# Confusion on Eigenvalues of Similar Matrices

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 – Martin Sleziak Dec 22 '12 at 8:38
Do you want to compute Jordan normal form or reduced row echelon form? (Your comment seems to indicate that you were using some kind of row and column operations.) – Martin Sleziak Dec 22 '12 at 9:03
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. – Marc van Leeuwen Dec 22 '12 at 10:18

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.
I'm still confused. I used $R_2-R_1\to C_2-C_1\to R_3-R_1 \to C_3-C_1.$ – Sugata Adhya Dec 22 '12 at 8:47
@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.) – Martin Sleziak Dec 22 '12 at 9:00
@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 $P$. – Sugata Adhya Dec 22 '12 at 9:06