Finding Minimal polynomial from Characteristic polynomial
If $T$ is a linear operator with characteristic polynomial $(x^2-1)^6$ such that $mathrm{rank}(T-I) = 9$, $mathrm{rank}(T-I)^2 = 7$, $mathrm{rank}(T +I) = 10$ and $mathrm{rank}(T +I)^2 = 9$, find the minimal polynomial of T?
Let $V$ be the vector space of functions on the real line spanned by the basis $B = lbrace sin x,cos x,xsin x,xcos xrbrace$. Let $T$ be defined by $Tf = f''$, where $f''$ denotes the second derivative of $f$ with respect to $x$. Determine the minimal polynomial of $T$?
I tried for both the problems and I am guessing (not 100% sure) that the Minimal polynomial will be $(x-1)^3(x+1)^2$ for Question 1.
For Question 2, characteristic polynomial will be $(-1-lambda)³(-1-lambda)$ and Minimal polynomial will be $(-1-λ)^2$.
Kindly verify and let me know if I am correct or wrong. Thanks!!
linear-algebra minimal-polynomials
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If $T$ is a linear operator with characteristic polynomial $(x^2-1)^6$ such that $mathrm{rank}(T-I) = 9$, $mathrm{rank}(T-I)^2 = 7$, $mathrm{rank}(T +I) = 10$ and $mathrm{rank}(T +I)^2 = 9$, find the minimal polynomial of T?
Let $V$ be the vector space of functions on the real line spanned by the basis $B = lbrace sin x,cos x,xsin x,xcos xrbrace$. Let $T$ be defined by $Tf = f''$, where $f''$ denotes the second derivative of $f$ with respect to $x$. Determine the minimal polynomial of $T$?
I tried for both the problems and I am guessing (not 100% sure) that the Minimal polynomial will be $(x-1)^3(x+1)^2$ for Question 1.
For Question 2, characteristic polynomial will be $(-1-lambda)³(-1-lambda)$ and Minimal polynomial will be $(-1-λ)^2$.
Kindly verify and let me know if I am correct or wrong. Thanks!!
linear-algebra minimal-polynomials
What is the minimal polynomial of a linear operator?
– Dionel Jaime
Dec 9 at 22:19
add a comment |
If $T$ is a linear operator with characteristic polynomial $(x^2-1)^6$ such that $mathrm{rank}(T-I) = 9$, $mathrm{rank}(T-I)^2 = 7$, $mathrm{rank}(T +I) = 10$ and $mathrm{rank}(T +I)^2 = 9$, find the minimal polynomial of T?
Let $V$ be the vector space of functions on the real line spanned by the basis $B = lbrace sin x,cos x,xsin x,xcos xrbrace$. Let $T$ be defined by $Tf = f''$, where $f''$ denotes the second derivative of $f$ with respect to $x$. Determine the minimal polynomial of $T$?
I tried for both the problems and I am guessing (not 100% sure) that the Minimal polynomial will be $(x-1)^3(x+1)^2$ for Question 1.
For Question 2, characteristic polynomial will be $(-1-lambda)³(-1-lambda)$ and Minimal polynomial will be $(-1-λ)^2$.
Kindly verify and let me know if I am correct or wrong. Thanks!!
linear-algebra minimal-polynomials
If $T$ is a linear operator with characteristic polynomial $(x^2-1)^6$ such that $mathrm{rank}(T-I) = 9$, $mathrm{rank}(T-I)^2 = 7$, $mathrm{rank}(T +I) = 10$ and $mathrm{rank}(T +I)^2 = 9$, find the minimal polynomial of T?
Let $V$ be the vector space of functions on the real line spanned by the basis $B = lbrace sin x,cos x,xsin x,xcos xrbrace$. Let $T$ be defined by $Tf = f''$, where $f''$ denotes the second derivative of $f$ with respect to $x$. Determine the minimal polynomial of $T$?
I tried for both the problems and I am guessing (not 100% sure) that the Minimal polynomial will be $(x-1)^3(x+1)^2$ for Question 1.
For Question 2, characteristic polynomial will be $(-1-lambda)³(-1-lambda)$ and Minimal polynomial will be $(-1-λ)^2$.
Kindly verify and let me know if I am correct or wrong. Thanks!!
linear-algebra minimal-polynomials
linear-algebra minimal-polynomials
edited Dec 9 at 19:36
Fakemistake
1,682815
1,682815
asked Dec 9 at 18:48
Ashis Jana
22
22
What is the minimal polynomial of a linear operator?
– Dionel Jaime
Dec 9 at 22:19
add a comment |
What is the minimal polynomial of a linear operator?
– Dionel Jaime
Dec 9 at 22:19
What is the minimal polynomial of a linear operator?
– Dionel Jaime
Dec 9 at 22:19
What is the minimal polynomial of a linear operator?
– Dionel Jaime
Dec 9 at 22:19
add a comment |
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What is the minimal polynomial of a linear operator?
– Dionel Jaime
Dec 9 at 22:19