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Math Help - linear algebra question

  1. #1
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    linear algebra question

    show that if f(t) is the characteristic polynomial of a diagonalizable linear op. T, then f(T)=T_0, the zero operator. We are given the fact that if T is a lin. op. on vector space V and g(t) is a polynomial with coefficients from F, then for a given eigenvector x of T with corresponding eigenvale t, g(T)(x) = g(t)x.
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  2. #2
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    Quote Originally Posted by squarerootof2 View Post
    show that if f(t) is the characteristic polynomial of a diagonalizable linear op. T, then f(T)=T_0, the zero operator. We are given the fact that if T is a lin. op. on vector space V and g(t) is a polynomial with coefficients from F, then for a given eigenvector x of T with corresponding eigenvale t, g(T)(x) = g(t)x.
    Let T: V\to V be a linear operator and let B be a basis for V. Since T is diagnolizable if we form [T]_B the matrix is a diagnolizable, therefore [T]_B = ADA^{-1} where D is a diagnol matrix whose entires consists of eigenvalues of [T]_B. Let \det(xI - [T]_B) = f(x) = x + a_{n-1}x^{n-1}+...+a_1x+a_0 be the characheristic polynomial. If k_1,...,k_n are (repeated) eigenvalues of [T]_B then f(k_1) = ... = f(k_n) = 0. The problem asks us to show that [T]_B satisfies X^n + a_{n-1}X^{n-1} + ... + a_1X + a_0I = \bold{0}. Note that [T]_B^k = (ADA^{-1})^k = AD^kA^{-1}. Therefore AD^nA^{-1} + a_{n-1}AD^{n-1}A^{-1} + ... + a_1ADA^{-1} + a_0AA^{-1} upon substituting this matrix into the equation. This becomes A(D^n+a_{n-1}D^{n-1}+...+a_1D+a_0I)A^{-1}. Behold, the middle expression. Computing D^k is simply raising each diagnol term to the k-th power. Therefore the middle expression consists of a diagnol matrix having its entries values to f(k_1),...,f(k_n). But f(k_1)=...=f(k_n) = 0 since they are eigenvalues. Thus, the middle matrix is the zero matrix. Thus, A(D^n+a_{n-1}D^{n-1}+...+a_1D+a_0I)A^{-1} = A\bold{0}A^{-1} = \bold{0}. And so f(T) is the zero operator.

    Note this theorem (called Cayley-Hamilton) is true even if T is a non-diagnolizable linear operation but that is much harder to prove for we cannot diagnolize.
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  3. #3
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    hmm one more question, did you change the notation of matrix representing the linear operator T to the matrix A? because that's what it seems like.
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    Quote Originally Posted by squarerootof2 View Post
    hmm one more question, did you change the notation of matrix representing the linear operator T to the matrix A? because that's what it seems like.
    I do not think so. I just wrote [T]_B = ADA^{-1}. This is of course possible since you are assuming diagnolizability*.

    *)Probably a made up word.
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  5. #5
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    oh oops, i guess you got the matrix A by assuming diagonalizability of T, i.e. that there exists A, an invertible matrix such that A^(-1)*[T]_B*A=D where D is a diagonal matrix, and just multiplied out to get [T]_B on one side by itself. is that right? i think i didnt get it at first because i'm used to using Q as the invertible matrix.

    Quote Originally Posted by ThePerfectHacker View Post
    Let T: V\to V be a linear operator and let B be a basis for V. Since T is diagnolizable if we form [T]_B the matrix is a diagnolizable, therefore [T]_B = ADA^{-1} where D is a diagnol matrix whose entires consists of eigenvalues of [T]_B. Let \det(xI - [T]_B) = f(x) = x + a_{n-1}x^{n-1}+...+a_1x+a_0 be the characheristic polynomial. If k_1,...,k_n are (repeated) eigenvalues of [T]_B then f(k_1) = ... = f(k_n) = 0. The problem asks us to show that [T]_B satisfies X^n + a_{n-1}X^{n-1} + ... + a_1X + a_0I = \bold{0}. Note that [T]_B^k = (ADA^{-1})^k = AD^kA^{-1}. Therefore AD^nA^{-1} + a_{n-1}AD^{n-1}A^{-1} + ... + a_1ADA^{-1} + a_0AA^{-1} upon substituting this matrix into the equation. This becomes A(D^n+a_{n-1}D^{n-1}+...+a_1D+a_0I)A^{-1}. Behold, the middle expression. Computing D^k is simply raising each diagnol term to the k-th power. Therefore the middle expression consists of a diagnol matrix having its entries values to f(k_1),...,f(k_n). But f(k_1)=...=f(k_n) = 0 since they are eigenvalues. Thus, the middle matrix is the zero matrix. Thus, A(D^n+a_{n-1}D^{n-1}+...+a_1D+a_0I)A^{-1} = A\bold{0}A^{-1} = \bold{0}. And so f(T) is the zero operator.

    Note this theorem (called Cayley-Hamilton) is true even if T is a non-diagnolizable linear operation but that is much harder to prove for we cannot diagnolize.
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  6. #6
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    Quote Originally Posted by squarerootof2 View Post
    oh oops, i guess you got the matrix A by assuming diagonalizability of T, i.e. that there exists A, an invertible matrix such that A^(-1)*[T]_B*A=D where D is a diagonal matrix, and just multiplied out to get [T]_B on one side by itself. is that right?
    If A^{-1} [T]_B A = D then [T]_B A = AD \implies [T]_B = ADA^{-1}.
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