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Counterexamples Invertibility [closed]
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If A is an invertible matrix,
is A^t always invertible? If t is a fractional exponent, would it guarantee invertibility?
linear-algebra matrices vector-spaces inverse
asked Jul 22 at 8:44
IvyGatekeeper
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closed as unclear what you're asking by Shailesh, Arnaud D., Tyrone, José Carlos Santos, Isaac Browne Jul 31 at 16:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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down vote
favorite
If A is an invertible matrix,
is A^t always invertible? If t is a fractional exponent, would it guarantee invertibility?
linear-algebra matrices vector-spaces inverse
asked Jul 22 at 8:44
IvyGatekeeper
1
closed as unclear what you're asking by Shailesh, Arnaud D., Tyrone, José Carlos Santos, Isaac Browne Jul 31 at 16:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
@Suzet can you clarify the introduction of B matrix? I was wondering based on A^t alone without introducing another matrix B. So for every t is A^t an invertible matrix given that A is already invertible
– IvyGatekeeper
Jul 22 at 8:46
My bad IvyGatekeeper, I erased my comment after I noticed I misread your question ($A^t$ made me think about the transposed matrix of $A$ actually). Please, do not mind what I wrote.
– Suzet
Jul 22 at 8:50
As Suzet says, $A^t$ for matrices almost always means the transpose, not a power.
– Empy2
Jul 22 at 12:06
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If A is an invertible matrix,
is A^t always invertible? If t is a fractional exponent, would it guarantee invertibility?
linear-algebra matrices vector-spaces inverse
asked Jul 22 at 8:44
IvyGatekeeper
1
If A is an invertible matrix,
is A^t always invertible? If t is a fractional exponent, would it guarantee invertibility?
linear-algebra matrices vector-spaces inverse
asked Jul 22 at 8:44
IvyGatekeeper
1
asked Jul 22 at 8:44
IvyGatekeeper
1
asked Jul 22 at 8:44
IvyGatekeeper
1
asked Jul 22 at 8:44
IvyGatekeeper
1
1
closed as unclear what you're asking by Shailesh, Arnaud D., Tyrone, José Carlos Santos, Isaac Browne Jul 31 at 16:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Shailesh, Arnaud D., Tyrone, José Carlos Santos, Isaac Browne Jul 31 at 16:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
@Suzet can you clarify the introduction of B matrix? I was wondering based on A^t alone without introducing another matrix B. So for every t is A^t an invertible matrix given that A is already invertible
– IvyGatekeeper
Jul 22 at 8:46
My bad IvyGatekeeper, I erased my comment after I noticed I misread your question ($A^t$ made me think about the transposed matrix of $A$ actually). Please, do not mind what I wrote.
– Suzet
Jul 22 at 8:50
As Suzet says, $A^t$ for matrices almost always means the transpose, not a power.
– Empy2
Jul 22 at 12:06
add a comment |Â
@Suzet can you clarify the introduction of B matrix? I was wondering based on A^t alone without introducing another matrix B. So for every t is A^t an invertible matrix given that A is already invertible
– IvyGatekeeper
Jul 22 at 8:46
My bad IvyGatekeeper, I erased my comment after I noticed I misread your question ($A^t$ made me think about the transposed matrix of $A$ actually). Please, do not mind what I wrote.
– Suzet
Jul 22 at 8:50
As Suzet says, $A^t$ for matrices almost always means the transpose, not a power.
– Empy2
Jul 22 at 12:06
@Suzet can you clarify the introduction of B matrix? I was wondering based on A^t alone without introducing another matrix B. So for every t is A^t an invertible matrix given that A is already invertible
– IvyGatekeeper
Jul 22 at 8:46
@Suzet can you clarify the introduction of B matrix? I was wondering based on A^t alone without introducing another matrix B. So for every t is A^t an invertible matrix given that A is already invertible
– IvyGatekeeper
Jul 22 at 8:46
My bad IvyGatekeeper, I erased my comment after I noticed I misread your question ($A^t$ made me think about the transposed matrix of $A$ actually). Please, do not mind what I wrote.
– Suzet
Jul 22 at 8:50
My bad IvyGatekeeper, I erased my comment after I noticed I misread your question ($A^t$ made me think about the transposed matrix of $A$ actually). Please, do not mind what I wrote.
– Suzet
Jul 22 at 8:50
As Suzet says, $A^t$ for matrices almost always means the transpose, not a power.
– Empy2
Jul 22 at 12:06
As Suzet says, $A^t$ for matrices almost always means the transpose, not a power.
– Empy2
Jul 22 at 12:06
add a comment |Â
1 Answer
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Let $t=p/q$.
Let $B = A^t$, i.e. $B^q = A^p$. Then, $B^q (A^-1)^p = A^p A^-p = I_n$, so $B(B^q-1A^-p) = I_n$, so $B$ is invertible, and $B^q-1A^-p$ is its inverse.
Caveat: in general, $A^p/q$ is not defined; if it is, it is not uniquely defined.
answered Jul 22 at 8:46
Kenny Lau
18.7k2157
I guess to clarify my question, will A^t, for any t, be invertible given that A is an invertible matrix. Thank You
– IvyGatekeeper
Jul 22 at 8:48
Did I not just prove that?
– Kenny Lau
Jul 22 at 8:49
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $t=p/q$.
Let $B = A^t$, i.e. $B^q = A^p$. Then, $B^q (A^-1)^p = A^p A^-p = I_n$, so $B(B^q-1A^-p) = I_n$, so $B$ is invertible, and $B^q-1A^-p$ is its inverse.
Caveat: in general, $A^p/q$ is not defined; if it is, it is not uniquely defined.
answered Jul 22 at 8:46
Kenny Lau
18.7k2157
I guess to clarify my question, will A^t, for any t, be invertible given that A is an invertible matrix. Thank You
– IvyGatekeeper
Jul 22 at 8:48
Did I not just prove that?
– Kenny Lau
Jul 22 at 8:49
add a comment |Â
up vote
1
down vote
Let $t=p/q$.
Let $B = A^t$, i.e. $B^q = A^p$. Then, $B^q (A^-1)^p = A^p A^-p = I_n$, so $B(B^q-1A^-p) = I_n$, so $B$ is invertible, and $B^q-1A^-p$ is its inverse.
Caveat: in general, $A^p/q$ is not defined; if it is, it is not uniquely defined.
answered Jul 22 at 8:46
Kenny Lau
18.7k2157
I guess to clarify my question, will A^t, for any t, be invertible given that A is an invertible matrix. Thank You
– IvyGatekeeper
Jul 22 at 8:48
Did I not just prove that?
– Kenny Lau
Jul 22 at 8:49
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $t=p/q$.
Let $B = A^t$, i.e. $B^q = A^p$. Then, $B^q (A^-1)^p = A^p A^-p = I_n$, so $B(B^q-1A^-p) = I_n$, so $B$ is invertible, and $B^q-1A^-p$ is its inverse.
Caveat: in general, $A^p/q$ is not defined; if it is, it is not uniquely defined.
answered Jul 22 at 8:46
Kenny Lau
18.7k2157
Let $t=p/q$.
Let $B = A^t$, i.e. $B^q = A^p$. Then, $B^q (A^-1)^p = A^p A^-p = I_n$, so $B(B^q-1A^-p) = I_n$, so $B$ is invertible, and $B^q-1A^-p$ is its inverse.
Caveat: in general, $A^p/q$ is not defined; if it is, it is not uniquely defined.
answered Jul 22 at 8:46
Kenny Lau
18.7k2157
answered Jul 22 at 8:46
Kenny Lau
18.7k2157
answered Jul 22 at 8:46
Kenny Lau
18.7k2157
answered Jul 22 at 8:46
Kenny Lau
18.7k2157
18.7k2157
I guess to clarify my question, will A^t, for any t, be invertible given that A is an invertible matrix. Thank You
– IvyGatekeeper
Jul 22 at 8:48
Did I not just prove that?
– Kenny Lau
Jul 22 at 8:49
add a comment |Â
I guess to clarify my question, will A^t, for any t, be invertible given that A is an invertible matrix. Thank You
– IvyGatekeeper
Jul 22 at 8:48
Did I not just prove that?
– Kenny Lau
Jul 22 at 8:49
I guess to clarify my question, will A^t, for any t, be invertible given that A is an invertible matrix. Thank You
– IvyGatekeeper
Jul 22 at 8:48
I guess to clarify my question, will A^t, for any t, be invertible given that A is an invertible matrix. Thank You
– IvyGatekeeper
Jul 22 at 8:48
Did I not just prove that?
– Kenny Lau
Jul 22 at 8:49
Did I not just prove that?
– Kenny Lau
Jul 22 at 8:49
add a comment |Â
@Suzet can you clarify the introduction of B matrix? I was wondering based on A^t alone without introducing another matrix B. So for every t is A^t an invertible matrix given that A is already invertible
– IvyGatekeeper
Jul 22 at 8:46
My bad IvyGatekeeper, I erased my comment after I noticed I misread your question ($A^t$ made me think about the transposed matrix of $A$ actually). Please, do not mind what I wrote.
– Suzet
Jul 22 at 8:50
As Suzet says, $A^t$ for matrices almost always means the transpose, not a power.
– Empy2
Jul 22 at 12:06