Let $;A;$ be a $;2times 2-$matrix with only one eigenvalue $;x=5.;$ Show that $;(5I −A)^2 = 0.$
$begingroup$
I know that every matrix is conjugate to an upper triangle form matrix and conjugate matrices have the same characteristic polynomial.
I then try to get the characteristic polynomial of the upper triangle form and find it to be zero, however am unsure where to go from there.
linear-algebra jordan-normal-form
edited Jan 22 at 9:48
user376343
3,8483829
asked Jan 22 at 9:05
RickRick
61
$endgroup$
add a comment |
$begingroup$
I know that every matrix is conjugate to an upper triangle form matrix and conjugate matrices have the same characteristic polynomial.
I then try to get the characteristic polynomial of the upper triangle form and find it to be zero, however am unsure where to go from there.
linear-algebra jordan-normal-form
edited Jan 22 at 9:48
user376343
3,8483829
asked Jan 22 at 9:05
RickRick
61
$endgroup$
1
$begingroup$
Welcome, Rick! Think about the Cayley-Hamilton theorem!
$endgroup$
– Jose Brox
Jan 22 at 9:27
add a comment |
$begingroup$
I know that every matrix is conjugate to an upper triangle form matrix and conjugate matrices have the same characteristic polynomial.
I then try to get the characteristic polynomial of the upper triangle form and find it to be zero, however am unsure where to go from there.
linear-algebra jordan-normal-form
edited Jan 22 at 9:48
user376343
3,8483829
asked Jan 22 at 9:05
RickRick
61
$endgroup$
I know that every matrix is conjugate to an upper triangle form matrix and conjugate matrices have the same characteristic polynomial.
I then try to get the characteristic polynomial of the upper triangle form and find it to be zero, however am unsure where to go from there.
linear-algebra jordan-normal-form
linear-algebra jordan-normal-form
edited Jan 22 at 9:48
user376343
3,8483829
asked Jan 22 at 9:05
RickRick
61
edited Jan 22 at 9:48
user376343
3,8483829
asked Jan 22 at 9:05
RickRick
61
edited Jan 22 at 9:48
user376343
3,8483829
edited Jan 22 at 9:48
user376343
3,8483829
edited Jan 22 at 9:48
user376343
3,8483829
3,8483829
asked Jan 22 at 9:05
RickRick
61
asked Jan 22 at 9:05
RickRick
61
asked Jan 22 at 9:05
RickRick
61
61
1
$begingroup$
Welcome, Rick! Think about the Cayley-Hamilton theorem!
$endgroup$
– Jose Brox
Jan 22 at 9:27
add a comment |
1
$begingroup$
Welcome, Rick! Think about the Cayley-Hamilton theorem!
$endgroup$
– Jose Brox
Jan 22 at 9:27
1
1
$begingroup$
Welcome, Rick! Think about the Cayley-Hamilton theorem!
$endgroup$
– Jose Brox
Jan 22 at 9:27
$begingroup$
Welcome, Rick! Think about the Cayley-Hamilton theorem!
$endgroup$
– Jose Brox
Jan 22 at 9:27
add a comment |
1 Answer
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$begingroup$
The characteristic polynomial has only one root, namely $5$. What does this tell you about the polynomial? The only polynomials of degree $2$ with $5$ as the only root are of the form $p(x)=c(x-5)^{2}$ right? Every square matrix satisfies its characteristic polynomial and that should give you the answer.
answered Jan 22 at 9:09
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
The only confusing part is the last line, "every square matrix satisfies its characteristic polynomial". Is this because each matrix can be written in upper triangular form, with its diagonal as the eigenvalues. Therefore its characteristic polynomial simplifies to (xI-5)^2. Thanks so much for your quick reply btw!
$endgroup$
– Rick
Jan 22 at 9:27
2
$begingroup$
@Rick See en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:33
$begingroup$
So would it be correct to then say substitution of A (square matrix) into the characteristic polynomial yields the result (5I-A)^2=0. I knew of Cayley-Hamilton theorem but never realized its utilization until now. If this is so, thanks so much
$endgroup$
– Rick
Jan 22 at 9:40
$begingroup$
@Rick Yes, that is the correct argument.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:41
add a comment |
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1 Answer
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$begingroup$
The characteristic polynomial has only one root, namely $5$. What does this tell you about the polynomial? The only polynomials of degree $2$ with $5$ as the only root are of the form $p(x)=c(x-5)^{2}$ right? Every square matrix satisfies its characteristic polynomial and that should give you the answer.
answered Jan 22 at 9:09
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
The only confusing part is the last line, "every square matrix satisfies its characteristic polynomial". Is this because each matrix can be written in upper triangular form, with its diagonal as the eigenvalues. Therefore its characteristic polynomial simplifies to (xI-5)^2. Thanks so much for your quick reply btw!
$endgroup$
– Rick
Jan 22 at 9:27
2
$begingroup$
@Rick See en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:33
$begingroup$
So would it be correct to then say substitution of A (square matrix) into the characteristic polynomial yields the result (5I-A)^2=0. I knew of Cayley-Hamilton theorem but never realized its utilization until now. If this is so, thanks so much
$endgroup$
– Rick
Jan 22 at 9:40
$begingroup$
@Rick Yes, that is the correct argument.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:41
add a comment |
$begingroup$
The characteristic polynomial has only one root, namely $5$. What does this tell you about the polynomial? The only polynomials of degree $2$ with $5$ as the only root are of the form $p(x)=c(x-5)^{2}$ right? Every square matrix satisfies its characteristic polynomial and that should give you the answer.
answered Jan 22 at 9:09
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
The only confusing part is the last line, "every square matrix satisfies its characteristic polynomial". Is this because each matrix can be written in upper triangular form, with its diagonal as the eigenvalues. Therefore its characteristic polynomial simplifies to (xI-5)^2. Thanks so much for your quick reply btw!
$endgroup$
– Rick
Jan 22 at 9:27
2
$begingroup$
@Rick See en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:33
$begingroup$
So would it be correct to then say substitution of A (square matrix) into the characteristic polynomial yields the result (5I-A)^2=0. I knew of Cayley-Hamilton theorem but never realized its utilization until now. If this is so, thanks so much
$endgroup$
– Rick
Jan 22 at 9:40
$begingroup$
@Rick Yes, that is the correct argument.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:41
add a comment |
$begingroup$
The characteristic polynomial has only one root, namely $5$. What does this tell you about the polynomial? The only polynomials of degree $2$ with $5$ as the only root are of the form $p(x)=c(x-5)^{2}$ right? Every square matrix satisfies its characteristic polynomial and that should give you the answer.
answered Jan 22 at 9:09
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
The characteristic polynomial has only one root, namely $5$. What does this tell you about the polynomial? The only polynomials of degree $2$ with $5$ as the only root are of the form $p(x)=c(x-5)^{2}$ right? Every square matrix satisfies its characteristic polynomial and that should give you the answer.
answered Jan 22 at 9:09
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 22 at 9:09
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 22 at 9:09
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 22 at 9:09
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
62.9k42362
$begingroup$
The only confusing part is the last line, "every square matrix satisfies its characteristic polynomial". Is this because each matrix can be written in upper triangular form, with its diagonal as the eigenvalues. Therefore its characteristic polynomial simplifies to (xI-5)^2. Thanks so much for your quick reply btw!
$endgroup$
– Rick
Jan 22 at 9:27
2
$begingroup$
@Rick See en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:33
$begingroup$
So would it be correct to then say substitution of A (square matrix) into the characteristic polynomial yields the result (5I-A)^2=0. I knew of Cayley-Hamilton theorem but never realized its utilization until now. If this is so, thanks so much
$endgroup$
– Rick
Jan 22 at 9:40
$begingroup$
@Rick Yes, that is the correct argument.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:41
add a comment |
$begingroup$
The only confusing part is the last line, "every square matrix satisfies its characteristic polynomial". Is this because each matrix can be written in upper triangular form, with its diagonal as the eigenvalues. Therefore its characteristic polynomial simplifies to (xI-5)^2. Thanks so much for your quick reply btw!
$endgroup$
– Rick
Jan 22 at 9:27
2
$begingroup$
@Rick See en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:33
$begingroup$
So would it be correct to then say substitution of A (square matrix) into the characteristic polynomial yields the result (5I-A)^2=0. I knew of Cayley-Hamilton theorem but never realized its utilization until now. If this is so, thanks so much
$endgroup$
– Rick
Jan 22 at 9:40
$begingroup$
@Rick Yes, that is the correct argument.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:41
$begingroup$
The only confusing part is the last line, "every square matrix satisfies its characteristic polynomial". Is this because each matrix can be written in upper triangular form, with its diagonal as the eigenvalues. Therefore its characteristic polynomial simplifies to (xI-5)^2. Thanks so much for your quick reply btw!
$endgroup$
– Rick
Jan 22 at 9:27
$begingroup$
The only confusing part is the last line, "every square matrix satisfies its characteristic polynomial". Is this because each matrix can be written in upper triangular form, with its diagonal as the eigenvalues. Therefore its characteristic polynomial simplifies to (xI-5)^2. Thanks so much for your quick reply btw!
$endgroup$
– Rick
Jan 22 at 9:27
2
2
$begingroup$
@Rick See en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:33
$begingroup$
@Rick See en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:33
$begingroup$
So would it be correct to then say substitution of A (square matrix) into the characteristic polynomial yields the result (5I-A)^2=0. I knew of Cayley-Hamilton theorem but never realized its utilization until now. If this is so, thanks so much
$endgroup$
– Rick
Jan 22 at 9:40
$begingroup$
So would it be correct to then say substitution of A (square matrix) into the characteristic polynomial yields the result (5I-A)^2=0. I knew of Cayley-Hamilton theorem but never realized its utilization until now. If this is so, thanks so much
$endgroup$
– Rick
Jan 22 at 9:40
$begingroup$
@Rick Yes, that is the correct argument.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:41
$begingroup$
@Rick Yes, that is the correct argument.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:41
add a comment |
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$begingroup$
Welcome, Rick! Think about the Cayley-Hamilton theorem!
$endgroup$
– Jose Brox
Jan 22 at 9:27