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How can I create a change in row entries of a matrix in which characteristic polynomial be same for two matrix even by this change to get nice arrangment? (for example: the second element of last row of matrix 3*3 is zero but I want to change this value.)

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Did I get it right? You want the characteristic polynomial to be independent of, say, the second entry of the third row. You need to make the corresponding cofactor to be zero. For example, make the entries of the third column on the 1st and 2nd rows to be zero. Or do you want something else? – Jyrki Lahtonen Jan 15 '12 at 17:26
@Jyrki Lahtonen: for example: $$ \begin{matrix} 2 & 0 & 1\\ 0 & 7 & 0\\ 3 & 0 & 3\\ \end{matrix} $$ when I have a matrix such as above, I can't obtain the Jordan matrix to get characteristic polynomial(because of existence zero entry at last row) but I can obtain that by other ways. – zeta Jan 15 '12 at 18:25
Sorry, I don't understand. Isn't it useful to have that zero, because your matrix is then sort of in a block diagonal form (one block on row/column 2 - the other on rows/columns 1 and 3)??? Why would it be easier to calculate the characteristic polynomial, if in place of that zero you had, say, a one? – Jyrki Lahtonen Jan 15 '12 at 18:35
Your welcome, my goal of what I said above is that I want to eliminate the zero at last row. (eg. by changing the position of second row and third row, the matrix will be $$ \begin{matrix} 2 & 0 & 1\\ 3 & 0 & 3\\ 0 & 7 & 0\\ \end{matrix} $$ that is obtained jordan matrix easier! but the characteristic polinomial is changed! ) – zeta Jan 15 '12 at 19:10
This does not explain why you want that entry to be non-zero. Could you edit your question to include an example calculation of characteristic polynomials showing, why you think it is beneficial to have something other than zero at that position, please? Then we may be able to give a more useful answer! – Jyrki Lahtonen Jan 16 '12 at 8:35

Matrices $A$ and $B^{-1}AB$ have the same characteristic polynomial, so given a matrix $A$ that you don't like because it has an annoying zero in it, you could just pick some random invertible matrix $B$ and then replace $A$ with $B^{-1}AB$. But I have to agree with Jyrki's comment; I don't see why having a zero in row 3, column 2 makes it hard to find the Jordan form.

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