Are Jordan “Formable” matrices closed under multiplication?
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After learning about Jordan Canonical Form, I began thinking about if you have two matrices $A, B in M_n(mathbb{R}),$ whether or not their product $AB$ will also have a Jordan Canonical Form. Trivially, if we consider polynomials in $ M_n(mathbb{C})$ then our polynomial will always have roots in the field. So, from what I understand, this is equivalent to asking whether or not the characteristic polynomial of $AB$ will have real roots.
I began playing around with the $2 times 2$ case and if we have that $AB$ has a negative determinant then we can't have that the eigenvalues of $AB$ are in $mathbb{C},$ since they must multiply to the determinant, but since the complex eigenvalues always come in a conjugate pair, that would mean that their product is always positive. I am not sure how I can break down the case if the determinant is positive.
Finally, my own intuition says matrices which have Jordan form will be closed under multiplication. This is because if we think about what matrices with strictly, real eigenvalues do, they simply reflect and stretch space. So I don't believe it is possible to find two matrices whose product matrix will be a rotation in the plane.
Also, I couldn't find this question asked elsewhere, but if it has an answer, I'd be happy to read it.
linear-algebra abstract-algebra
New contributor
add a comment |
up vote
5
down vote
favorite
After learning about Jordan Canonical Form, I began thinking about if you have two matrices $A, B in M_n(mathbb{R}),$ whether or not their product $AB$ will also have a Jordan Canonical Form. Trivially, if we consider polynomials in $ M_n(mathbb{C})$ then our polynomial will always have roots in the field. So, from what I understand, this is equivalent to asking whether or not the characteristic polynomial of $AB$ will have real roots.
I began playing around with the $2 times 2$ case and if we have that $AB$ has a negative determinant then we can't have that the eigenvalues of $AB$ are in $mathbb{C},$ since they must multiply to the determinant, but since the complex eigenvalues always come in a conjugate pair, that would mean that their product is always positive. I am not sure how I can break down the case if the determinant is positive.
Finally, my own intuition says matrices which have Jordan form will be closed under multiplication. This is because if we think about what matrices with strictly, real eigenvalues do, they simply reflect and stretch space. So I don't believe it is possible to find two matrices whose product matrix will be a rotation in the plane.
Also, I couldn't find this question asked elsewhere, but if it has an answer, I'd be happy to read it.
linear-algebra abstract-algebra
New contributor
3
Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
– I like Serena
Nov 15 at 22:33
2
@IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
– Travis
Nov 16 at 0:13
2
Thank you @Travis. Done.
– I like Serena
Nov 16 at 0:20
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
After learning about Jordan Canonical Form, I began thinking about if you have two matrices $A, B in M_n(mathbb{R}),$ whether or not their product $AB$ will also have a Jordan Canonical Form. Trivially, if we consider polynomials in $ M_n(mathbb{C})$ then our polynomial will always have roots in the field. So, from what I understand, this is equivalent to asking whether or not the characteristic polynomial of $AB$ will have real roots.
I began playing around with the $2 times 2$ case and if we have that $AB$ has a negative determinant then we can't have that the eigenvalues of $AB$ are in $mathbb{C},$ since they must multiply to the determinant, but since the complex eigenvalues always come in a conjugate pair, that would mean that their product is always positive. I am not sure how I can break down the case if the determinant is positive.
Finally, my own intuition says matrices which have Jordan form will be closed under multiplication. This is because if we think about what matrices with strictly, real eigenvalues do, they simply reflect and stretch space. So I don't believe it is possible to find two matrices whose product matrix will be a rotation in the plane.
Also, I couldn't find this question asked elsewhere, but if it has an answer, I'd be happy to read it.
linear-algebra abstract-algebra
New contributor
After learning about Jordan Canonical Form, I began thinking about if you have two matrices $A, B in M_n(mathbb{R}),$ whether or not their product $AB$ will also have a Jordan Canonical Form. Trivially, if we consider polynomials in $ M_n(mathbb{C})$ then our polynomial will always have roots in the field. So, from what I understand, this is equivalent to asking whether or not the characteristic polynomial of $AB$ will have real roots.
I began playing around with the $2 times 2$ case and if we have that $AB$ has a negative determinant then we can't have that the eigenvalues of $AB$ are in $mathbb{C},$ since they must multiply to the determinant, but since the complex eigenvalues always come in a conjugate pair, that would mean that their product is always positive. I am not sure how I can break down the case if the determinant is positive.
Finally, my own intuition says matrices which have Jordan form will be closed under multiplication. This is because if we think about what matrices with strictly, real eigenvalues do, they simply reflect and stretch space. So I don't believe it is possible to find two matrices whose product matrix will be a rotation in the plane.
Also, I couldn't find this question asked elsewhere, but if it has an answer, I'd be happy to read it.
linear-algebra abstract-algebra
linear-algebra abstract-algebra
New contributor
New contributor
New contributor
asked Nov 15 at 21:00
Josabanks
384
384
New contributor
New contributor
3
Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
– I like Serena
Nov 15 at 22:33
2
@IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
– Travis
Nov 16 at 0:13
2
Thank you @Travis. Done.
– I like Serena
Nov 16 at 0:20
add a comment |
3
Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
– I like Serena
Nov 15 at 22:33
2
@IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
– Travis
Nov 16 at 0:13
2
Thank you @Travis. Done.
– I like Serena
Nov 16 at 0:20
3
3
Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
– I like Serena
Nov 15 at 22:33
Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
– I like Serena
Nov 15 at 22:33
2
2
@IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
– Travis
Nov 16 at 0:13
@IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
– Travis
Nov 16 at 0:13
2
2
Thank you @Travis. Done.
– I like Serena
Nov 16 at 0:20
Thank you @Travis. Done.
– I like Serena
Nov 16 at 0:20
add a comment |
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
This is false already in the $2 times 2$ case. Consider the matrices
$$A = left[ begin{array}{cc} -1 & 2 \ -2 & 3 end{array} right], B = left[ begin{array}{cc} 3 & 0 \ 0 & 1 end{array} right]$$
$B$ has eigenvalues $3, 1$, and the characteristic polynomial of $A$ is $t^2 - 2t + 1 = (t - 1)^2$, so $A$ has eigenvalues $1, 1$. But
$$AB = left[ begin{array}{cc} -3 & 2 \ -6 & 3 end{array} right]$$
has characteristic polynomial $t^2 + 3$, so has eigenvalues $pm i sqrt{3}$.
I found these matrices by first taking $B$ to be diagonal with diagonal entries $s, 1$ and then looked at the effect multiplication by $B$ had on the characteristic polynomial of a generic $A$. It's possible to choose $s$ so that $AB$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.
add a comment |
up vote
7
down vote
Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.
So the product $AB$ of two real matrices $A$ and $B$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.
1
Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
– Daniel Schepler
Nov 16 at 1:35
I've clarified that it's 'generally'.
– I like Serena
Nov 16 at 10:28
add a comment |
up vote
3
down vote
Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).
If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Consequently, every square matrix is a product of two symmetric matrices (because $A=(SA^T)S^{-1}$) and one of them is non-singular.
It follows immediately that the answer to your question is negative unless $n=1$.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
This is false already in the $2 times 2$ case. Consider the matrices
$$A = left[ begin{array}{cc} -1 & 2 \ -2 & 3 end{array} right], B = left[ begin{array}{cc} 3 & 0 \ 0 & 1 end{array} right]$$
$B$ has eigenvalues $3, 1$, and the characteristic polynomial of $A$ is $t^2 - 2t + 1 = (t - 1)^2$, so $A$ has eigenvalues $1, 1$. But
$$AB = left[ begin{array}{cc} -3 & 2 \ -6 & 3 end{array} right]$$
has characteristic polynomial $t^2 + 3$, so has eigenvalues $pm i sqrt{3}$.
I found these matrices by first taking $B$ to be diagonal with diagonal entries $s, 1$ and then looked at the effect multiplication by $B$ had on the characteristic polynomial of a generic $A$. It's possible to choose $s$ so that $AB$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.
add a comment |
up vote
6
down vote
accepted
This is false already in the $2 times 2$ case. Consider the matrices
$$A = left[ begin{array}{cc} -1 & 2 \ -2 & 3 end{array} right], B = left[ begin{array}{cc} 3 & 0 \ 0 & 1 end{array} right]$$
$B$ has eigenvalues $3, 1$, and the characteristic polynomial of $A$ is $t^2 - 2t + 1 = (t - 1)^2$, so $A$ has eigenvalues $1, 1$. But
$$AB = left[ begin{array}{cc} -3 & 2 \ -6 & 3 end{array} right]$$
has characteristic polynomial $t^2 + 3$, so has eigenvalues $pm i sqrt{3}$.
I found these matrices by first taking $B$ to be diagonal with diagonal entries $s, 1$ and then looked at the effect multiplication by $B$ had on the characteristic polynomial of a generic $A$. It's possible to choose $s$ so that $AB$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
This is false already in the $2 times 2$ case. Consider the matrices
$$A = left[ begin{array}{cc} -1 & 2 \ -2 & 3 end{array} right], B = left[ begin{array}{cc} 3 & 0 \ 0 & 1 end{array} right]$$
$B$ has eigenvalues $3, 1$, and the characteristic polynomial of $A$ is $t^2 - 2t + 1 = (t - 1)^2$, so $A$ has eigenvalues $1, 1$. But
$$AB = left[ begin{array}{cc} -3 & 2 \ -6 & 3 end{array} right]$$
has characteristic polynomial $t^2 + 3$, so has eigenvalues $pm i sqrt{3}$.
I found these matrices by first taking $B$ to be diagonal with diagonal entries $s, 1$ and then looked at the effect multiplication by $B$ had on the characteristic polynomial of a generic $A$. It's possible to choose $s$ so that $AB$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.
This is false already in the $2 times 2$ case. Consider the matrices
$$A = left[ begin{array}{cc} -1 & 2 \ -2 & 3 end{array} right], B = left[ begin{array}{cc} 3 & 0 \ 0 & 1 end{array} right]$$
$B$ has eigenvalues $3, 1$, and the characteristic polynomial of $A$ is $t^2 - 2t + 1 = (t - 1)^2$, so $A$ has eigenvalues $1, 1$. But
$$AB = left[ begin{array}{cc} -3 & 2 \ -6 & 3 end{array} right]$$
has characteristic polynomial $t^2 + 3$, so has eigenvalues $pm i sqrt{3}$.
I found these matrices by first taking $B$ to be diagonal with diagonal entries $s, 1$ and then looked at the effect multiplication by $B$ had on the characteristic polynomial of a generic $A$. It's possible to choose $s$ so that $AB$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.
edited Nov 16 at 0:23
Trevor Gunn
13.6k32045
13.6k32045
answered Nov 15 at 22:25
Qiaochu Yuan
274k32578914
274k32578914
add a comment |
add a comment |
up vote
7
down vote
Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.
So the product $AB$ of two real matrices $A$ and $B$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.
1
Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
– Daniel Schepler
Nov 16 at 1:35
I've clarified that it's 'generally'.
– I like Serena
Nov 16 at 10:28
add a comment |
up vote
7
down vote
Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.
So the product $AB$ of two real matrices $A$ and $B$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.
1
Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
– Daniel Schepler
Nov 16 at 1:35
I've clarified that it's 'generally'.
– I like Serena
Nov 16 at 10:28
add a comment |
up vote
7
down vote
up vote
7
down vote
Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.
So the product $AB$ of two real matrices $A$ and $B$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.
Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.
So the product $AB$ of two real matrices $A$ and $B$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.
edited Nov 16 at 10:27
answered Nov 16 at 0:19
I like Serena
3,1981718
3,1981718
1
Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
– Daniel Schepler
Nov 16 at 1:35
I've clarified that it's 'generally'.
– I like Serena
Nov 16 at 10:28
add a comment |
1
Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
– Daniel Schepler
Nov 16 at 1:35
I've clarified that it's 'generally'.
– I like Serena
Nov 16 at 10:28
1
1
Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
– Daniel Schepler
Nov 16 at 1:35
Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
– Daniel Schepler
Nov 16 at 1:35
I've clarified that it's 'generally'.
– I like Serena
Nov 16 at 10:28
I've clarified that it's 'generally'.
– I like Serena
Nov 16 at 10:28
add a comment |
up vote
3
down vote
Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).
If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Consequently, every square matrix is a product of two symmetric matrices (because $A=(SA^T)S^{-1}$) and one of them is non-singular.
It follows immediately that the answer to your question is negative unless $n=1$.
add a comment |
up vote
3
down vote
Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).
If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Consequently, every square matrix is a product of two symmetric matrices (because $A=(SA^T)S^{-1}$) and one of them is non-singular.
It follows immediately that the answer to your question is negative unless $n=1$.
add a comment |
up vote
3
down vote
up vote
3
down vote
Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).
If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Consequently, every square matrix is a product of two symmetric matrices (because $A=(SA^T)S^{-1}$) and one of them is non-singular.
It follows immediately that the answer to your question is negative unless $n=1$.
Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).
If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Consequently, every square matrix is a product of two symmetric matrices (because $A=(SA^T)S^{-1}$) and one of them is non-singular.
It follows immediately that the answer to your question is negative unless $n=1$.
edited Nov 16 at 15:44
answered Nov 16 at 10:28
user1551
70.2k566125
70.2k566125
add a comment |
add a comment |
Josabanks is a new contributor. Be nice, and check out our Code of Conduct.
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3
Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
– I like Serena
Nov 15 at 22:33
2
@IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
– Travis
Nov 16 at 0:13
2
Thank you @Travis. Done.
– I like Serena
Nov 16 at 0:20