Prove if the product of $k$ matrices $A_1$ … $A_k$ is nonsingular, then each matrix $A_i$ is nonsingular.
$begingroup$
I'm having trouble proving this without using determinants. I know how to prove it with the product of just two matrices, but I'm not sure how to generalize this to a product of k matrices. Is there a way to do this proof without determinants?
To clarify the question, each matrix is an $n$ by $n$ matrix.
For example, I know that if the determinant of the product is nonzero, then all of the determinants of the individual matrices must also be nonzero.
I also know that if a product of two matrices $A$ and $B$, $AB$ is nonsingular, then there exists a matrix $C$ so that $C(AB) = I$ and $(AB)C = I$, and so $(CA)B = I$ and $A(BC) = I$, so $A$ and $B$ are both invertible, and thus nonsingular.
I'm looking for a way to generalize this. Or just any other way to prove this without determinants.
Thanks so much!
linear-algebra matrices
asked Dec 5 '18 at 4:06
music127music127
61
$endgroup$
add a comment |
$begingroup$
I'm having trouble proving this without using determinants. I know how to prove it with the product of just two matrices, but I'm not sure how to generalize this to a product of k matrices. Is there a way to do this proof without determinants?
To clarify the question, each matrix is an $n$ by $n$ matrix.
For example, I know that if the determinant of the product is nonzero, then all of the determinants of the individual matrices must also be nonzero.
I also know that if a product of two matrices $A$ and $B$, $AB$ is nonsingular, then there exists a matrix $C$ so that $C(AB) = I$ and $(AB)C = I$, and so $(CA)B = I$ and $A(BC) = I$, so $A$ and $B$ are both invertible, and thus nonsingular.
I'm looking for a way to generalize this. Or just any other way to prove this without determinants.
Thanks so much!
linear-algebra matrices
asked Dec 5 '18 at 4:06
music127music127
61
$endgroup$
1
$begingroup$
Take your argument for two and induct.
$endgroup$
– Randall
Dec 5 '18 at 4:20
1
$begingroup$
Induction is how you would prove this for all $k geq 2$. I would also suggest proving the contrapositive of the statement
$endgroup$
– JavaMan
Dec 5 '18 at 4:32
add a comment |
$begingroup$
I'm having trouble proving this without using determinants. I know how to prove it with the product of just two matrices, but I'm not sure how to generalize this to a product of k matrices. Is there a way to do this proof without determinants?
To clarify the question, each matrix is an $n$ by $n$ matrix.
For example, I know that if the determinant of the product is nonzero, then all of the determinants of the individual matrices must also be nonzero.
I also know that if a product of two matrices $A$ and $B$, $AB$ is nonsingular, then there exists a matrix $C$ so that $C(AB) = I$ and $(AB)C = I$, and so $(CA)B = I$ and $A(BC) = I$, so $A$ and $B$ are both invertible, and thus nonsingular.
I'm looking for a way to generalize this. Or just any other way to prove this without determinants.
Thanks so much!
linear-algebra matrices
asked Dec 5 '18 at 4:06
music127music127
61
$endgroup$
I'm having trouble proving this without using determinants. I know how to prove it with the product of just two matrices, but I'm not sure how to generalize this to a product of k matrices. Is there a way to do this proof without determinants?
To clarify the question, each matrix is an $n$ by $n$ matrix.
For example, I know that if the determinant of the product is nonzero, then all of the determinants of the individual matrices must also be nonzero.
I also know that if a product of two matrices $A$ and $B$, $AB$ is nonsingular, then there exists a matrix $C$ so that $C(AB) = I$ and $(AB)C = I$, and so $(CA)B = I$ and $A(BC) = I$, so $A$ and $B$ are both invertible, and thus nonsingular.
I'm looking for a way to generalize this. Or just any other way to prove this without determinants.
Thanks so much!
linear-algebra matrices
linear-algebra matrices
asked Dec 5 '18 at 4:06
music127music127
61
asked Dec 5 '18 at 4:06
music127music127
61
asked Dec 5 '18 at 4:06
music127music127
61
asked Dec 5 '18 at 4:06
music127music127
61
asked Dec 5 '18 at 4:06
music127music127
61
61
1
$begingroup$
Take your argument for two and induct.
$endgroup$
– Randall
Dec 5 '18 at 4:20
1
$begingroup$
Induction is how you would prove this for all $k geq 2$. I would also suggest proving the contrapositive of the statement
$endgroup$
– JavaMan
Dec 5 '18 at 4:32
add a comment |
1
$begingroup$
Take your argument for two and induct.
$endgroup$
– Randall
Dec 5 '18 at 4:20
1
$begingroup$
Induction is how you would prove this for all $k geq 2$. I would also suggest proving the contrapositive of the statement
$endgroup$
– JavaMan
Dec 5 '18 at 4:32
1
1
$begingroup$
Take your argument for two and induct.
$endgroup$
– Randall
Dec 5 '18 at 4:20
$begingroup$
Take your argument for two and induct.
$endgroup$
– Randall
Dec 5 '18 at 4:20
1
1
$begingroup$
Induction is how you would prove this for all $k geq 2$. I would also suggest proving the contrapositive of the statement
$endgroup$
– JavaMan
Dec 5 '18 at 4:32
$begingroup$
Induction is how you would prove this for all $k geq 2$. I would also suggest proving the contrapositive of the statement
$endgroup$
– JavaMan
Dec 5 '18 at 4:32
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since each $A_j : mathbb F^n to mathbb F^n$ ($mathbb F$ can be $mathbb R$ or $mathbb C$), $A_j$ is invertible if and only if $A_j$ is injective by Rank-Nullity Theorem.
Now if the product $A_1 dots A_k$ is invertible, then $A_1 dots A_k$ is injective and it follows for every $j$, $A_j$ must be injective. So $A_j$ is invertible for each $j$.
answered Dec 5 '18 at 4:30
user1101010user1101010
7751730
$endgroup$
add a comment |
$begingroup$
Another way to present it: A matrix $A$ is invertible if and only if image of $R^n$ by $A$ is equal to $R^n$. It follows that if a given $A_i$ is singular, then the product is singular
answered Dec 5 '18 at 10:20
DamienDamien
60714
$endgroup$
add a comment |
$begingroup$
It's already stated in comment, but I think it deserve an answer.
You say you are able to do it with 2 matrix $A$ and $B$. So, if $k=2$, you're done. Now, use induction to prove it's true for every $k$.
First, let P(k)="if the product of $k$ matrices $A_1$ ... $A_k$ is nonsingular, then each matrix $A_i$ is nonsingular"
To use induction, you need to prove:
- $exists k in mathbb N, P(k)$
- $forall n in mathbb N, P(n) implies P(n+1)$
In your case, this is how you do it:
- You can show that P(2) is true as you did in the question body, but you can even start with P(1), which read as "If $A_1$ is nonsingular, then $A_1$ is non singular".
- Suppose P(n) and try to proove that P(n+1) holds. To do this, consider $A=prod_{i in {1ldots n}}{A_i}$ and $B=A_{n+1}$, and use the same technic as in your question. You'll end up with 2 nonsingular matrix. One being $A_{n+1}$, and the other one being a nonsingular product of $n$ matrix. But since you supposed $P(n)$, this means that every matrix in the product is nonsingular itself.
answered Dec 5 '18 at 12:15
F.CaretteF.Carette
1,22612
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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
$begingroup$
Since each $A_j : mathbb F^n to mathbb F^n$ ($mathbb F$ can be $mathbb R$ or $mathbb C$), $A_j$ is invertible if and only if $A_j$ is injective by Rank-Nullity Theorem.
Now if the product $A_1 dots A_k$ is invertible, then $A_1 dots A_k$ is injective and it follows for every $j$, $A_j$ must be injective. So $A_j$ is invertible for each $j$.
answered Dec 5 '18 at 4:30
user1101010user1101010
7751730
$endgroup$
add a comment |
$begingroup$
Since each $A_j : mathbb F^n to mathbb F^n$ ($mathbb F$ can be $mathbb R$ or $mathbb C$), $A_j$ is invertible if and only if $A_j$ is injective by Rank-Nullity Theorem.
Now if the product $A_1 dots A_k$ is invertible, then $A_1 dots A_k$ is injective and it follows for every $j$, $A_j$ must be injective. So $A_j$ is invertible for each $j$.
answered Dec 5 '18 at 4:30
user1101010user1101010
7751730
$endgroup$
add a comment |
$begingroup$
Since each $A_j : mathbb F^n to mathbb F^n$ ($mathbb F$ can be $mathbb R$ or $mathbb C$), $A_j$ is invertible if and only if $A_j$ is injective by Rank-Nullity Theorem.
Now if the product $A_1 dots A_k$ is invertible, then $A_1 dots A_k$ is injective and it follows for every $j$, $A_j$ must be injective. So $A_j$ is invertible for each $j$.
answered Dec 5 '18 at 4:30
user1101010user1101010
7751730
$endgroup$
Since each $A_j : mathbb F^n to mathbb F^n$ ($mathbb F$ can be $mathbb R$ or $mathbb C$), $A_j$ is invertible if and only if $A_j$ is injective by Rank-Nullity Theorem.
Now if the product $A_1 dots A_k$ is invertible, then $A_1 dots A_k$ is injective and it follows for every $j$, $A_j$ must be injective. So $A_j$ is invertible for each $j$.
answered Dec 5 '18 at 4:30
user1101010user1101010
7751730
answered Dec 5 '18 at 4:30
user1101010user1101010
7751730
answered Dec 5 '18 at 4:30
user1101010user1101010
7751730
answered Dec 5 '18 at 4:30
user1101010user1101010
7751730
7751730
add a comment |
add a comment |
$begingroup$
Another way to present it: A matrix $A$ is invertible if and only if image of $R^n$ by $A$ is equal to $R^n$. It follows that if a given $A_i$ is singular, then the product is singular
answered Dec 5 '18 at 10:20
DamienDamien
60714
$endgroup$
add a comment |
$begingroup$
Another way to present it: A matrix $A$ is invertible if and only if image of $R^n$ by $A$ is equal to $R^n$. It follows that if a given $A_i$ is singular, then the product is singular
answered Dec 5 '18 at 10:20
DamienDamien
60714
$endgroup$
add a comment |
$begingroup$
Another way to present it: A matrix $A$ is invertible if and only if image of $R^n$ by $A$ is equal to $R^n$. It follows that if a given $A_i$ is singular, then the product is singular
answered Dec 5 '18 at 10:20
DamienDamien
60714
$endgroup$
Another way to present it: A matrix $A$ is invertible if and only if image of $R^n$ by $A$ is equal to $R^n$. It follows that if a given $A_i$ is singular, then the product is singular
answered Dec 5 '18 at 10:20
DamienDamien
60714
answered Dec 5 '18 at 10:20
DamienDamien
60714
answered Dec 5 '18 at 10:20
DamienDamien
60714
answered Dec 5 '18 at 10:20
DamienDamien
60714
60714
add a comment |
add a comment |
$begingroup$
It's already stated in comment, but I think it deserve an answer.
You say you are able to do it with 2 matrix $A$ and $B$. So, if $k=2$, you're done. Now, use induction to prove it's true for every $k$.
First, let P(k)="if the product of $k$ matrices $A_1$ ... $A_k$ is nonsingular, then each matrix $A_i$ is nonsingular"
To use induction, you need to prove:
- $exists k in mathbb N, P(k)$
- $forall n in mathbb N, P(n) implies P(n+1)$
In your case, this is how you do it:
- You can show that P(2) is true as you did in the question body, but you can even start with P(1), which read as "If $A_1$ is nonsingular, then $A_1$ is non singular".
- Suppose P(n) and try to proove that P(n+1) holds. To do this, consider $A=prod_{i in {1ldots n}}{A_i}$ and $B=A_{n+1}$, and use the same technic as in your question. You'll end up with 2 nonsingular matrix. One being $A_{n+1}$, and the other one being a nonsingular product of $n$ matrix. But since you supposed $P(n)$, this means that every matrix in the product is nonsingular itself.
answered Dec 5 '18 at 12:15
F.CaretteF.Carette
1,22612
$endgroup$
add a comment |
$begingroup$
It's already stated in comment, but I think it deserve an answer.
You say you are able to do it with 2 matrix $A$ and $B$. So, if $k=2$, you're done. Now, use induction to prove it's true for every $k$.
First, let P(k)="if the product of $k$ matrices $A_1$ ... $A_k$ is nonsingular, then each matrix $A_i$ is nonsingular"
To use induction, you need to prove:
- $exists k in mathbb N, P(k)$
- $forall n in mathbb N, P(n) implies P(n+1)$
In your case, this is how you do it:
- You can show that P(2) is true as you did in the question body, but you can even start with P(1), which read as "If $A_1$ is nonsingular, then $A_1$ is non singular".
- Suppose P(n) and try to proove that P(n+1) holds. To do this, consider $A=prod_{i in {1ldots n}}{A_i}$ and $B=A_{n+1}$, and use the same technic as in your question. You'll end up with 2 nonsingular matrix. One being $A_{n+1}$, and the other one being a nonsingular product of $n$ matrix. But since you supposed $P(n)$, this means that every matrix in the product is nonsingular itself.
answered Dec 5 '18 at 12:15
F.CaretteF.Carette
1,22612
$endgroup$
add a comment |
$begingroup$
It's already stated in comment, but I think it deserve an answer.
You say you are able to do it with 2 matrix $A$ and $B$. So, if $k=2$, you're done. Now, use induction to prove it's true for every $k$.
First, let P(k)="if the product of $k$ matrices $A_1$ ... $A_k$ is nonsingular, then each matrix $A_i$ is nonsingular"
To use induction, you need to prove:
- $exists k in mathbb N, P(k)$
- $forall n in mathbb N, P(n) implies P(n+1)$
In your case, this is how you do it:
- You can show that P(2) is true as you did in the question body, but you can even start with P(1), which read as "If $A_1$ is nonsingular, then $A_1$ is non singular".
- Suppose P(n) and try to proove that P(n+1) holds. To do this, consider $A=prod_{i in {1ldots n}}{A_i}$ and $B=A_{n+1}$, and use the same technic as in your question. You'll end up with 2 nonsingular matrix. One being $A_{n+1}$, and the other one being a nonsingular product of $n$ matrix. But since you supposed $P(n)$, this means that every matrix in the product is nonsingular itself.
answered Dec 5 '18 at 12:15
F.CaretteF.Carette
1,22612
$endgroup$
It's already stated in comment, but I think it deserve an answer.
You say you are able to do it with 2 matrix $A$ and $B$. So, if $k=2$, you're done. Now, use induction to prove it's true for every $k$.
First, let P(k)="if the product of $k$ matrices $A_1$ ... $A_k$ is nonsingular, then each matrix $A_i$ is nonsingular"
To use induction, you need to prove:
- $exists k in mathbb N, P(k)$
- $forall n in mathbb N, P(n) implies P(n+1)$
In your case, this is how you do it:
- You can show that P(2) is true as you did in the question body, but you can even start with P(1), which read as "If $A_1$ is nonsingular, then $A_1$ is non singular".
- Suppose P(n) and try to proove that P(n+1) holds. To do this, consider $A=prod_{i in {1ldots n}}{A_i}$ and $B=A_{n+1}$, and use the same technic as in your question. You'll end up with 2 nonsingular matrix. One being $A_{n+1}$, and the other one being a nonsingular product of $n$ matrix. But since you supposed $P(n)$, this means that every matrix in the product is nonsingular itself.
answered Dec 5 '18 at 12:15
F.CaretteF.Carette
1,22612
answered Dec 5 '18 at 12:15
F.CaretteF.Carette
1,22612
answered Dec 5 '18 at 12:15
F.CaretteF.Carette
1,22612
answered Dec 5 '18 at 12:15
F.CaretteF.Carette
1,22612
1,22612
add a comment |
add a comment |
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1
$begingroup$
Take your argument for two and induct.
$endgroup$
– Randall
Dec 5 '18 at 4:20
1
$begingroup$
Induction is how you would prove this for all $k geq 2$. I would also suggest proving the contrapositive of the statement
$endgroup$
– JavaMan
Dec 5 '18 at 4:32