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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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    A more general question: [Eigenvalues of a matrix of $1$'s](http://math.stackexchange.com/questions/153457/eigenvalues-of-a-matrix-of-1s)2012-12-22
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    Do you want to compute [Jordan normal form](http://en.wikipedia.org/wiki/Jordan_normal_form) or [reduced row echelon form](http://en.wikipedia.org/wiki/Reduced_row_echelon_form)? (Your [comment](http://math.stackexchange.com/questions/263595/confusion-on-eigenvalues-of-similar-matrices/263599#comment575867_263599) seems to indicate that you were using some kind of row and column operations.)2012-12-22
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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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    Which can be easily checked using WolframAlpha: http://tinyurl.com/bmsfo5q2012-12-22
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    I'm still confused. I used $R_2-R_1\to C_2-C_1\to R_3-R_1 \to C_3-C_1.$2012-12-22
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    @MartinSleziak: Please help !2012-12-22
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    @SugataAdhya I don't know what more should be said here. The correct result for Jordan normal form is $\begin{pmatrix} 3 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$ and not $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$. (As you can check in the link I gave you.)2012-12-22
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    @SugataAdhya Are you using *both* row and column operations?2012-12-22
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    @MartinSleziak: Oh No !! I've confused **transpose** with **inverse**. Here $B=PAP^t$ for some non-singular $P$, not $B=PAP^{-1}$ for some non-singular $P$.2012-12-22
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    @hwhm: Ya, I was wrongly trying to find out congruent of $A$ !!2012-12-22