###### Example4.3.4

We will try to find a diagonalization of $$A = \left[\begin{array}{rr} -5 \amp 6 \\ -3 \amp 4 \\ \end{array}\right] \text{.}$$

First, we find the eigenvalues of $$A$$ by solving the characteristic equation

\begin{equation*} \det(A-\lambda I) = (-5-\lambda)(4-\lambda)+18 = (-2-\lambda)(1-\lambda) = 0\text{.} \end{equation*}

This shows that the eigenvalues of $$A$$ are $$\lambda_1 = -2$$ and $$\lambda_2 = 1\text{.}$$

By constructing $$\nul(A-(-2)I)\text{,}$$ we find a basis for $$E_{-2}$$ consisting of the vector $$\vvec_1 = \twovec{2}{1}\text{.}$$ Similarly, a basis for $$E_1$$ consists of the vector $$\vvec_2 = \twovec{1}{1}\text{.}$$ This shows that we can construct a basis $$\bcal=\{\vvec_1,\vvec_2\}$$ of $$\real^2$$ consisting of eigenvectors of $$A\text{.}$$

We now form the matrices

\begin{equation*} D = \left[\begin{array}{rr} -2 \amp 0 \\ 0 \amp 1 \\ \end{array}\right],\qquad P = \left[\begin{array}{cc} \vvec_1 \amp \vvec_2 \end{array}\right] = \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 1 \\ \end{array}\right] \end{equation*}

and verify that

\begin{equation*} PDP^{-1} = \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 1 \\ \end{array}\right] \left[\begin{array}{rr} -2 \amp 0 \\ 0 \amp 1 \\ \end{array}\right] \left[\begin{array}{rr} 1 \amp -1 \\ -1 \amp 2 \\ \end{array}\right] = \left[\begin{array}{rr} -5 \amp 6 \\ -3 \amp 4 \\ \end{array}\right] = A\text{.} \end{equation*}

There are, of course, many ways to diagonalize $$A\text{.}$$ For instance, we could change the order of the eigenvalues and eigenvectors and write

\begin{equation*} D = \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp -2 \\ \end{array}\right],\qquad P = \left[\begin{array}{cc} \vvec_2 \amp \vvec_1 \end{array}\right] = \left[\begin{array}{rr} 1 \amp 2 \\ 1 \amp 1 \\ \end{array}\right]\text{.} \end{equation*}

If we choose a different basis for the eigenspaces, we will also find a different matrix $$P$$ that diagonalizes $$A\text{.}$$ The point is that there are many ways in which $$A$$ can be written in the form $$A=PDP^{-1}\text{.}$$

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