# Let $A \in Mat_{n,n}(\mathbb C)$ (diagonalizable) and $u_1, \ldots, u_k$ be the different eigenvalues of $A$. Show $(A-u_1I_n) \dots (A-u_k I_n) = 0$.

Let $A \in Mat_{n,n}(\mathbb C)$ (diagonalizable) and suppose $p_A(\lambda) = (-1)^n(\lambda -\lambda_1 ) \dots (\lambda - \lambda_n)$ (characteristic polynomial of $A$), where $\lambda_i$ is an eigenvalue of $A$.

By Cayley-Hamilton we know $p_A(A) = 0$.

Let $u_1, \ldots, u_k$ be the different eigenvalues among $\lambda_1, \ldots, \lambda_n$.

How can I see that $(A-u_1I_n) \dots (A-u_k I_n) = 0$ ?

• The statement is not true in general. In particular, if the minimum polynomial contains a factor that is not linear, this is not true. May 15, 2014 at 10:58
• It is just stated in some notes of mine, in a section regarding the Putzer-algorithm, that if $u_1, \ldots, u_p$ are the different values of $\lambda_1, \ldots, \lambda_n$ then $(A-u_1 I_n) \dots (A-u_n I_n)$, so we can make the Putzer-algorithm more efficient, since after (and including) $P_p = \prod^p_{j=1} (A-\lambda_j I_n)$, we have that $P_{p+1}, \ldots,$ will be $0$, but I guess it is a mistake then? May 15, 2014 at 11:03
• If $A$ is diagonalizable it is true, but as in the first answer below that you have received you can see that it is not true in general. Maybe you missed a '$A$ is diagonalizable' when taking the notes? May 15, 2014 at 11:11
• You're probably correct. Can you show it is true when $A$ is diagonalizable ? May 15, 2014 at 11:22
• yes, you just write each factor as $(PDP^{-1}-u_iPP^{-1})$, then take the $P$'s and $P^{-1}$ outside the brackets, simplify, and then you have multiplication of a series of diagonal matrices with 0's in different positions, when you multiply them out you will get the zero matrix. Sorry I don't have time now to do it explicitly May 15, 2014 at 11:27

Put $$A=\begin{pmatrix} 1&1\\0&1 \end{pmatrix}.$$ Then $\lambda_{12}=1$ but $$A-I_2=\begin{pmatrix} 0&1\\0&0 \end{pmatrix}\neq 0.$$