- #1

iamsmooth

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## Homework Statement

Let [tex] A = \left[ \begin{array}{cc} -6 & 0.25 \\ 7 & -3 \end{array} \right][/tex]

Find an invertible S and a diagonal D such that [tex]S^{-1}AS=D[/tex]

## Homework Equations

...

## The Attempt at a Solution

So first I need to get eigenvalues so I can get the eigenvectors which will give me the invertible S (or so I believe).

Anyways, here is my work so far:

[tex] A = \left[ \begin{array}{cc} -6 & 0.25 \\ 7 & -3 \end{array} \right] = \left[ \begin{array}{cc} \lambda+6 & -0.25 \\ -7 & \lambda+3 \end{array} \right]

[/tex]

[tex]

(\lambda+6)(\lambda+3)-1.75[/tex]

[tex]

\lambda^2+3\lamda+6\lambda+18-1.75[/tex]

[tex]\lambda^2+9\lambda+16.25[/tex]

Using quadratic equation or factoring, the roots are:

[tex]\lambda=-2.5[/tex]

[tex]\lambda=-6.5[/tex]

From here, subbing lambda in, I get:

[tex]

\left[ \begin{array}{cc} -2.5+6 & -0.25 \\ -7 & -2.5+3 \end{array} \right] \left[ \begin{array}{cc} X_1 \\ X_2 \\ \end{array} \right][/tex]

[tex]

\left[ \begin{array}{cc} -6.5+6 & -0.25 \\ -7 & -6.5+3 \end{array} \right] \left[ \begin{array}{cc} X_1 \\ X_2 \end{array} \right]

[/tex]

Solving for these systems I get:

[tex]

E_{-2.5} = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right][/tex]

[tex]

E_{-6.5} = \left[ \begin{array}{cc} 1 & 0.5 \\ 0 & 0 \end{array} \right][/tex]

From here, how do I get the eigenvectors? I think I'm close (if my work is correct so far)

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