I saw a method for finding eigenvalues by using Gauss elimination to find an upper triangular matrix, then just taking the diagonal elements as the eigenvalues. It seems to work except for this case:

\begin{bmatrix} 1&1\\ 1&2 \end{bmatrix}

if i make it upper triangular by subtracting the first row from the second to get:

\begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}

then I would assume the eigenvalues should be 1, but they are not. What am I doing wrong?


You are doing nothing wrong. Gaussian elimination does not preserve eigenvalues, and this method simply does not work.

If you are lucky enough to start with an upper triangular matrix, then it is true that the diagonals are the eigenvalues. This is very convenient. Otherwise, you have to go and actually compute them.

For two by two matrices, there is a sort of shortcut (maybe). The eigenvalues completely determine the trace and the determinant. In particular, $\lambda_1 + \lambda_2 = \text{Trace} A$, and $\lambda_1\lambda_2 = \text{Det}A$. So here, we know that $\lambda_1 + \lambda_2 = 3$ and $\lambda_1 \lambda_2 = 1$.

So $\lambda_1 = 1/\lambda_2$, which leads to $\lambda_1^2 - 3\lambda_1 + 1 = 0$. And so $\lambda_1,\lambda_2$ are the two roots of the polynomial $x^2 - 3x + 1$, which are $\frac{3 \pm \sqrt{5}}{2}$. Is that easier? Sometimes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.