Parker Pitts

2022-10-01

If M is a linear operator on ${\mathbb{R}}^{3}$ with unique and real eigenvalues ${\lambda}_{1}<{\lambda}_{2}<{\lambda}_{3}$, such that $\mathrm{\exists}x\in {\mathbb{R}}^{3}\setminus \{0\}$, satisfying the condition $\underset{n\to \mathrm{\infty}}{lim}||{M}^{n}x||=0$. What are the possible values of ${\lambda}_{1}$?

Lamar Esparza

Beginner2022-10-02Added 8 answers

That's not as straightforward of an answer as it seems.

If x is a linear multiple of ${\zeta}_{1}$, then we can affirm that $|{\lambda}_{1}|<1$

Otherwise, we have $x={c}_{1}{\zeta}_{1}+{c}_{2}{\zeta}_{2}+{c}_{3}{\zeta}_{3}$ and that means ${M}^{n}x={c}_{1}{\lambda}_{1}^{n}{\zeta}_{2}+{c}_{2}{\lambda}_{2}^{n}{\zeta}_{2}+{c}_{3}{\lambda}_{3}^{n}{\zeta}_{3}$, implying all ${\lambda}_{i}$ should have magnitude less than 1 to satisfy that property for any arbitrary vector x.

If x is a linear multiple of ${\zeta}_{1}$, then we can affirm that $|{\lambda}_{1}|<1$

Otherwise, we have $x={c}_{1}{\zeta}_{1}+{c}_{2}{\zeta}_{2}+{c}_{3}{\zeta}_{3}$ and that means ${M}^{n}x={c}_{1}{\lambda}_{1}^{n}{\zeta}_{2}+{c}_{2}{\lambda}_{2}^{n}{\zeta}_{2}+{c}_{3}{\lambda}_{3}^{n}{\zeta}_{3}$, implying all ${\lambda}_{i}$ should have magnitude less than 1 to satisfy that property for any arbitrary vector x.

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