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Please help me to identify where I went wrong:

The completely reduced normal form of the real matrix $A= $\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{pmatrix}

is the following matrix $B= $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} Clearly the eigenvalues of $B$ are $1,0,0$ so should the eigenvalues of $A$ since similar matrices have the same characteristic polynomial. But when I'm trying to formally evaluate the eigenvalues of $A$ the roots of $\chi_A$ become $0,0,3$. $\begin{vmatrix} 1-x & 1 & 1 \\ 1 & 1-x & 1 \\ 1 & 1 & 1-x \\ \end{vmatrix}=0$ $\implies (1-x)[(1-x)^2-1]-1[(1-x)-1]+1[1-(1-x)]=0$

$\implies (1-x)[x^2-2x]-1(-x)+x=0$

$\implies (1-x)[x^2-2x]+2x=0$

$\implies x^2-2x-x^3+2x^2+2x=0$

$\implies 3x^2-x^3=0$

$\implies x^2(3-x)=0$

So the eigenvalues of $A$ are $0,0,3.$

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    What do you mean by "completely reduced normal form"? You must specify what operations are allowed. Every operation changes a matrix (a tautology) so if you want to apply operations you must know which class of matrix you want to remain in. Since here you want a _similarity_ class, the only operations possible are _conjugations_, I don't think you limited yourself to that.2012-12-22

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From your comment to the answer by jakncoke, I gather that you applied row and column operations to $A$. Those operations do not give you similar matrices, just equivalent matrices. Therefore you can conclude that the rank of $A$ is equal to the rank of $B$ (indeed both are $1$) and that is all. The matrices $A$ and $B$ are not similar. If you want the characteristic polynomial, you can do row and column operations on the polynomial matrix $ XI_3-A = \pmatrix{X-1&-1&-1\\-1&X-1&-1\\-1&-1&X-1} $ before taking the determinant (the determinant is invariant under row and column operations if you take care to throw in a sign $-1$ for transpositions of two rows or two columns), but this is a lot harder than to do row and column operations on the matrix $A$. Deciding similarity is not so easy.

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Your reduced normal form for the matrix A is wrong.

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    @hwhm: Ya, I was wrongly trying to find out congruent of $A$ !!2012-12-22