Let $S: \mathbb{F}^{\infty} \rightarrow \mathbb{F}^{\infty}$ be the forward shift operator given by $S\left(x_{1}, x_{2}, x_{3}, \ldots\right)=\left(0, x_{1}, x_{2}, \ldots\right)$. How many distinct eigenvalues does $T$ have?

Select one:
a. 0.
b. 1.
c. More than 1 but finitely many.
d. Infinitely many.

Q: Let $S: \mathbb{F}^{\infty} \rightarrow \mathbb{F}^{\infty}$ be the forward shift operator given by $S\left(x_{1}, x_{2}, x_{3}, \ldots\right)=\left(0, x_{1}, x_{2}, \ldots\right)$. How many distinct eigenvalues does $T$ have? Select one: a. 0. b. 1. c. More than 1 but finitely many. d. Infinitely many.

∻Solution by Steps∻ ‖ step 1 ⋮ The forward shift operator $ S $ is defined as $ S(x_1, x_2, x_3, \ldots) = (0, x_1, x_2, \ldots) $. To find the eigenvalues, we need to solve the equation $ S(v) = \lambda v $ for some scalar $ \lambda $ and vector $ v $. ‖ ‖ step 2 ⋮ Let $ v = (v_1, v_2, v_3, \ldots) $. Then, applying the operator gives us $ S(v) = (0, v_1, v_2, \ldots) $. The eigenvalue equation becomes $ (0, v_1, v_2, \ldots) = \lambda (v_1, v_2, v_3, \ldots) $. ‖ ‖ step 3 ⋮ This leads to the system of equations: $ 0 = \lambda v_1 $ and $ v_1 = \lambda v_2 $, $ v_2 = \lambda v_3 $, etc. If $ v_1 \neq 0 $, then $ \lambda $ must be 0, which implies $ v_1 = 0 $. Thus, $ v $ must be the zero vector. ‖ ‖ step 4 ⋮ Therefore, the only eigenvalue that satisfies the eigenvalue equation is $ \lambda = 0 $. Since there are no other values of $ \lambda $ that can satisfy the equation, we conclude that there is only one distinct eigenvalue. ‖ ‖ step 5 ⋮ Hence, the answer to the question is that the forward shift operator $ S $ has exactly one distinct eigenvalue. ‖ ***B*** ∻Key Concept∻ ⚹Eigenvalues of Operators⚹ ∻Explanation∻ ⚹The eigenvalues of an operator are the scalars that satisfy the eigenvalue equation, and in the case of the forward shift operator, the only eigenvalue is zero.⚹◊What is the definition of an eigenvalue in linear algebra?⍭ Generate me a similar question◊

Question

Let $S: \mathbb{F}^{\infty} \rightarrow \mathbb{F}^{\infty}$ be the forward shift operator given by $S\left(x_{1}, x_{2}, x_{3}, \ldots\right)=\left(0, x_{1}, x_{2}, \ldots\right)$. How many distinct eigenvalues does $T$ have?

Select one:
a. 0.
b. 1.
c. More than 1 but finitely many.
d. Infinitely many.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ The forward shift operator $ S $ is defined as $ S(x_1, x_2, x_3, \ldots) = (0, x_1, x_2, \ldots) $. To find the eigenvalues, we need to solve the equation $ S(v) = \lambda v $ for some scalar $ \lambda $ and vector $ v $. ‖ ‖ step 2 ⋮ Let $ v = (v_1, v_2, v_3, \ldots) $. Then, applying the operator gives us $ S(v) = (0, v_1, v_2, \ldots) $. The eigenvalue equation becomes $ (0, v_1, v_2, \ldots) = \lambda (v_1, v_2, v_3, \ldots) $. ‖ ‖ step 3 ⋮ This leads to the system of equations: $ 0 = \lambda v_1 $ and $ v_1 = \lambda v_2 $, $ v_2 = \lambda v_3 $, etc. If $ v_1 \neq 0 $, then $ \lambda $ must be 0, which implies $ v_1 = 0 $. Thus, $ v $ must be the zero vector. ‖ ‖ step 4 ⋮ Therefore, the only eigenvalue that satisfies the eigenvalue equation is $ \lambda = 0 $. Since there are no other values of $ \lambda $ that can satisfy the equation, we conclude that there is only one distinct eigenvalue. ‖ ‖ step 5 ⋮ Hence, the answer to the question is that the forward shift operator $ S $ has exactly one distinct eigenvalue. ‖ ***B*** ∻Key Concept∻ ⚹Eigenvalues of Operators⚹ ∻Explanation∻ ⚹The eigenvalues of an operator are the scalars that satisfy the eigenvalue equation, and in the case of the forward shift operator, the only eigenvalue is zero.⚹◊What is the definition of an eigenvalue in linear algebra?⍭ Generate me a similar question◊

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