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If given $A^n = alpha I$ prove $A$ is diagonizable
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraMinimal Polynomial of a scalar multiple of a MatrixProperties of a matrix with the minimal polynomial $m_A(t) = t^3+2t^2+t+1$?What can we conclude from the equality $m_A(x) = m_B(x)$?If $f_A(x) ne m_A(x)$ and $A^3=I$ then $A=I$?Minimal polynomial for an invertible matrix and its determinantLet $A in mathbbM_n(F)$. Every polynomial$(p)$ of degree n such that $p(A)=0$ then how $p$ can be expressed?How do you show that if a block matrix is diagonalizable, then its diagonal entries are diagonalizable?Calculate the determinant of $A-2A^-1$ given the characteristic and minimal polynomials of $A$Prove that $P^-1v$ is an eigenvector of $B$ corresponding to the eigenvalue $lambda$.Condition for linear transformation being invertible
$begingroup$
Given matrix $ A inM_n(mathbbC), 0nealphainmathbbC$ such that $A^n = alpha I$ prove that $A$ is diagonizable.
I've tried proving that the minimal polynomial($m_A$) splits into linear polynomials therefore proving that $A$ is diagonizable but haven't had great success with that.
I've also found that $|A| = alpha$ even though that didn't really give me a good starting place.
How can I prove that $A$ is diagonizable?
linear-algebra matrices diagonalization
edited yesterday
YuiTo Cheng
2,4064937
asked yesterday
Guysudai1Guysudai1
296111
$endgroup$
add a comment |
$begingroup$
Given matrix $ A inM_n(mathbbC), 0nealphainmathbbC$ such that $A^n = alpha I$ prove that $A$ is diagonizable.
I've tried proving that the minimal polynomial($m_A$) splits into linear polynomials therefore proving that $A$ is diagonizable but haven't had great success with that.
I've also found that $|A| = alpha$ even though that didn't really give me a good starting place.
How can I prove that $A$ is diagonizable?
linear-algebra matrices diagonalization
edited yesterday
YuiTo Cheng
2,4064937
asked yesterday
Guysudai1Guysudai1
296111
$endgroup$
1
$begingroup$
Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
$endgroup$
– mihaild
yesterday
1
$begingroup$
I haven't learned the Jordan form yet, sorry
$endgroup$
– Guysudai1
yesterday
add a comment |
$begingroup$
Given matrix $ A inM_n(mathbbC), 0nealphainmathbbC$ such that $A^n = alpha I$ prove that $A$ is diagonizable.
I've tried proving that the minimal polynomial($m_A$) splits into linear polynomials therefore proving that $A$ is diagonizable but haven't had great success with that.
I've also found that $|A| = alpha$ even though that didn't really give me a good starting place.
How can I prove that $A$ is diagonizable?
linear-algebra matrices diagonalization
edited yesterday
YuiTo Cheng
2,4064937
asked yesterday
Guysudai1Guysudai1
296111
$endgroup$
Given matrix $ A inM_n(mathbbC), 0nealphainmathbbC$ such that $A^n = alpha I$ prove that $A$ is diagonizable.
I've tried proving that the minimal polynomial($m_A$) splits into linear polynomials therefore proving that $A$ is diagonizable but haven't had great success with that.
I've also found that $|A| = alpha$ even though that didn't really give me a good starting place.
How can I prove that $A$ is diagonizable?
linear-algebra matrices diagonalization
linear-algebra matrices diagonalization
edited yesterday
YuiTo Cheng
2,4064937
asked yesterday
Guysudai1Guysudai1
296111
edited yesterday
YuiTo Cheng
2,4064937
asked yesterday
Guysudai1Guysudai1
296111
edited yesterday
YuiTo Cheng
2,4064937
edited yesterday
YuiTo Cheng
2,4064937
edited yesterday
YuiTo Cheng
2,4064937
2,4064937
asked yesterday
Guysudai1Guysudai1
296111
asked yesterday
Guysudai1Guysudai1
296111
asked yesterday
Guysudai1Guysudai1
296111
296111
1
$begingroup$
Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
$endgroup$
– mihaild
yesterday
1
$begingroup$
I haven't learned the Jordan form yet, sorry
$endgroup$
– Guysudai1
yesterday
add a comment |
1
$begingroup$
Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
$endgroup$
– mihaild
yesterday
1
$begingroup$
I haven't learned the Jordan form yet, sorry
$endgroup$
– Guysudai1
yesterday
1
1
$begingroup$
Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
$endgroup$
– mihaild
yesterday
$begingroup$
Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
$endgroup$
– mihaild
yesterday
1
1
$begingroup$
I haven't learned the Jordan form yet, sorry
$endgroup$
– Guysudai1
yesterday
$begingroup$
I haven't learned the Jordan form yet, sorry
$endgroup$
– Guysudai1
yesterday
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Small remark: Over $mathbbC$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.
Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.
answered yesterday
DirkDirk
4,658219
$endgroup$
2
$begingroup$
(+1) Your answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
If $A^n=alphaoperatornameId$, with $alphaneq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$beginbmatrixlambda&1&0&0&ldots&0\0&lambda&1&0&ldots&0\0&0&lambda&1&ldots&0\vdots&vdots&vdots&vdots&ddots&vdots\0&0&0&0&cdots&lambdaendbmatrix,$$with $lambdaneq0$. But no power of such a matrix is diagonlizable.
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
add a comment |
$begingroup$
Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...
answered yesterday
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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votes
active
oldest
votes
$begingroup$
Small remark: Over $mathbbC$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.
Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.
answered yesterday
DirkDirk
4,658219
$endgroup$
2
$begingroup$
(+1) Your answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
Small remark: Over $mathbbC$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.
Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.
answered yesterday
DirkDirk
4,658219
$endgroup$
2
$begingroup$
(+1) Your answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
Small remark: Over $mathbbC$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.
Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.
answered yesterday
DirkDirk
4,658219
$endgroup$
Small remark: Over $mathbbC$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.
Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.
answered yesterday
DirkDirk
4,658219
answered yesterday
DirkDirk
4,658219
answered yesterday
DirkDirk
4,658219
answered yesterday
DirkDirk
4,658219
4,658219
2
$begingroup$
(+1) Your answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
2
$begingroup$
(+1) Your answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
2
2
$begingroup$
(+1) Your answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
(+1) Your answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
If $A^n=alphaoperatornameId$, with $alphaneq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$beginbmatrixlambda&1&0&0&ldots&0\0&lambda&1&0&ldots&0\0&0&lambda&1&ldots&0\vdots&vdots&vdots&vdots&ddots&vdots\0&0&0&0&cdots&lambdaendbmatrix,$$with $lambdaneq0$. But no power of such a matrix is diagonlizable.
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
add a comment |
$begingroup$
If $A^n=alphaoperatornameId$, with $alphaneq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$beginbmatrixlambda&1&0&0&ldots&0\0&lambda&1&0&ldots&0\0&0&lambda&1&ldots&0\vdots&vdots&vdots&vdots&ddots&vdots\0&0&0&0&cdots&lambdaendbmatrix,$$with $lambdaneq0$. But no power of such a matrix is diagonlizable.
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
add a comment |
$begingroup$
If $A^n=alphaoperatornameId$, with $alphaneq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$beginbmatrixlambda&1&0&0&ldots&0\0&lambda&1&0&ldots&0\0&0&lambda&1&ldots&0\vdots&vdots&vdots&vdots&ddots&vdots\0&0&0&0&cdots&lambdaendbmatrix,$$with $lambdaneq0$. But no power of such a matrix is diagonlizable.
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
If $A^n=alphaoperatornameId$, with $alphaneq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$beginbmatrixlambda&1&0&0&ldots&0\0&lambda&1&0&ldots&0\0&0&lambda&1&ldots&0\vdots&vdots&vdots&vdots&ddots&vdots\0&0&0&0&cdots&lambdaendbmatrix,$$with $lambdaneq0$. But no power of such a matrix is diagonlizable.
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23133243
edited yesterday
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23133243
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23133243
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23133243
174k23133243
add a comment |
add a comment |
$begingroup$
Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...
answered yesterday
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
$endgroup$
add a comment |
$begingroup$
Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...
answered yesterday
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
$endgroup$
add a comment |
$begingroup$
Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...
answered yesterday
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
$endgroup$
Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...
answered yesterday
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
answered yesterday
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
answered yesterday
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
answered yesterday
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
35.5k42972
add a comment |
add a comment |
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-diagonalization, linear-algebra, matrices
1
$begingroup$
Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
$endgroup$
– mihaild
yesterday
1
$begingroup$
I haven't learned the Jordan form yet, sorry
$endgroup$
– Guysudai1
yesterday