prove that the matrix A is diagonalizableBlock Diagonal Matrix DiagonalizableNew proof about normal matrix is diagonalizable.Show that matrix $A$ is NOT diagonalizable.Prove a matrix is not diagonalizableHow to use inner products in C(n) to prove normal matrix is unitarily diagonalizable after knowing that normal matrix is diagonalizable?Is the Matrix Diagonalizable if $A^2=4I$Prove that $A$ is diagonalizable.Prove that a general matrix is diagonalizableDetermine $a$ to make matrix $A$ diagonalizableDiagonalizable block-diagonal matrix
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prove that the matrix A is diagonalizable
Block Diagonal Matrix DiagonalizableNew proof about normal matrix is diagonalizable.Show that matrix $A$ is NOT diagonalizable.Prove a matrix is not diagonalizableHow to use inner products in C(n) to prove normal matrix is unitarily diagonalizable after knowing that normal matrix is diagonalizable?Is the Matrix Diagonalizable if $A^2=4I$Prove that $A$ is diagonalizable.Prove that a general matrix is diagonalizableDetermine $a$ to make matrix $A$ diagonalizableDiagonalizable block-diagonal matrix
$begingroup$
We have :
$A^3-3A^2-A+3I_n = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked 1 hour ago
JoshuaKJoshuaK
254
$endgroup$
add a comment |
$begingroup$
We have :
$A^3-3A^2-A+3I_n = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked 1 hour ago
JoshuaKJoshuaK
254
$endgroup$
1
$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
1 hour ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
1 hour ago
add a comment |
$begingroup$
We have :
$A^3-3A^2-A+3I_n = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked 1 hour ago
JoshuaKJoshuaK
254
$endgroup$
We have :
$A^3-3A^2-A+3I_n = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked 1 hour ago
JoshuaKJoshuaK
254
asked 1 hour ago
JoshuaKJoshuaK
254
edited 1 hour ago
JoshuaK
asked 1 hour ago
JoshuaKJoshuaK
254
asked 1 hour ago
JoshuaKJoshuaK
254
asked 1 hour ago
JoshuaKJoshuaK
254
254
1
$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
1 hour ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
1 hour ago
add a comment |
1
$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
1 hour ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
1 hour ago
1
1
$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
1 hour ago
$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
1 hour ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
1 hour ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.
answered 1 hour ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 59 mins ago
EricEric
513
$endgroup$
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 1 hour ago
GSoferGSofer
8631313
$endgroup$
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
43 mins ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.
answered 1 hour ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
add a comment |
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.
answered 1 hour ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
add a comment |
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.
answered 1 hour ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.
answered 1 hour ago
TheSilverDoeTheSilverDoe
5,324215
answered 1 hour ago
TheSilverDoeTheSilverDoe
5,324215
answered 1 hour ago
TheSilverDoeTheSilverDoe
5,324215
answered 1 hour ago
TheSilverDoeTheSilverDoe
5,324215
5,324215
add a comment |
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 59 mins ago
EricEric
513
$endgroup$
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 59 mins ago
EricEric
513
$endgroup$
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 59 mins ago
EricEric
513
$endgroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 59 mins ago
EricEric
513
answered 59 mins ago
EricEric
513
answered 59 mins ago
EricEric
513
answered 59 mins ago
EricEric
513
513
add a comment |
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 1 hour ago
GSoferGSofer
8631313
$endgroup$
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
43 mins ago
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 1 hour ago
GSoferGSofer
8631313
$endgroup$
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
43 mins ago
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 1 hour ago
GSoferGSofer
8631313
$endgroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 1 hour ago
GSoferGSofer
8631313
answered 1 hour ago
GSoferGSofer
8631313
answered 1 hour ago
GSoferGSofer
8631313
answered 1 hour ago
GSoferGSofer
8631313
8631313
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
43 mins ago
add a comment |
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
43 mins ago
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
43 mins ago
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
43 mins ago
add a comment |
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$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
1 hour ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
1 hour ago