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$A=\begin{bmatrix}3 & -2 & 5\\ 1 & 0 & 7\\ 0 & 0 & 2\end{bmatrix}$, Find the eigenvalues of A.

I realized that if I swap columns I and II then I can make it an upper triangular matrix. Then the detrminant would be the product of the elements of the main diagonal. And then I would just need to find the roots of that.

However I know that swapping columns flips the sign of the determinant, but I don't know how that will effect finding the eigenvalues.

So I tried it anyways and got determinant of $(x+2)(x-1)(x-2)$ which has roots -2, 1, and 2. But I know that this is incorrect because the answers are supposed to be $\lambda=1,2,2$.

What did I do wrong?

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marked as duplicate by Stella Biderman, Fabian, Vlad, Claude Leibovici, G-man Dec 21 '15 at 10:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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To see that you can't swap columns, try finding the eigenvalues of $\pmatrix{1&0\cr0&1\cr}$ and $\pmatrix{0&1\cr1&0\cr}$, or of $\pmatrix{1&0\cr0&-1\cr}$ and $\pmatrix{0&1\cr-1&0\cr}$. – Gerry Myerson May 3 '12 at 6:33

If you don't see the shortcut, you can at least compute the polynomial:

$det(Ix-A)=det(\begin{bmatrix}x-3 & +2 & -5\\ -1 & x & -7\\ 0 & 0 & x-2\end{bmatrix})$ you can expand along the last row to save time and get:

$(x-2)(x(x-3)+2)=(x-2)(x^2-3x+2)=(x-2)(x-2)(x-1)$

I would venture you might have made a sign error. This would be especially likely if you expanded the determinant any other way.

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Unfortunately swapping two columns changes the eigenvalues of the matrix!

However you can think of your matrix as block upper triangular. The upper left $2 \times 2$ block will give you two eigenvalues, and the lower right entry gives the third eigenvalue, namely 2.

Do you know how to find the eigenvalues of the $2 \times 2$ block?

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How do you know that the lower right entry is one of the eigenvalues? – Kyle V. May 3 '12 at 2:26
    
It is a theorem that eigenvalues of a block upper-triangular matrix are the combined eigenvalues of the blocks. It's easy to verify that $A$ has a corresponding left eigenvector, $(0~0~1)A = (0~0~2)$. It follows that there must be a right eigenvector as well, though it's not immediately evident what it might be. – hardmath May 3 '12 at 4:47

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