Finding eigenvectors for diagonalization

[iamsmooth]

Homework Statement

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

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

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:

A = \left[ \begin{array}{cc} -6 &amp; 0.25 \\ 7 &amp; -3 \end{array} \right] = \left[ \begin{array}{cc} \lambda+6 &amp; -0.25 \\ -7 &amp; \lambda+3 \end{array} \right]<br />

<br /> (\lambda+6)(\lambda+3)-1.75

<br /> \lambda^2+3\lamda+6\lambda+18-1.75

\lambda^2+9\lambda+16.25

Using quadratic equation or factoring, the roots are:

\lambda=-2.5

\lambda=-6.5

From here, subbing lambda in, I get:

<br /> \left[ \begin{array}{cc} -2.5+6 &amp; -0.25 \\ -7 &amp; -2.5+3 \end{array} \right] \left[ \begin{array}{cc} X_1 \\ X_2 \\ \end{array} \right]

<br /> \left[ \begin{array}{cc} -6.5+6 &amp; -0.25 \\ -7 &amp; -6.5+3 \end{array} \right] \left[ \begin{array}{cc} X_1 \\ X_2 \end{array} \right]<br />

Solving for these systems I get:

<br /> E_{-2.5} = \left[ \begin{array}{cc} 1 &amp; 0 \\ 0 &amp; 1 \end{array} \right]

<br /> E_{-6.5} = \left[ \begin{array}{cc} 1 &amp; 0.5 \\ 0 &amp; 0 \end{array} \right]

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

 

[iamsmooth]

I guess my eigenspace is probably wrong, can someone help me figure that out? If I get that, I know how to get the diagonal.

Thanks.

 

[iamsmooth]

So I used a matrix calculator to get the vectors (since I can't get them) and it gave me 1, -2 for \lambda of -6.25, and -2, 14 for \lambda of-2.5

So with this, the eigenspace should be

<br /> S = \left[ \begin{array}{cc} 1 &amp; 1 \\ -2 &amp; 14 \end{array} \right]<br />

And when I calculate it out as S^{-1}AS = D

<br /> D = \left[ \begin{array}{cc} -104 &amp; 0 \\ 0 &amp; -40 \end{array} \right]<br />

Which seems like it works since it's now diagonal, but the answer is wrong for some reason. I've tried switching the signs and it's still wrong. Really not sure what I'm doing wrong. This is the last question I have for practice and this is getting frustrating. I'm starting to think maybe there's an error with the program.

 

[HallsofIvy]

Homework Statement

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

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

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:

A = \left[ \begin{array}{cc} -6 &amp; 0.25 \\ 7 &amp; -3 \end{array} \right] = \left[ \begin{array}{cc} \lambda+6 &amp; -0.25 \\ -7 &amp; \lambda+3 \end{array} \right]<br />

<br /> (\lambda+6)(\lambda+3)-1.75

<br /> \lambda^2+3\lamda+6\lambda+18-1.75

\lambda^2+9\lambda+16.25

Using quadratic equation or factoring, the roots are:

\lambda=-2.5

\lambda=-6.5

From here, subbing lambda in, I get:

<br /> \left[ \begin{array}{cc} -2.5+6 &amp; -0.25 \\ -7 &amp; -2.5+3 \end{array} \right] \left[ \begin{array}{cc} X_1 \\ X_2 \\ \end{array} \right]

<br /> \left[ \begin{array}{cc} -6.5+6 &amp; -0.25 \\ -7 &amp; -6.5+3 \end{array} \right] \left[ \begin{array}{cc} X_1 \\ X_2 \end{array} \right]<br />

Solving for these systems I get:

<br /> E_{-2.5} = \left[ \begin{array}{cc} 1 &amp; 0 \\ 0 &amp; 1 \end{array} \right]

<br /> E_{-6.5} = \left[ \begin{array}{cc} 1 &amp; 0.5 \\ 0 &amp; 0 \end{array} \right]

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

Your error is in subtracting the two different eigenvalues from the same matrix. To find an eigenvalue corresponding to x= -2.5 Look at the equation
\begin{bmatrix}-6 &amp; .25 \\ 7 &amp; -3\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= -2.5\begin{bmatrix}x \\ y\end{bmatrix}
or, equivalently
\begin{bmatrix}6+ 2.5 &amp; -.25 \\ -7 &amp; 3+ 2.5\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}0 \\ 0\end{bmatrix}
That reduces to the two equations 6x- 2.5y= 0 and -7x+ 2.5y= 0 which both reduce to y= 34x. Any eigenvector, corresponding to eigenvalue -2.5, must be of the form
x\begin{bmatrix}1 \\ 34\end{bmatrix}
and you can take x to be 1, say, to get a specific eigenvalue, or 1/\sqrt{34^2+ 1} for a unit eigenvalue.

Similarly, to find an eigenvector corresponding to eigenvalue -6.5, solve
\begin{bmatrix}-6 &amp; .25 \\ 7 &amp; -3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= -6.5\begin{bmatrix}x \\ y \end{bmatrix}
or, equivalently,
\begin{bmatrix}-6+6.5 &amp; .25 \\ 7 &amp; -3+ 6.5\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} 0.5 &amp; .25\\ 7 &amp; 3.5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}

 

[iamsmooth]

\begin{bmatrix}6+ 2.5 &amp; -.25 \\ -7 &amp; 3+ 2.5\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}0 \\ 0\end{bmatrix}
That reduces to the two equations 6x- 2.5y= 0 and -7x+ 2.5y= 0 which both reduce to y= 34x. Any eigenvector, corresponding to eigenvalue -2.5, must be of the form
x\begin{bmatrix}1 \\ 34\end{bmatrix}
and you can take x to be 1, say, to get a specific eigenvalue, or 1/\sqrt{34^2+ 1} for a unit eigenvalue.

I don't understand where you get the two equations from that matrix. Even if I accept it, I don't understand how it reduces to y=34x.