Prove or disprove: If $A$ is $ntimes n$ and $exists;min BbbN:;A^m=I_n$, then $A$ is invertible. The Next CEO of Stack OverflowDeterminants of matrices; Does the equation $det M=det(I_n-M)$ have a solution?If $A^2+2A+I_n=O_n$ then $A$ is invertibleDeterminant of a $2 times 2$ block matrixprove that if a square matrix $A$ is invertible then $AA^T$ is invertible.Matrices such that $M^2+M^T=I_n$ are invertibleIf a matrix is row equivalent to some invertible matrix then it is invertibleProve that $A+I_n$ is invertible, where $leftlVert ArightrVert<1$Is it true that $m=nimplies A$ is invertible, for an $mtimes n$ matrix satisfying $(AA^T)^r=I$Help with proof or counterexample: $A^3=0 implies I_n+A$ is invertibleThe matrix $I_n - v^t x$ is invertible when $langle v, x rangle neq 1$
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Prove or disprove: If $A$ is $ntimes n$ and $exists;min BbbN:;A^m=I_n$, then $A$ is invertible.
The Next CEO of Stack OverflowDeterminants of matrices; Does the equation $det M=det(I_n-M)$ have a solution?If $A^2+2A+I_n=O_n$ then $A$ is invertibleDeterminant of a $2 times 2$ block matrixprove that if a square matrix $A$ is invertible then $AA^T$ is invertible.Matrices such that $M^2+M^T=I_n$ are invertibleIf a matrix is row equivalent to some invertible matrix then it is invertibleProve that $A+I_n$ is invertible, where $leftlVert ArightrVert<1$Is it true that $m=nimplies A$ is invertible, for an $mtimes n$ matrix satisfying $(AA^T)^r=I$Help with proof or counterexample: $A^3=0 implies I_n+A$ is invertibleThe matrix $I_n - v^t x$ is invertible when $langle v, x rangle neq 1$
$begingroup$
Is this statement true? If $A$ is an $ntimes n$ matrix and $A^m=I_n$ for some $min BbbN$, then $A$ is invertible.
My trial
Let $nin BbbN$ be fixed. Then, $$[det(A)]^m=det(A^m)=I_n=1.$$
Hence, $$det(A)=1neq 0.$$
Thus, $A$ is invertible since $det(A)neq 0.$. I'm I right or is there a counter-example?
matrices algebra-precalculus matrix-equations
asked Mar 18 at 11:29
Omojola MichealOmojola Micheal
1,999424
$endgroup$
add a comment |
$begingroup$
Is this statement true? If $A$ is an $ntimes n$ matrix and $A^m=I_n$ for some $min BbbN$, then $A$ is invertible.
My trial
Let $nin BbbN$ be fixed. Then, $$[det(A)]^m=det(A^m)=I_n=1.$$
Hence, $$det(A)=1neq 0.$$
Thus, $A$ is invertible since $det(A)neq 0.$. I'm I right or is there a counter-example?
matrices algebra-precalculus matrix-equations
asked Mar 18 at 11:29
Omojola MichealOmojola Micheal
1,999424
$endgroup$
2
$begingroup$
Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
$endgroup$
– Chrystomath
Mar 18 at 11:32
add a comment |
$begingroup$
Is this statement true? If $A$ is an $ntimes n$ matrix and $A^m=I_n$ for some $min BbbN$, then $A$ is invertible.
My trial
Let $nin BbbN$ be fixed. Then, $$[det(A)]^m=det(A^m)=I_n=1.$$
Hence, $$det(A)=1neq 0.$$
Thus, $A$ is invertible since $det(A)neq 0.$. I'm I right or is there a counter-example?
matrices algebra-precalculus matrix-equations
asked Mar 18 at 11:29
Omojola MichealOmojola Micheal
1,999424
$endgroup$
Is this statement true? If $A$ is an $ntimes n$ matrix and $A^m=I_n$ for some $min BbbN$, then $A$ is invertible.
My trial
Let $nin BbbN$ be fixed. Then, $$[det(A)]^m=det(A^m)=I_n=1.$$
Hence, $$det(A)=1neq 0.$$
Thus, $A$ is invertible since $det(A)neq 0.$. I'm I right or is there a counter-example?
matrices algebra-precalculus matrix-equations
matrices algebra-precalculus matrix-equations
asked Mar 18 at 11:29
Omojola MichealOmojola Micheal
1,999424
asked Mar 18 at 11:29
Omojola MichealOmojola Micheal
1,999424
edited Mar 18 at 11:34
Omojola Micheal
asked Mar 18 at 11:29
Omojola MichealOmojola Micheal
1,999424
asked Mar 18 at 11:29
Omojola MichealOmojola Micheal
1,999424
asked Mar 18 at 11:29
Omojola MichealOmojola Micheal
1,999424
1,999424
2
$begingroup$
Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
$endgroup$
– Chrystomath
Mar 18 at 11:32
add a comment |
2
$begingroup$
Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
$endgroup$
– Chrystomath
Mar 18 at 11:32
2
2
$begingroup$
Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
$endgroup$
– Chrystomath
Mar 18 at 11:32
$begingroup$
Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
$endgroup$
– Chrystomath
Mar 18 at 11:32
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The right conclusion is $det(A) ne 0$, hence it is invertible.
Notice that we can't conclude that $det(A)=1$. After all, it can take value $-1$.
answered Mar 18 at 11:32
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
Sorry, I'll edit!
$endgroup$
– Omojola Micheal
Mar 18 at 11:33
$begingroup$
Thanks for the prompt reply.
$endgroup$
– Omojola Micheal
Mar 18 at 11:34
add a comment |
$begingroup$
I think it is easier to see it this way: what is $AA^m-1 = A^m-1A$?
answered Mar 18 at 11:33
MariahMariah
2,1431718
$endgroup$
add a comment |
$begingroup$
It's invertible since it has a inverse, $A^m-1$ that is. By the way, $det(A)$ could also be -1.
answered Mar 18 at 11:46
Sander KortewegSander Korteweg
566
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The right conclusion is $det(A) ne 0$, hence it is invertible.
Notice that we can't conclude that $det(A)=1$. After all, it can take value $-1$.
answered Mar 18 at 11:32
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
Sorry, I'll edit!
$endgroup$
– Omojola Micheal
Mar 18 at 11:33
$begingroup$
Thanks for the prompt reply.
$endgroup$
– Omojola Micheal
Mar 18 at 11:34
add a comment |
$begingroup$
The right conclusion is $det(A) ne 0$, hence it is invertible.
Notice that we can't conclude that $det(A)=1$. After all, it can take value $-1$.
answered Mar 18 at 11:32
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
Sorry, I'll edit!
$endgroup$
– Omojola Micheal
Mar 18 at 11:33
$begingroup$
Thanks for the prompt reply.
$endgroup$
– Omojola Micheal
Mar 18 at 11:34
add a comment |
$begingroup$
The right conclusion is $det(A) ne 0$, hence it is invertible.
Notice that we can't conclude that $det(A)=1$. After all, it can take value $-1$.
answered Mar 18 at 11:32
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
The right conclusion is $det(A) ne 0$, hence it is invertible.
Notice that we can't conclude that $det(A)=1$. After all, it can take value $-1$.
answered Mar 18 at 11:32
Siong Thye GohSiong Thye Goh
103k1468119
answered Mar 18 at 11:32
Siong Thye GohSiong Thye Goh
103k1468119
answered Mar 18 at 11:32
Siong Thye GohSiong Thye Goh
103k1468119
answered Mar 18 at 11:32
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
$begingroup$
Sorry, I'll edit!
$endgroup$
– Omojola Micheal
Mar 18 at 11:33
$begingroup$
Thanks for the prompt reply.
$endgroup$
– Omojola Micheal
Mar 18 at 11:34
add a comment |
$begingroup$
Sorry, I'll edit!
$endgroup$
– Omojola Micheal
Mar 18 at 11:33
$begingroup$
Thanks for the prompt reply.
$endgroup$
– Omojola Micheal
Mar 18 at 11:34
$begingroup$
Sorry, I'll edit!
$endgroup$
– Omojola Micheal
Mar 18 at 11:33
$begingroup$
Sorry, I'll edit!
$endgroup$
– Omojola Micheal
Mar 18 at 11:33
$begingroup$
Thanks for the prompt reply.
$endgroup$
– Omojola Micheal
Mar 18 at 11:34
$begingroup$
Thanks for the prompt reply.
$endgroup$
– Omojola Micheal
Mar 18 at 11:34
add a comment |
$begingroup$
I think it is easier to see it this way: what is $AA^m-1 = A^m-1A$?
answered Mar 18 at 11:33
MariahMariah
2,1431718
$endgroup$
add a comment |
$begingroup$
I think it is easier to see it this way: what is $AA^m-1 = A^m-1A$?
answered Mar 18 at 11:33
MariahMariah
2,1431718
$endgroup$
add a comment |
$begingroup$
I think it is easier to see it this way: what is $AA^m-1 = A^m-1A$?
answered Mar 18 at 11:33
MariahMariah
2,1431718
$endgroup$
I think it is easier to see it this way: what is $AA^m-1 = A^m-1A$?
answered Mar 18 at 11:33
MariahMariah
2,1431718
answered Mar 18 at 11:33
MariahMariah
2,1431718
answered Mar 18 at 11:33
MariahMariah
2,1431718
answered Mar 18 at 11:33
MariahMariah
2,1431718
2,1431718
add a comment |
add a comment |
$begingroup$
It's invertible since it has a inverse, $A^m-1$ that is. By the way, $det(A)$ could also be -1.
answered Mar 18 at 11:46
Sander KortewegSander Korteweg
566
$endgroup$
add a comment |
$begingroup$
It's invertible since it has a inverse, $A^m-1$ that is. By the way, $det(A)$ could also be -1.
answered Mar 18 at 11:46
Sander KortewegSander Korteweg
566
$endgroup$
add a comment |
$begingroup$
It's invertible since it has a inverse, $A^m-1$ that is. By the way, $det(A)$ could also be -1.
answered Mar 18 at 11:46
Sander KortewegSander Korteweg
566
$endgroup$
It's invertible since it has a inverse, $A^m-1$ that is. By the way, $det(A)$ could also be -1.
answered Mar 18 at 11:46
Sander KortewegSander Korteweg
566
answered Mar 18 at 11:46
Sander KortewegSander Korteweg
566
answered Mar 18 at 11:46
Sander KortewegSander Korteweg
566
answered Mar 18 at 11:46
Sander KortewegSander Korteweg
566
566
add a comment |
add a comment |
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$begingroup$
Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
$endgroup$
– Chrystomath
Mar 18 at 11:32