Calculating Eigenvectors and Eigenvalues for a Given Matrix

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Homework Statement


Find the eigenvalues and the eigenvectors for the given matrix.

Homework Equations


\[<br /> A =<br /> \left[ {\begin{array}{ccc}<br /> -1 &amp; 6 &amp; 2 \\<br /> 0 &amp; 5 &amp; -6 \\<br /> 1 &amp; 0 &amp; -2 \\<br /> \end{array} } \right]<br /> \]

The Attempt at a Solution


I solved A-\lambda I = 0 and got eigenvalues of -4 and 3, which I've confirmed as correct. After that, to solve for the eigenvalue of -4:
\[<br /> A+4I =<br /> \left[ {\begin{array}{ccc}<br /> 3 &amp; 6 &amp; 2 \\<br /> 0 &amp; 9 &amp; -6 \\<br /> 1 &amp; 0 &amp; 2 \\<br /> \end{array} } \right]<br /> \] ~ \[<br /> \left[ {\begin{array}{ccc}<br /> 1 &amp; 0 &amp; 2 \\<br /> 0 &amp; 1 &amp; -2/3 \\<br /> 0 &amp; 0 &amp; 0 \\<br /> \end{array} } \right]<br /> \]

so \vec x = a \[<br /> \left[ {\begin{array}{c}<br /> -2 \\<br /> 2/3 \\<br /> 1 \\<br /> \end{array} } \right]<br /> \] for all a≠0.

...however, according to two different calculators, \vec x = a \[<br /> \left[ {\begin{array}{c}<br /> 1.1872 \\<br /> -.3957 \\<br /> -.5936 \\<br /> \end{array} } \right]<br /> \]

...which I've verified as being a working solution, while with mine:

( A + 4 I ) \vec x =\[<br /> \left[ {\begin{array}{c}<br /> -2 \\<br /> 6 \\<br /> -2 \\<br /> \end{array} } \right]<br /> \] - all values should equal zero, which they do with the calculated solution. What am I doing wrong?
 
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I think you've just made a calculation error while checking your work. I just did the multiplication and (A+4I)x=0. Somehow when you multiplied the matrix by x you came up with just the negative of the third column in (A+4I). Try it again and if you're still having the same problem try googling "Matrix Multiplication" and make sure you're using the correct process.
 
Your eigenvector works for the eigenvalue -4. You need to do the same for eigen value 3, I don't think it's possible to find 3 L.I. eigenvectors though!


\left(\begin{array}{c}<br /> 1 \\<br /> \frac{3}{2} \\<br /> 0<br /> \end{array} \right)

that is a "nice" vector to put in P though!
 
Doh. I entered 0 in my calculator instead of 1 for the last value of x. Thanks anyways :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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