diagonalizable question
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Let $V$ be a ï¬Ânite dimensional complex vector space. Let $T : V → V$ be a diagonalizable, linear map. For each positive integer $k$ we write $T^k = T ◦···◦T$, the $k$-fold composition of the map $T$. With this notation, we can deï¬Âne $p(T) := a_0 I_V + a_1T +···+ a_nT^n$, whenever $p(x) = a_0 + a_1x + ··· + a_nx^n$ is a polynomial with complex coefficients $a_0,dots,a_n$.
Question: I already showed that $p(λ)$ is an eigenvalue of $p(T)$. How can I prove that if $β$ is a basis of $V$ then $p([T])$ respect from $β$ to $β$ is diagonalizable for any polynomial $p(x)$ with complex coefficients. Deduce that $p(T)$ is diagonalizable.
linear-algebra
edited Aug 9 at 19:01
FakeAnalyst56
175
asked Aug 9 at 18:04
Zhenqing Xu
615
add a comment |Â
up vote
1
down vote
favorite
Let $V$ be a ï¬Ânite dimensional complex vector space. Let $T : V → V$ be a diagonalizable, linear map. For each positive integer $k$ we write $T^k = T ◦···◦T$, the $k$-fold composition of the map $T$. With this notation, we can deï¬Âne $p(T) := a_0 I_V + a_1T +···+ a_nT^n$, whenever $p(x) = a_0 + a_1x + ··· + a_nx^n$ is a polynomial with complex coefficients $a_0,dots,a_n$.
Question: I already showed that $p(λ)$ is an eigenvalue of $p(T)$. How can I prove that if $β$ is a basis of $V$ then $p([T])$ respect from $β$ to $β$ is diagonalizable for any polynomial $p(x)$ with complex coefficients. Deduce that $p(T)$ is diagonalizable.
linear-algebra
edited Aug 9 at 19:01
FakeAnalyst56
175
asked Aug 9 at 18:04
Zhenqing Xu
615
$T$ diagonalizable $Rightarrow$ there is a basis of eigenvectors for $T$, but each eigenvector for $T$ is an eigenvector for $p(T)$ (with the eigenvalue $p(lambda)$) $Rightarrow$ there is a basis of eigenvectors for $p(T)$ $Rightarrow$ $p(T)$ diagonalizable.
– A.Γ.
Aug 9 at 19:20
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
Let $V$ be a ï¬Ânite dimensional complex vector space. Let $T : V → V$ be a diagonalizable, linear map. For each positive integer $k$ we write $T^k = T ◦···◦T$, the $k$-fold composition of the map $T$. With this notation, we can deï¬Âne $p(T) := a_0 I_V + a_1T +···+ a_nT^n$, whenever $p(x) = a_0 + a_1x + ··· + a_nx^n$ is a polynomial with complex coefficients $a_0,dots,a_n$.
Question: I already showed that $p(λ)$ is an eigenvalue of $p(T)$. How can I prove that if $β$ is a basis of $V$ then $p([T])$ respect from $β$ to $β$ is diagonalizable for any polynomial $p(x)$ with complex coefficients. Deduce that $p(T)$ is diagonalizable.
linear-algebra
edited Aug 9 at 19:01
FakeAnalyst56
175
asked Aug 9 at 18:04
Zhenqing Xu
615
Let $V$ be a ï¬Ânite dimensional complex vector space. Let $T : V → V$ be a diagonalizable, linear map. For each positive integer $k$ we write $T^k = T ◦···◦T$, the $k$-fold composition of the map $T$. With this notation, we can deï¬Âne $p(T) := a_0 I_V + a_1T +···+ a_nT^n$, whenever $p(x) = a_0 + a_1x + ··· + a_nx^n$ is a polynomial with complex coefficients $a_0,dots,a_n$.
Question: I already showed that $p(λ)$ is an eigenvalue of $p(T)$. How can I prove that if $β$ is a basis of $V$ then $p([T])$ respect from $β$ to $β$ is diagonalizable for any polynomial $p(x)$ with complex coefficients. Deduce that $p(T)$ is diagonalizable.
linear-algebra
edited Aug 9 at 19:01
FakeAnalyst56
175
asked Aug 9 at 18:04
Zhenqing Xu
615
edited Aug 9 at 19:01
FakeAnalyst56
175
edited Aug 9 at 19:01
FakeAnalyst56
175
edited Aug 9 at 19:01
FakeAnalyst56
175
175
asked Aug 9 at 18:04
Zhenqing Xu
615
asked Aug 9 at 18:04
Zhenqing Xu
615
asked Aug 9 at 18:04
Zhenqing Xu
615
615
$T$ diagonalizable $Rightarrow$ there is a basis of eigenvectors for $T$, but each eigenvector for $T$ is an eigenvector for $p(T)$ (with the eigenvalue $p(lambda)$) $Rightarrow$ there is a basis of eigenvectors for $p(T)$ $Rightarrow$ $p(T)$ diagonalizable.
– A.Γ.
Aug 9 at 19:20
add a comment |Â
$T$ diagonalizable $Rightarrow$ there is a basis of eigenvectors for $T$, but each eigenvector for $T$ is an eigenvector for $p(T)$ (with the eigenvalue $p(lambda)$) $Rightarrow$ there is a basis of eigenvectors for $p(T)$ $Rightarrow$ $p(T)$ diagonalizable.
– A.Γ.
Aug 9 at 19:20
$T$ diagonalizable $Rightarrow$ there is a basis of eigenvectors for $T$, but each eigenvector for $T$ is an eigenvector for $p(T)$ (with the eigenvalue $p(lambda)$) $Rightarrow$ there is a basis of eigenvectors for $p(T)$ $Rightarrow$ $p(T)$ diagonalizable.
– A.Γ.
Aug 9 at 19:20
$T$ diagonalizable $Rightarrow$ there is a basis of eigenvectors for $T$, but each eigenvector for $T$ is an eigenvector for $p(T)$ (with the eigenvalue $p(lambda)$) $Rightarrow$ there is a basis of eigenvectors for $p(T)$ $Rightarrow$ $p(T)$ diagonalizable.
– A.Γ.
Aug 9 at 19:20
add a comment |Â
1 Answer
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Let $A$ be the matrix $[T]_beta to beta$. With the notation $p(A) = a_0 I + a_1 A + a_2 A^2 + cdots + a_n A^n$, your question is asking you to show $p(A)$ is diagonalizable.
By assumption $A$ is diagonalizable, so it can be written as $PDP^-1$ for some invertible $P$ and diagonal $D$. Then $A^k = P D^k P^-1$ for any positive integer $k$. Do you see how this implies $p(A)$ is diagonalizable?
answered Aug 9 at 18:11
angryavian
34.8k12874
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $A$ be the matrix $[T]_beta to beta$. With the notation $p(A) = a_0 I + a_1 A + a_2 A^2 + cdots + a_n A^n$, your question is asking you to show $p(A)$ is diagonalizable.
By assumption $A$ is diagonalizable, so it can be written as $PDP^-1$ for some invertible $P$ and diagonal $D$. Then $A^k = P D^k P^-1$ for any positive integer $k$. Do you see how this implies $p(A)$ is diagonalizable?
answered Aug 9 at 18:11
angryavian
34.8k12874
add a comment |Â
up vote
1
down vote
Let $A$ be the matrix $[T]_beta to beta$. With the notation $p(A) = a_0 I + a_1 A + a_2 A^2 + cdots + a_n A^n$, your question is asking you to show $p(A)$ is diagonalizable.
By assumption $A$ is diagonalizable, so it can be written as $PDP^-1$ for some invertible $P$ and diagonal $D$. Then $A^k = P D^k P^-1$ for any positive integer $k$. Do you see how this implies $p(A)$ is diagonalizable?
answered Aug 9 at 18:11
angryavian
34.8k12874
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $A$ be the matrix $[T]_beta to beta$. With the notation $p(A) = a_0 I + a_1 A + a_2 A^2 + cdots + a_n A^n$, your question is asking you to show $p(A)$ is diagonalizable.
By assumption $A$ is diagonalizable, so it can be written as $PDP^-1$ for some invertible $P$ and diagonal $D$. Then $A^k = P D^k P^-1$ for any positive integer $k$. Do you see how this implies $p(A)$ is diagonalizable?
answered Aug 9 at 18:11
angryavian
34.8k12874
Let $A$ be the matrix $[T]_beta to beta$. With the notation $p(A) = a_0 I + a_1 A + a_2 A^2 + cdots + a_n A^n$, your question is asking you to show $p(A)$ is diagonalizable.
By assumption $A$ is diagonalizable, so it can be written as $PDP^-1$ for some invertible $P$ and diagonal $D$. Then $A^k = P D^k P^-1$ for any positive integer $k$. Do you see how this implies $p(A)$ is diagonalizable?
answered Aug 9 at 18:11
angryavian
34.8k12874
answered Aug 9 at 18:11
angryavian
34.8k12874
answered Aug 9 at 18:11
angryavian
34.8k12874
answered Aug 9 at 18:11
angryavian
34.8k12874
34.8k12874
add a comment |Â
add a comment |Â
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$T$ diagonalizable $Rightarrow$ there is a basis of eigenvectors for $T$, but each eigenvector for $T$ is an eigenvector for $p(T)$ (with the eigenvalue $p(lambda)$) $Rightarrow$ there is a basis of eigenvectors for $p(T)$ $Rightarrow$ $p(T)$ diagonalizable.
– A.Γ.
Aug 9 at 19:20