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### Let $A,B$ be $2$ square matrices such that $A+AB-2B=0$. Prove $2$ isn't an eigenvalue of $A$.

$A+AB-2B=0$ $A+AB-2B-2I=-2I$ $(A-2I)(B+I)=-2I$ $\therefore A-2I$ is non-singular. $\therefore$ for any $v \neq 0,$ $(A-2I)v \neq 0$ $Av \neq 2v$ $\therefore 2$ cannot be an eigenvalue.
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### Let $A,B$ be $2$ square matrices such that $A+AB-2B=0$. Prove $2$ isn't an eigenvalue of $A$.

Recall that since the matrices are square, $A$ and $A^T$ have the same eigenvalues. In particular for $v\neq 0$ such that $A^Tv=2v$ then $$A^Tv+B^TA^Tv-2B^Tv=2v+2B^Tv-2B^Tv=2v\neq 0$$ which is a ...
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### (Linear Independence) Show that for two vectors $v,w \in \mathbb{R}^n$, the conditions (i), (ii), (iii) are equivalent

(i) $\implies (ii)$ if $w\neq 0$ and $v=\rho w$ since $\Bbb R$ is a field exist the inverse of $\rho$ namely $\frac{1}{\rho}$ but this lead to a contradiction since \$\frac{1}{\rho} \in \Bbb R, w = \...
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### Are all nilpotent matrices strictly upper triangularizable (over the complex and real fields)?

This answer would fit better as a comment to Eric Wofsey but it does give a very elementary proof to the original question by showing that if two real matrices are conjugates over complex matrices, ...
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