Mixing
Proof of statement about invertible matrices raised to the power k
$begingroup$
True or false?
Let $A$ be a matrix of size $n times n$. If there is a natural number $k$ such that $A^k$ is invertible, then $A$ is invertible.
I intuitively understand that this is correct, but I would like to know what the formal proof for this statement is…
Thank you!
linear-algebra matrices inverse
edited Feb 2 at 14:14
Mostafa Ayaz
18.1k31040
asked Feb 2 at 13:50
daltadalta
1578
$endgroup$
add a comment |
$begingroup$
True or false?
Let $A$ be a matrix of size $n times n$. If there is a natural number $k$ such that $A^k$ is invertible, then $A$ is invertible.
I intuitively understand that this is correct, but I would like to know what the formal proof for this statement is…
Thank you!
linear-algebra matrices inverse
edited Feb 2 at 14:14
Mostafa Ayaz
18.1k31040
asked Feb 2 at 13:50
daltadalta
1578
$endgroup$
$begingroup$
One idea is to use determinants, noting that $det(A^k) = det(A)^k$
$endgroup$
– Omnomnomnom
Feb 2 at 13:52
1
$begingroup$
Hint: $I = A^kB = AA^{k-1}B$.
$endgroup$
– Ethan Bolker
Feb 2 at 13:53
add a comment |
$begingroup$
True or false?
Let $A$ be a matrix of size $n times n$. If there is a natural number $k$ such that $A^k$ is invertible, then $A$ is invertible.
I intuitively understand that this is correct, but I would like to know what the formal proof for this statement is…
Thank you!
linear-algebra matrices inverse
edited Feb 2 at 14:14
Mostafa Ayaz
18.1k31040
asked Feb 2 at 13:50
daltadalta
1578
$endgroup$
True or false?
Let $A$ be a matrix of size $n times n$. If there is a natural number $k$ such that $A^k$ is invertible, then $A$ is invertible.
I intuitively understand that this is correct, but I would like to know what the formal proof for this statement is…
Thank you!
linear-algebra matrices inverse
linear-algebra matrices inverse
edited Feb 2 at 14:14
Mostafa Ayaz
18.1k31040
asked Feb 2 at 13:50
daltadalta
1578
edited Feb 2 at 14:14
Mostafa Ayaz
18.1k31040
asked Feb 2 at 13:50
daltadalta
1578
edited Feb 2 at 14:14
Mostafa Ayaz
18.1k31040
edited Feb 2 at 14:14
Mostafa Ayaz
18.1k31040
edited Feb 2 at 14:14
Mostafa Ayaz
18.1k31040
18.1k31040
asked Feb 2 at 13:50
daltadalta
1578
asked Feb 2 at 13:50
daltadalta
1578
asked Feb 2 at 13:50
daltadalta
1578
1578
$begingroup$
One idea is to use determinants, noting that $det(A^k) = det(A)^k$
$endgroup$
– Omnomnomnom
Feb 2 at 13:52
1
$begingroup$
Hint: $I = A^kB = AA^{k-1}B$.
$endgroup$
– Ethan Bolker
Feb 2 at 13:53
add a comment |
$begingroup$
One idea is to use determinants, noting that $det(A^k) = det(A)^k$
$endgroup$
– Omnomnomnom
Feb 2 at 13:52
1
$begingroup$
Hint: $I = A^kB = AA^{k-1}B$.
$endgroup$
– Ethan Bolker
Feb 2 at 13:53
$begingroup$
One idea is to use determinants, noting that $det(A^k) = det(A)^k$
$endgroup$
– Omnomnomnom
Feb 2 at 13:52
$begingroup$
One idea is to use determinants, noting that $det(A^k) = det(A)^k$
$endgroup$
– Omnomnomnom
Feb 2 at 13:52
1
1
$begingroup$
Hint: $I = A^kB = AA^{k-1}B$.
$endgroup$
– Ethan Bolker
Feb 2 at 13:53
$begingroup$
Hint: $I = A^kB = AA^{k-1}B$.
$endgroup$
– Ethan Bolker
Feb 2 at 13:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint
If the eigenvalues of the matrix $A$ are $lambda_1,cdots ,lambda_n$, then the eigenvalues of matrix $A^k$ would be $lambda_1^k,cdots ,lambda_n^k$
answered Feb 2 at 14:15
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
If $A^k$ is invertible then $det (A^k)ne 0$. But $det(A^k)=det(A)^k$ (since a field has no zero divisors, so $det(A)^kne 0$ or $det(A)ne 0$, meaning $A$ is invertible.
answered Feb 2 at 14:26
Yuval GatYuval Gat
9641213
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
Hint
If the eigenvalues of the matrix $A$ are $lambda_1,cdots ,lambda_n$, then the eigenvalues of matrix $A^k$ would be $lambda_1^k,cdots ,lambda_n^k$
answered Feb 2 at 14:15
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
If the eigenvalues of the matrix $A$ are $lambda_1,cdots ,lambda_n$, then the eigenvalues of matrix $A^k$ would be $lambda_1^k,cdots ,lambda_n^k$
answered Feb 2 at 14:15
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
If the eigenvalues of the matrix $A$ are $lambda_1,cdots ,lambda_n$, then the eigenvalues of matrix $A^k$ would be $lambda_1^k,cdots ,lambda_n^k$
answered Feb 2 at 14:15
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Hint
If the eigenvalues of the matrix $A$ are $lambda_1,cdots ,lambda_n$, then the eigenvalues of matrix $A^k$ would be $lambda_1^k,cdots ,lambda_n^k$
answered Feb 2 at 14:15
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 2 at 14:15
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 2 at 14:15
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 2 at 14:15
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
$begingroup$
If $A^k$ is invertible then $det (A^k)ne 0$. But $det(A^k)=det(A)^k$ (since a field has no zero divisors, so $det(A)^kne 0$ or $det(A)ne 0$, meaning $A$ is invertible.
answered Feb 2 at 14:26
Yuval GatYuval Gat
9641213
$endgroup$
add a comment |
$begingroup$
If $A^k$ is invertible then $det (A^k)ne 0$. But $det(A^k)=det(A)^k$ (since a field has no zero divisors, so $det(A)^kne 0$ or $det(A)ne 0$, meaning $A$ is invertible.
answered Feb 2 at 14:26
Yuval GatYuval Gat
9641213
$endgroup$
add a comment |
$begingroup$
If $A^k$ is invertible then $det (A^k)ne 0$. But $det(A^k)=det(A)^k$ (since a field has no zero divisors, so $det(A)^kne 0$ or $det(A)ne 0$, meaning $A$ is invertible.
answered Feb 2 at 14:26
Yuval GatYuval Gat
9641213
$endgroup$
If $A^k$ is invertible then $det (A^k)ne 0$. But $det(A^k)=det(A)^k$ (since a field has no zero divisors, so $det(A)^kne 0$ or $det(A)ne 0$, meaning $A$ is invertible.
answered Feb 2 at 14:26
Yuval GatYuval Gat
9641213
answered Feb 2 at 14:26
Yuval GatYuval Gat
9641213
answered Feb 2 at 14:26
Yuval GatYuval Gat
9641213
answered Feb 2 at 14:26
Yuval GatYuval Gat
9641213
9641213
add a comment |
add a comment |
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$begingroup$
One idea is to use determinants, noting that $det(A^k) = det(A)^k$
$endgroup$
– Omnomnomnom
Feb 2 at 13:52
1
$begingroup$
Hint: $I = A^kB = AA^{k-1}B$.
$endgroup$
– Ethan Bolker
Feb 2 at 13:53