# Tag Info

2

Take $$A= \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \ \ \ \ \ \ \ \ \mbox{ and } \ \ \ \ \ \ \ \ B= ( \begin{matrix} 1 & 1 \end{matrix} )$$

2

Using the definition of matrix multiplication, you can just check directly that $De_n$ ($e_n$ is the unit vector with $1$ at the $n$-th position) is $\kappa e_n$.

1

If $$A = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix} \right) \quad B = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right)$$ then  A B = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \quad B A = \left( ...

1

This isn't actually correct. One can have $A = \left[ \begin{array}{cc} 1 & -1 \\ 3 & -2 \end{array} \right]$, so that $A^3 = I$, but the characteristic polynomial of $A$ is $\lambda^2 + \lambda + 1$, whose roots are complex. What is guaranteed is that the eigenvalues will be roots of the polynomial $\lambda^3 - 1,$ one root of which is \$\lambda = ...

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