Question

The operator of differentiation on $P_m(\mathbb{R})$ is nilpotent because the $(m+1)^{st}$ derivative of every polynomial of degree at most $m$ equals $0$. Note that on this space of dimension $m+1$, we need to raise the nilpotent operator to the power $m+1$ to get the $0$ operator.

Name: The operator of differentiation on Pm(ℝ) is nilpotent because the (m+1)^st derivative of every polynomial of degree at most m equals 0. Note that on this space of dimension m+1, we need to raise the nilpotent operator to the power m+1 to get the 0 operator.
Uploaded: 2023-05-22T15:04:36-08:00
Duration: 1 min 24 s
Channel: Hoan Nguyen
Description: The operator of differentiation on Pm(ℝ) is nilpotent because the (m+1)^st derivative of every polynomial of degree at most m equals 0. Note that on this space of dimension m+1, we need to raise the nilpotent operator to the power m+1 to get the 0 operator.

          The operator of differentiation on $P_m(\mathbb{R})$ is nilpotent because the $(m+1)^{st}$ derivative of every polynomial of degree at most $m$ equals $0$. Note that on this space of dimension $m+1$, we need to raise the nilpotent operator to the power $m+1$ to get the $0$ operator.

The operator of differentiation on Pm(ℝ) is nilpotent because the (m+1)^st derivative of every polynomial of degree at most m equals 0. Note that on this space of dimension m+1, we need to raise the nilpotent operator to the power m+1 to get the 0 operator.

Added by Lori P.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Hoan Nguyen

05/22/2023

Step 1

Step 1: Consider the space Pm(R) which consists of all polynomials of degree at most m. Show more…

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The operator of differentiation on P_(m)(R) is nilpotent because the (m+1)^(st ) derivative of every polynomial of degree at most m equals 0 . Note that on this space of dimension m+1, we need to raise the nilpotent operator to the power m+1 to get the 0 operator Explain the way of solve The operator of differentiation on Pm(R) is nilpotent because the (m + 1)st derivative of every polynomial of degree at most m equals 0 Note that on this space of dimension m + 1, we need to raise the nilpotent operator to the power m + 1 to get the 0 operator.