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3. For a polynomial $g(x) = a_0 + a_1x + cdots + a_mx^m in P(mathbb{F})$ and $T in mathcal{L}(V)$, we define $g(T) = a_0I_V + a_1T + cdots + a_mT^m in mathcal{L}(V)$. (a) Let $v in V$ be an eigenvector of $T$ with an eigenvalue $lambda$. Prove that $g(T)(v) = g(lambda)v$. (b) Assume that $T$ is a diagonalizable operator, and $p_T$ is the characteristic polynomial of $T$. Prove that $p_T(T) = 0$.

          3. For a polynomial $g(x) = a_0 + a_1x + cdots + a_mx^m in P(mathbb{F})$ and $T in mathcal{L}(V)$, we define $g(T) = a_0I_V + a_1T + cdots + a_mT^m in mathcal{L}(V)$.
(a) Let $v in V$ be an eigenvector of $T$ with an eigenvalue $lambda$. Prove that $g(T)(v) = g(lambda)v$.
(b) Assume that $T$ is a diagonalizable operator, and $p_T$ is the characteristic polynomial of $T$. Prove that $p_T(T) = 0$.

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3. For a polynomial g(x) = a0 + a1x + cdots + amx^m in P(mathbbF) and T in mathcalL(V), we define g(T) = a0IV + a1T + cdots + amT^m in mathcalL(V).
(a) Let v in V be an eigenvector of T with an eigenvalue lambda. Prove that g(T)(v) = g(lambda)v.
(b) Assume that T is a diagonalizable operator, and pT is the characteristic polynomial of T. Prove that pT(T) = 0.

Added by Rhonda D.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Adi S

07/13/2023

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Then, we can write g(T)(v) as: g(T)(v) = a0v + a1Tv + ... + amTmv = a0v + a1Av + ... + amAmv (substituting Tv = Av) = g(A)v (since g(A) = a0 + a1A + ... + amAm) Show more…

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For a polynomial g(x) = a0 + a1x + ... + amx^m in P(F) and T in L(V), we define g(T) = a0Iv + a1T + ... + amT^m in L(V). (a) Let v in V be an eigenvector of T with an eigenvalue lambda. Prove that g(T)(v) = g(lambda)v. (b) Assume that T is a diagonalizable operator, and pT is the characteristic polynomial of T. Prove that pT(T) = 0.

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