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Determine the jordan form of $A = \begin{pmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 4 \end{pmatrix} $

First, I find the characteristic polynomial. $C_A(x)=(x-1)(x-4)^2$. Therefore, the minimal polynomial will be $m_A(x)=(x-1)(x-4)^n, n\leq2$.

Next, I find the eigenspace of 4 ($E_4$): $Null(A-4I) = \begin{pmatrix} -3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & 0 \end{pmatrix}$. hence, $dim(E_4)=1 \implies m_A(x)=(x-1)(x-4)$. therefore the jordan form will be : $\begin{pmatrix} 1 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 4 \end{pmatrix}$

the problem my solution does not agree with the key.

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marked as duplicate by Batominovski, Joel Reyes Noche, wythagoras, muaddib, colormegone Aug 2 '15 at 15:52

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    $\begingroup$ Since $dim(E_4)=1$, you will have a $1$ in the upper right of the two $4$ of your Jordan form... $\endgroup$ – Martigan May 28 '15 at 15:34
  • $\begingroup$ I thought the largest jordan block is determined by the power of (x-4) in my minimal polynomial. this is false? $\endgroup$ – user3382078 May 28 '15 at 15:37
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    $\begingroup$ If the order of your minimal polynomial is inferior to the dimension, the matrix cannot be diagonalizable. $\endgroup$ – Martigan May 28 '15 at 15:54
  • $\begingroup$ @Martigan then I think my minimal polynomial must be incorrect. n should be 2 $\endgroup$ – user3382078 May 28 '15 at 16:56