Trace, determinant, and eigenvalues 3x3

[Dustinsfl]

Use the trace and determinant to compute eigenvalues.

I know how to do this with a 2x2 but not sure how to do it with a matrix of nxn where n>2.

\begin{bmatrix}<br /> \frac{1}{2} &amp; \frac{1}{3} &amp; \frac{1}{5}\\ <br /> \frac{1}{4} &amp; \frac{1}{3} &amp; \frac{2}{5}\\ <br /> \frac{1}{4} &amp; \frac{1}{3} &amp; \frac{2}{5}<br /> \end{bmatrix} the det=0 and the trace=\frac{37}{30}

 

[kof9595995]

I don't think it's possible for n>2 cases, you can probably work out infinite number of matrices with the same trace & det, but with different eigenvalues

 

[hgfalling]

You can't use only the determinant and trace to find the eigenvalues of a 3x3 matrix the way you can with a 2x2 matrix. For example, suppose that det(A) = 0 and tr(A) = t. Then any matrix of the form:

<br /> \begin{bmatrix}<br /> a &amp; 0 &amp; 0\\ <br /> 0 &amp; t-a &amp; 0\\ <br /> 0 &amp; 0 &amp; 0\\ <br /> \end{bmatrix}<br />

has trace = t and determinant 0 with eigenvalues a and t-a. So you'll have to go back to the matrix to find the eigenvalues.

 

[Dustinsfl]

I don't think it's possible for n>2 cases, you can probably work out infinite number of matrices with the same trace & det, but with different eigenvalues

According to the book it can be done with this matrix.

 

[Dustinsfl]

Epiphany: since the determinant is 0, one of the eigenvalues has to be 0.

 

[hgfalling]

Now if we could just divide the determinant of the matrix by zero and get back \lambda_1 \lambda_2!

 

[kof9595995]

According to the book it can be done with this matrix.

Then I got no clue, sorry.