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I have a question on determining eigenvalues for a given matrix A:

$$ A= \begin{bmatrix} 2 & 1 & 2 \\ 0 & 2 & -1 \\ 0 & 1 & 0 \\ \end{bmatrix} $$

Here's what I have so far:

\begin{align} & det(A-\lambda I_3) = 0\\ & =det \begin{bmatrix} 2-\lambda & 1 & 2 \\ 0 & 2-\lambda & -1 \\ 0 & 1 & 0-\lambda \\ \end{bmatrix} = 0 \\ & = (2-\lambda) \begin{vmatrix} 2-\lambda & -1 \\ 1 & 0-\lambda \\ \end{vmatrix} \\ & = (2-\lambda)(-2\lambda+\lambda^2+1) \\ & = (2-\lambda)(\lambda-1)^2 = 0\end{align}

So my eigenvalues are $2$ and $1$. I am almost postitive this is correct. Here's my confusion/question. I know that the determinant of an upper triangular matrix is the product of the terms along the diagonal.

Some of my classmates used a different approach where from the beginning they row reduced $A$ into an upper triangular matrix by eliminating the $1$ from the $3$rd row, then subtracted $\lambda I$ and then instead of solving the determinant like I did they just multiplied the diagonal terms together.

The problem I see with this is that you can obtain different matrices depending on how you choose to row reduce, which means you can get different eigenvalues. Is it possible to solve for the eigenvalues by using the approach of making $A$ into an upper triangular matrix first? (So that you can solve the determinant by multiplying the diagonal entries) I hope that makes sense. Thanks!

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  • $\begingroup$ The best "trick" is your way because it is standard and easy. Recognizing the zero's under the $2$ in the upper left corner was the "trick". Now can you find corresponding eigen vectors too? $\endgroup$ – imranfat Apr 26 '16 at 20:42
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    $\begingroup$ Row operations may not preserve eigenvalues. $\endgroup$ – peterwhy Apr 26 '16 at 20:47
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    $\begingroup$ In fact, most of the times they do not preserve eigenvalues +1, so it should not be done, but the OP's question is conceptionally a great question... $\endgroup$ – imranfat Apr 26 '16 at 20:50
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    $\begingroup$ Some row operations will change the determinant (and the eigenvalues) of A and some will not. If you are careful and know what you are doing, you can find eigenvalues this way. However, I would generally advise against it. $\endgroup$ – Doug M Apr 26 '16 at 20:58
  • $\begingroup$ Don't do like your classmates! Your method is sound, theirs isn't. $\endgroup$ – egreg Apr 26 '16 at 21:33
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The answer to your question is no. That is, your friends are wrong. In general, applying row-reduction to your matrix will change its eigenvalues.

We can see what goes wrong if we take this approach slightly differently. In particular: $$ \pmatrix{ 2 & 1 & 2 \\ 0 & 2 & -1 \\ 0 & 1 & 0 \\ } \to \pmatrix{ 2 & 1 & 2 \\ 0 & 1 & 0 \\ 0 & 2 & -1 \\ } \to \pmatrix{2&1&2\\0&1&0\\0&0&-1} $$ however, $-1$ is not an eigenvalue of $A$.

What's even more extreme: if we keep row-reducing, we can go from $A$ to $I$. It is certainly not the case, however, that $A$ has only the eigenvalue $1$.

If you insist on calculating the determinant via row-operations, subtract $\lambda I$ first: $$ \det(A - \lambda I) = \det \pmatrix{ 2 - \lambda & 1 & 2 \\ 0 & 2 - \lambda & -1 \\ 0 & 1 & -\lambda \\ } = \\ \det\pmatrix{ 1&0&0\\ 0&1&0\\ 0&-\frac{1}{2 - \lambda} & 1 }^{-1} \det \pmatrix{ 2 - \lambda & 1 & 2 \\ 0 & 2 - \lambda & -1 \\ 0 & 0 & -\lambda + \frac{1}{2 - \lambda} \\ } = \\ 1 \cdot \det \pmatrix{ 2 - \lambda & 1 & 2 \\ 0 & 2 - \lambda & -1 \\ 0 & 0 & -\lambda + \frac{1}{2 - \lambda} \\ } $$

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