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# 7.6: Diagonalization of $$2\times 2$$ matrices and Applications

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Let $$A = \begin{bmatrix} a&b\\ c&d \end{bmatrix} \in \mathbb{F}^{2\times 2}$$, and recall that we can define a linear operator $$T \in \mathcal{L}(\mathbb{F}^{2})$$ on $$\mathbb{F}^{2}$$ by setting $$T(v) = A v$$ for each $$v = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \in \mathbb{F}^2$$.

One method for finding the eigen-information of $$T$$ is to analyze the solutions of the matrix equation $$A v = \lambda v$$ for $$\lambda \in \mathbb{F}$$ and $$v \in \mathbb{F}^{2}$$. In particular, using the definition of eigenvector and eigenvalue, $$v$$ is an eigenvector associated to the eigenvalue $$\lambda$$ if and only if $$A v = T(v) = \lambda v$$.

A simpler method involves the equivalent matrix equation $$(A - \lambda I)v = 0$$, where $$I$$ denotes the identity map on $$\mathbb{F}^{2}$$. In particular, $$0 \neq v \in \mathbb{F}^{2}$$ is an eigenvector for $$T$$ associated to the eigenvalue $$\lambda \in \mathbb{F}$$ if and only if the system of linear equations

\begin{equation}
\left.
\begin{array}{rrrrr}
(a - \lambda) v_{1} & + & b v_{2} & = & 0 \\
c v_{1} & + & (d - \lambda) v_{2} & = & 0
\end{array}
\right\} \label{7.6.1}
\end{equation}

has a non-trivial solution. Moreover, System \ref{7.6.1} has a non-trivial solution if and only if the polynomial $$p(\lambda) = (a - \lambda)(d - \lambda) - bc$$ evaluates to zero. (See Proof-writing Exercise 12 in Exercises for Chapter 7.)

In other words, the eigenvalues for $$T$$ are exactly the $$\lambda \in \mathbb{F}$$ for which $$p(\lambda) = 0$$, and the eigenvectors for $$T$$ associated to an eigenvalue $$\lambda$$ are exactly the non-zero vectors $$v = \begin{bmatrix} v_{1} \\ v_{2} \end{bmatrix} \in \mathbb{F}^2$$ that satisfy System \ref{7.6.1}.

Example $$\PageIndex{1}$$

Let $$A = \begin{bmatrix} -2 & -1 \\ 5 & 2 \end{bmatrix}$$. Then $$p(\lambda) = (-2 -\lambda)(2 - \lambda) - (-1)(5) = \lambda^{2} + 1$$, which is equal to zero exactly when $$\lambda = \pm i$$. Moreover, if $$\lambda = i$$, then the System(7.6.1) becomes

$\left. \begin{array}{rrrrr} (-2 - i) v_{1} & - & v_{2} & = & 0 \\ 5 v_{1} & + & (2 - i) v_{2} & = & 0 \end{array} \right\},$

which is satisfied by any vector $$v = \begin{bmatrix} v_1\\ v_2 \end{bmatrix}\in \mathbb{C}^2$$ such that $$v_{2} = (-2 - i) v_{1}$$. Similarly, if $$\lambda = -i$$, then the System \ref{7.6.1} becomes

$\left. \begin{array}{rrrrr} (-2 + i) v_{1} & - & v_{2} & = & 0 \\ 5 v_{1} & + & (2 + i) v_{2} & = & 0 \end{array} \right\},$

which is satisfied by any vector $$v = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \in \mathbb{C}^2$$ such that $$v_{2} = (-2 + i) v_{1}$$.

It follows that, given $$A = \begin{bmatrix} -2 & -1 \\ 5 & 2 \end{bmatrix}$$, the linear operator on $$\mathbb{C}^{2}$$defined by $$T(v) = A v$$ has eigenvalues $$\lambda = \pm i$$, with associated eigenvectors as described above.

Example $$\PageIndex{2}$$

Take the rotation $$R_\theta:\mathbb{R}^2 \to \mathbb{R}^2$$ by an angle $$\theta \in [0,2\pi)$$ given by the matrix

\begin{equation*}
R_\theta = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}
\end{equation*}

Then we obtain the eigenvalues by solving the polynomial equation

\begin{equation*}
\begin{split}
p(\lambda) &= (\cos \theta -\lambda)^2 + \sin^2 \theta\\
&= \lambda^2-2\lambda \cos \theta + 1 =0,
\end{split}
\end{equation*}

where we have used the fact that $$\sin^2 \theta + \cos^2 \theta =1$$. Solving for $$\lambda$$in $$\mathbb{C}$$, we obtain

\begin{equation*}
\lambda = \cos \theta \pm \sqrt{\cos^2 \theta -1} = \cos\theta \pm \sqrt{-\sin^2 \theta}
= \cos \theta \pm i \sin \theta = e^{\pm i \theta}.
\end{equation*}

We see that, as an operator over the real vector space $$\mathbb{R}^2$$, the operator $$R_\theta$$ only has eigenvalues when $$\theta=0$$ or $$\theta=\pi$$. However, if we interpret the vector $$\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \in \mathbb{R}^2$$as a complex number $$z=x_1+ix_2$$, then $$z$$is an eigenvector if $$R_\theta:\mathbb{C}\to\mathbb{C}$$ maps $$z\mapsto \lambda z=e^{\pm i \theta}z$$. Moreover, from Section 2.3, we know that multiplication by $$e^{\pm i \theta}$$corresponds to rotation by the angle $$\pm\theta$$.

## Contributors

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