eigenvalue and independence Let $B$ be a $5\times 5$ real matrix and assume:


*

*$B$ has eigenvalues 2 and 3 with corresponding eigenvectors $p_1$ and $p_3$, respectively.

*$B$ has generalized eigenvectors $p_2,p_4$ and $p_5$ satisfying
$Bp_2=p_1+2p_2,Bp_4=p_3+3p_4,Bp_5=p_4+3p_5$.


Prove that $\{p_1,p_2,p_3,p_4,p_5\}$ is linearly independent set.

One can do a direct proof. Here is the beginning:
Suppose $0=\lambda_1p_1+\dots+\lambda_5p_5$. Then 
$$\begin{align*}
0&=(B-3I)^3(\lambda_1p_1+\dots+\lambda_5p_5)\\
&=(B-3I)^3(\lambda_1p_1+\lambda_2p_2)\\
&=(B-2I-I)^3(\lambda_1p_1+\lambda_2p_2)\text{ now one can apply binomial theorem as $B-2I$ and $-I$ commute}\\
&= (B-2I)^3(\lambda_1p_1+\lambda_2p_2)+3(B-2I)^2(-I)(\lambda_1p_1+\lambda_2p_2)+3(B-2I)(-I)^2(\lambda_1p_1+\lambda_2p_2)+(-I)^3(\lambda_1p_1+\lambda_2p_2)\\
&=(3\lambda_2-\lambda_1)p_1-\lambda_2p_2.
\end{align*}$$
Now a similar application of $(B-2I)$ yields $\lambda_1=0$ and $\lambda_2=0$. Again similar applications of $(B-3I)^k$ for appropriate $k$ yield the vanishing of the other coefficients. I'll leave that to the reader.
