# find Jordan form [duplicate]

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.

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

• Since $dim(E_4)=1$, you will have a $1$ in the upper right of the two $4$ of your Jordan form... – Martigan May 28 '15 at 15:34