If $p$ & $q$ are eigenvalues of $A$ & $B$ respectively, then are $p+q$ & $pq$ that of $A+B$ & $AB$ respectively?
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I really can't get it, either a proof or a way to disprove. Suppose $AB=BA$? Does this result hold then? Here, $A$ & $B$ are any $ntimes n$ matrices.
If so then how? Please help.
linear-algebra
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up vote
0
down vote
favorite
I really can't get it, either a proof or a way to disprove. Suppose $AB=BA$? Does this result hold then? Here, $A$ & $B$ are any $ntimes n$ matrices.
If so then how? Please help.
linear-algebra
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I really can't get it, either a proof or a way to disprove. Suppose $AB=BA$? Does this result hold then? Here, $A$ & $B$ are any $ntimes n$ matrices.
If so then how? Please help.
linear-algebra
I really can't get it, either a proof or a way to disprove. Suppose $AB=BA$? Does this result hold then? Here, $A$ & $B$ are any $ntimes n$ matrices.
If so then how? Please help.
linear-algebra
linear-algebra
edited Sep 8 at 15:22
GoodDeeds
10.2k21335
10.2k21335
asked Sep 8 at 3:21
Priya Dey
453
453
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add a comment |Â
1 Answer
1
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up vote
2
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For $A+B$, say $A=beginpmatrix2&0\0&0endpmatrix$ and $B=beginpmatrix0&0\0&1endpmatrix$. Is $3$ an e-value of $A+B$?
Secondly, using the same $A$ and $B$, say. $AB=0$. Hence $2cdot1$ is not an e-value...
is there any condition under which the above is true?
â Priya Dey
Sep 8 at 6:48
I guess if $A$ and $B$ were the identity matrix, off the top of my head... maybe even multiples of the identity. I'd guess there might be other examples (where it's true).
â Chris Custer
Sep 8 at 6:58
If $p$ and $q$ were e-values with a common e-vector, say.
â Chris Custer
Sep 8 at 7:15
this means commutativity of A &B also fails to have the above property?
â Priya Dey
Sep 8 at 7:24
It won't always have it. Notice in my example above $A$ and $B$ commute. In fact $AB=BA=0$.
â Chris Custer
Sep 8 at 7:40
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
For $A+B$, say $A=beginpmatrix2&0\0&0endpmatrix$ and $B=beginpmatrix0&0\0&1endpmatrix$. Is $3$ an e-value of $A+B$?
Secondly, using the same $A$ and $B$, say. $AB=0$. Hence $2cdot1$ is not an e-value...
is there any condition under which the above is true?
â Priya Dey
Sep 8 at 6:48
I guess if $A$ and $B$ were the identity matrix, off the top of my head... maybe even multiples of the identity. I'd guess there might be other examples (where it's true).
â Chris Custer
Sep 8 at 6:58
If $p$ and $q$ were e-values with a common e-vector, say.
â Chris Custer
Sep 8 at 7:15
this means commutativity of A &B also fails to have the above property?
â Priya Dey
Sep 8 at 7:24
It won't always have it. Notice in my example above $A$ and $B$ commute. In fact $AB=BA=0$.
â Chris Custer
Sep 8 at 7:40
 |Â
show 1 more comment
up vote
2
down vote
For $A+B$, say $A=beginpmatrix2&0\0&0endpmatrix$ and $B=beginpmatrix0&0\0&1endpmatrix$. Is $3$ an e-value of $A+B$?
Secondly, using the same $A$ and $B$, say. $AB=0$. Hence $2cdot1$ is not an e-value...
is there any condition under which the above is true?
â Priya Dey
Sep 8 at 6:48
I guess if $A$ and $B$ were the identity matrix, off the top of my head... maybe even multiples of the identity. I'd guess there might be other examples (where it's true).
â Chris Custer
Sep 8 at 6:58
If $p$ and $q$ were e-values with a common e-vector, say.
â Chris Custer
Sep 8 at 7:15
this means commutativity of A &B also fails to have the above property?
â Priya Dey
Sep 8 at 7:24
It won't always have it. Notice in my example above $A$ and $B$ commute. In fact $AB=BA=0$.
â Chris Custer
Sep 8 at 7:40
 |Â
show 1 more comment
up vote
2
down vote
up vote
2
down vote
For $A+B$, say $A=beginpmatrix2&0\0&0endpmatrix$ and $B=beginpmatrix0&0\0&1endpmatrix$. Is $3$ an e-value of $A+B$?
Secondly, using the same $A$ and $B$, say. $AB=0$. Hence $2cdot1$ is not an e-value...
For $A+B$, say $A=beginpmatrix2&0\0&0endpmatrix$ and $B=beginpmatrix0&0\0&1endpmatrix$. Is $3$ an e-value of $A+B$?
Secondly, using the same $A$ and $B$, say. $AB=0$. Hence $2cdot1$ is not an e-value...
answered Sep 8 at 3:42
Chris Custer
6,6192622
6,6192622
is there any condition under which the above is true?
â Priya Dey
Sep 8 at 6:48
I guess if $A$ and $B$ were the identity matrix, off the top of my head... maybe even multiples of the identity. I'd guess there might be other examples (where it's true).
â Chris Custer
Sep 8 at 6:58
If $p$ and $q$ were e-values with a common e-vector, say.
â Chris Custer
Sep 8 at 7:15
this means commutativity of A &B also fails to have the above property?
â Priya Dey
Sep 8 at 7:24
It won't always have it. Notice in my example above $A$ and $B$ commute. In fact $AB=BA=0$.
â Chris Custer
Sep 8 at 7:40
 |Â
show 1 more comment
is there any condition under which the above is true?
â Priya Dey
Sep 8 at 6:48
I guess if $A$ and $B$ were the identity matrix, off the top of my head... maybe even multiples of the identity. I'd guess there might be other examples (where it's true).
â Chris Custer
Sep 8 at 6:58
If $p$ and $q$ were e-values with a common e-vector, say.
â Chris Custer
Sep 8 at 7:15
this means commutativity of A &B also fails to have the above property?
â Priya Dey
Sep 8 at 7:24
It won't always have it. Notice in my example above $A$ and $B$ commute. In fact $AB=BA=0$.
â Chris Custer
Sep 8 at 7:40
is there any condition under which the above is true?
â Priya Dey
Sep 8 at 6:48
is there any condition under which the above is true?
â Priya Dey
Sep 8 at 6:48
I guess if $A$ and $B$ were the identity matrix, off the top of my head... maybe even multiples of the identity. I'd guess there might be other examples (where it's true).
â Chris Custer
Sep 8 at 6:58
I guess if $A$ and $B$ were the identity matrix, off the top of my head... maybe even multiples of the identity. I'd guess there might be other examples (where it's true).
â Chris Custer
Sep 8 at 6:58
If $p$ and $q$ were e-values with a common e-vector, say.
â Chris Custer
Sep 8 at 7:15
If $p$ and $q$ were e-values with a common e-vector, say.
â Chris Custer
Sep 8 at 7:15
this means commutativity of A &B also fails to have the above property?
â Priya Dey
Sep 8 at 7:24
this means commutativity of A &B also fails to have the above property?
â Priya Dey
Sep 8 at 7:24
It won't always have it. Notice in my example above $A$ and $B$ commute. In fact $AB=BA=0$.
â Chris Custer
Sep 8 at 7:40
It won't always have it. Notice in my example above $A$ and $B$ commute. In fact $AB=BA=0$.
â Chris Custer
Sep 8 at 7:40
 |Â
show 1 more comment
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