Mixing
Inverses of matrices and its properties.
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I'm reading in my textbook this theorem about the properties of inverse matrices:
I don't follow how 1) and 3) follow from the idea that inverses are unique.
Why does the fact that a matrix only has one inverse matter in these theorems?
This is the theorem about uniqueness that I read:
Quick question, why does $CI = C$ in the proof for uniqueness?
linear-algebra matrices inverse
edited Jul 18 at 16:20
José Carlos Santos
114k1698177
asked Jul 18 at 15:33
Jwan622
1,61211224
add a comment |Â
up vote
1
down vote
favorite
I'm reading in my textbook this theorem about the properties of inverse matrices:
I don't follow how 1) and 3) follow from the idea that inverses are unique.
Why does the fact that a matrix only has one inverse matter in these theorems?
This is the theorem about uniqueness that I read:
Quick question, why does $CI = C$ in the proof for uniqueness?
linear-algebra matrices inverse
edited Jul 18 at 16:20
José Carlos Santos
114k1698177
asked Jul 18 at 15:33
Jwan622
1,61211224
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm reading in my textbook this theorem about the properties of inverse matrices:
I don't follow how 1) and 3) follow from the idea that inverses are unique.
Why does the fact that a matrix only has one inverse matter in these theorems?
This is the theorem about uniqueness that I read:
Quick question, why does $CI = C$ in the proof for uniqueness?
linear-algebra matrices inverse
edited Jul 18 at 16:20
José Carlos Santos
114k1698177
asked Jul 18 at 15:33
Jwan622
1,61211224
I'm reading in my textbook this theorem about the properties of inverse matrices:
I don't follow how 1) and 3) follow from the idea that inverses are unique.
Why does the fact that a matrix only has one inverse matter in these theorems?
This is the theorem about uniqueness that I read:
Quick question, why does $CI = C$ in the proof for uniqueness?
linear-algebra matrices inverse
edited Jul 18 at 16:20
José Carlos Santos
114k1698177
asked Jul 18 at 15:33
Jwan622
1,61211224
edited Jul 18 at 16:20
José Carlos Santos
114k1698177
edited Jul 18 at 16:20
José Carlos Santos
114k1698177
edited Jul 18 at 16:20
José Carlos Santos
114k1698177
114k1698177
asked Jul 18 at 15:33
Jwan622
1,61211224
asked Jul 18 at 15:33
Jwan622
1,61211224
asked Jul 18 at 15:33
Jwan622
1,61211224
1,61211224
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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up vote
2
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accepted
Identity matrix is neutral with respect to multiplication, i.e. for any square matrix $P$ you have $PI=IP=P$. You can easily show this property, just put the definition of $I$ in the the definition of matrix multiplication.
For (1), suppose that $AB=BA=I$. By definition, it means that $A = B^-1$. Then again, if you put $B = A^-1$, then our hypothesis holds ($AA^-1=A^-1A=I$), hence $A = B^-1 = (A^-1)^-1$.
The (3) is done likewise.
answered Jul 18 at 15:41
TZakrevskiy
19.8k12253
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:12
add a comment |Â
up vote
2
down vote
The answer to the quick question is: because that's a property of the identity matrix. Just do the computation.
Concerning 1), you have $A.A^-1=operatornameId$. Since the inverse of $A^-1$ is the only matrix whose product by $A^-1$ is $operatornameId$ and since $A$ has the property, $(A^-1)^-1=A$.
The case of 3) is similar. You have $left(frac1cA^-1right).cA=frac1ccA^-1A=operatornameId$. Since the inverse of $cA$ is the only matrix whose product by $cA$ is $operatornameId$ and since $frac1cA^-1$ has the property, $(cA)^-1=frac1cA^-a$.
answered Jul 18 at 15:41
José Carlos Santos
114k1698177
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:13
Consider property 1), for instance. I did not actually prove that $A$ is the inverse of $A^-1$. All I proved was that $A.A^-1=operatornameId$. But then,since $A^-1$ has one and only one inverse, that inverse can only be $A$.
– José Carlos Santos
Jul 19 at 14:17
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Identity matrix is neutral with respect to multiplication, i.e. for any square matrix $P$ you have $PI=IP=P$. You can easily show this property, just put the definition of $I$ in the the definition of matrix multiplication.
For (1), suppose that $AB=BA=I$. By definition, it means that $A = B^-1$. Then again, if you put $B = A^-1$, then our hypothesis holds ($AA^-1=A^-1A=I$), hence $A = B^-1 = (A^-1)^-1$.
The (3) is done likewise.
answered Jul 18 at 15:41
TZakrevskiy
19.8k12253
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:12
add a comment |Â
up vote
2
down vote
accepted
Identity matrix is neutral with respect to multiplication, i.e. for any square matrix $P$ you have $PI=IP=P$. You can easily show this property, just put the definition of $I$ in the the definition of matrix multiplication.
For (1), suppose that $AB=BA=I$. By definition, it means that $A = B^-1$. Then again, if you put $B = A^-1$, then our hypothesis holds ($AA^-1=A^-1A=I$), hence $A = B^-1 = (A^-1)^-1$.
The (3) is done likewise.
answered Jul 18 at 15:41
TZakrevskiy
19.8k12253
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:12
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Identity matrix is neutral with respect to multiplication, i.e. for any square matrix $P$ you have $PI=IP=P$. You can easily show this property, just put the definition of $I$ in the the definition of matrix multiplication.
For (1), suppose that $AB=BA=I$. By definition, it means that $A = B^-1$. Then again, if you put $B = A^-1$, then our hypothesis holds ($AA^-1=A^-1A=I$), hence $A = B^-1 = (A^-1)^-1$.
The (3) is done likewise.
answered Jul 18 at 15:41
TZakrevskiy
19.8k12253
Identity matrix is neutral with respect to multiplication, i.e. for any square matrix $P$ you have $PI=IP=P$. You can easily show this property, just put the definition of $I$ in the the definition of matrix multiplication.
For (1), suppose that $AB=BA=I$. By definition, it means that $A = B^-1$. Then again, if you put $B = A^-1$, then our hypothesis holds ($AA^-1=A^-1A=I$), hence $A = B^-1 = (A^-1)^-1$.
The (3) is done likewise.
answered Jul 18 at 15:41
TZakrevskiy
19.8k12253
answered Jul 18 at 15:41
TZakrevskiy
19.8k12253
answered Jul 18 at 15:41
TZakrevskiy
19.8k12253
answered Jul 18 at 15:41
TZakrevskiy
19.8k12253
19.8k12253
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:12
add a comment |Â
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:12
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:12
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:12
add a comment |Â
up vote
2
down vote
The answer to the quick question is: because that's a property of the identity matrix. Just do the computation.
Concerning 1), you have $A.A^-1=operatornameId$. Since the inverse of $A^-1$ is the only matrix whose product by $A^-1$ is $operatornameId$ and since $A$ has the property, $(A^-1)^-1=A$.
The case of 3) is similar. You have $left(frac1cA^-1right).cA=frac1ccA^-1A=operatornameId$. Since the inverse of $cA$ is the only matrix whose product by $cA$ is $operatornameId$ and since $frac1cA^-1$ has the property, $(cA)^-1=frac1cA^-a$.
answered Jul 18 at 15:41
José Carlos Santos
114k1698177
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:13
Consider property 1), for instance. I did not actually prove that $A$ is the inverse of $A^-1$. All I proved was that $A.A^-1=operatornameId$. But then,since $A^-1$ has one and only one inverse, that inverse can only be $A$.
– José Carlos Santos
Jul 19 at 14:17
add a comment |Â
up vote
2
down vote
The answer to the quick question is: because that's a property of the identity matrix. Just do the computation.
Concerning 1), you have $A.A^-1=operatornameId$. Since the inverse of $A^-1$ is the only matrix whose product by $A^-1$ is $operatornameId$ and since $A$ has the property, $(A^-1)^-1=A$.
The case of 3) is similar. You have $left(frac1cA^-1right).cA=frac1ccA^-1A=operatornameId$. Since the inverse of $cA$ is the only matrix whose product by $cA$ is $operatornameId$ and since $frac1cA^-1$ has the property, $(cA)^-1=frac1cA^-a$.
answered Jul 18 at 15:41
José Carlos Santos
114k1698177
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:13
Consider property 1), for instance. I did not actually prove that $A$ is the inverse of $A^-1$. All I proved was that $A.A^-1=operatornameId$. But then,since $A^-1$ has one and only one inverse, that inverse can only be $A$.
– José Carlos Santos
Jul 19 at 14:17
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The answer to the quick question is: because that's a property of the identity matrix. Just do the computation.
Concerning 1), you have $A.A^-1=operatornameId$. Since the inverse of $A^-1$ is the only matrix whose product by $A^-1$ is $operatornameId$ and since $A$ has the property, $(A^-1)^-1=A$.
The case of 3) is similar. You have $left(frac1cA^-1right).cA=frac1ccA^-1A=operatornameId$. Since the inverse of $cA$ is the only matrix whose product by $cA$ is $operatornameId$ and since $frac1cA^-1$ has the property, $(cA)^-1=frac1cA^-a$.
answered Jul 18 at 15:41
José Carlos Santos
114k1698177
The answer to the quick question is: because that's a property of the identity matrix. Just do the computation.
Concerning 1), you have $A.A^-1=operatornameId$. Since the inverse of $A^-1$ is the only matrix whose product by $A^-1$ is $operatornameId$ and since $A$ has the property, $(A^-1)^-1=A$.
The case of 3) is similar. You have $left(frac1cA^-1right).cA=frac1ccA^-1A=operatornameId$. Since the inverse of $cA$ is the only matrix whose product by $cA$ is $operatornameId$ and since $frac1cA^-1$ has the property, $(cA)^-1=frac1cA^-a$.
answered Jul 18 at 15:41
José Carlos Santos
114k1698177
answered Jul 18 at 15:41
José Carlos Santos
114k1698177
answered Jul 18 at 15:41
José Carlos Santos
114k1698177
answered Jul 18 at 15:41
José Carlos Santos
114k1698177
114k1698177
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:13
Consider property 1), for instance. I did not actually prove that $A$ is the inverse of $A^-1$. All I proved was that $A.A^-1=operatornameId$. But then,since $A^-1$ has one and only one inverse, that inverse can only be $A$.
– José Carlos Santos
Jul 19 at 14:17
add a comment |Â
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:13
Consider property 1), for instance. I did not actually prove that $A$ is the inverse of $A^-1$. All I proved was that $A.A^-1=operatornameId$. But then,since $A^-1$ has one and only one inverse, that inverse can only be $A$.
– José Carlos Santos
Jul 19 at 14:17
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:13
But why does this all follow from the idea that the inverse matrices are unique?
– Jwan622
Jul 19 at 14:13
Consider property 1), for instance. I did not actually prove that $A$ is the inverse of $A^-1$. All I proved was that $A.A^-1=operatornameId$. But then,since $A^-1$ has one and only one inverse, that inverse can only be $A$.
– José Carlos Santos
Jul 19 at 14:17
Consider property 1), for instance. I did not actually prove that $A$ is the inverse of $A^-1$. All I proved was that $A.A^-1=operatornameId$. But then,since $A^-1$ has one and only one inverse, that inverse can only be $A$.
– José Carlos Santos
Jul 19 at 14:17
add a comment |Â
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