1. ## 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.

2. Originally Posted by squarerootof2
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.

3. 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.

4. Originally Posted by squarerootof2
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*.

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.
If $A^{-1} [T]_B A = D$ then $[T]_B A = AD \implies [T]_B = ADA^{-1}$.