^{300} & 0\\ 0& 2^{400} \end{bmatrix} S^{-1} \\[6pt] &=S\begin{bmatrix} w^3 & 0\\ 0& w^4 \end{bmatrix} S^{-1}, \end{align*} where we put w=2^{100}. Hence we have \begin{align*} B^{100}&=\begin{bmatrix} 1 & 1\\ -1& 2 \end{bmatrix} \begin{bmatrix} w^3 & 0\\ 0& w^4 \end{bmatrix} \frac{1}{3}\begin{bmatrix} 2 & -1\\ 1& 1 \end{bmatrix}\\[6pt] &=\frac{w^3}{3}\begin{bmatrix} 2+w & -1+w\\ -2+2w& 1-2w \end{bmatrix}. \end{align*} Therefore, the result is \[(A^3-5A^2+3A+I)^{100}=\frac{w^3}{3}\begin{bmatrix} 2+w & -1+w\\ -2+2w& 1-2w \end{bmatrix}.

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1. 06/21/2017

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