Proving associativity of matrix multiplication
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I'm trying to prove that matrix multiplication is associative, but seem to be making mistakes in each of my past write-ups, so hopefully someone can check over my work.
Theorem. Let $A$ be $alpha times beta$, $B$ be $beta times gamma$, and $C$ be $gamma times delta$. Prove that $(AB)C = A(BC)$.
Proof. Define general entries of the matrices $A$, $B$, and $C$ by $a_g,h$, $b_i,j$, and $c_k,m$, respectively. Then, for the LHS:
beginalign*
& (AB)_alpha, gamma = sumlimits_p=1^beta a_alpha,p b_p,gamma \
& left((AB)Cright)_alpha, delta = sumlimits_n=1^gamma left(ABright)_alpha, n c_n, delta = sumlimits_n=1^gamma left(sumlimits_p=1^beta a_alpha,p b_p,n right) c_n, delta = sumlimits_n=1^gamma sumlimits_p=1^beta left(a_alpha,p b_p,nright) c_n, delta.
endalign*
For the RHS:
beginalign*
& left(BCright)_beta, delta = sumlimits_n=1^gamma b_beta, n c_n, delta \
& left(Aleft(BCright)right)_alpha,delta = sumlimits_p=1^beta a_alpha,p (BC)_p, delta = sumlimits_p=1^beta a_alpha,p left(sumlimits_n=1^gamma b_p, n c_n, delta right) = sumlimits_p=1^beta sumlimits_n=1^gamma a_alpha,p left(b_p, n c_n, delta right).
endalign*
Assuming I have written these correctly, we can make two observations: first, the summands are equivalent, as multiplication is associative. Second, the order of the summations doesn't matter when we're summing a finite number of entries. Thus, $(AB)C = A(BC)$.
How does this look?
linear-algebra proof-verification
asked Sep 11 at 4:30
Matt.P
1,015414
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I'm trying to prove that matrix multiplication is associative, but seem to be making mistakes in each of my past write-ups, so hopefully someone can check over my work.
Theorem. Let $A$ be $alpha times beta$, $B$ be $beta times gamma$, and $C$ be $gamma times delta$. Prove that $(AB)C = A(BC)$.
Proof. Define general entries of the matrices $A$, $B$, and $C$ by $a_g,h$, $b_i,j$, and $c_k,m$, respectively. Then, for the LHS:
beginalign*
& (AB)_alpha, gamma = sumlimits_p=1^beta a_alpha,p b_p,gamma \
& left((AB)Cright)_alpha, delta = sumlimits_n=1^gamma left(ABright)_alpha, n c_n, delta = sumlimits_n=1^gamma left(sumlimits_p=1^beta a_alpha,p b_p,n right) c_n, delta = sumlimits_n=1^gamma sumlimits_p=1^beta left(a_alpha,p b_p,nright) c_n, delta.
endalign*
For the RHS:
beginalign*
& left(BCright)_beta, delta = sumlimits_n=1^gamma b_beta, n c_n, delta \
& left(Aleft(BCright)right)_alpha,delta = sumlimits_p=1^beta a_alpha,p (BC)_p, delta = sumlimits_p=1^beta a_alpha,p left(sumlimits_n=1^gamma b_p, n c_n, delta right) = sumlimits_p=1^beta sumlimits_n=1^gamma a_alpha,p left(b_p, n c_n, delta right).
endalign*
Assuming I have written these correctly, we can make two observations: first, the summands are equivalent, as multiplication is associative. Second, the order of the summations doesn't matter when we're summing a finite number of entries. Thus, $(AB)C = A(BC)$.
How does this look?
linear-algebra proof-verification
asked Sep 11 at 4:30
Matt.P
1,015414
2
$checkmark,!$
– David
Sep 11 at 4:37
1
Just a point on notation: you are using $alpha$ as both the number of rows of $A$ and as a general row-index; and same for$beta$, $gamma$ and $delta$. You should clean that up for a completely correct proof.
– ancientmathematician
Sep 11 at 6:56
This is very true and I hadn't realized it. Thank you. A general entry of $(AB)C$ and $A(BC)$ should have the indices $g,m$, I believe.
– Matt.P
Sep 11 at 14:23
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to prove that matrix multiplication is associative, but seem to be making mistakes in each of my past write-ups, so hopefully someone can check over my work.
Theorem. Let $A$ be $alpha times beta$, $B$ be $beta times gamma$, and $C$ be $gamma times delta$. Prove that $(AB)C = A(BC)$.
Proof. Define general entries of the matrices $A$, $B$, and $C$ by $a_g,h$, $b_i,j$, and $c_k,m$, respectively. Then, for the LHS:
beginalign*
& (AB)_alpha, gamma = sumlimits_p=1^beta a_alpha,p b_p,gamma \
& left((AB)Cright)_alpha, delta = sumlimits_n=1^gamma left(ABright)_alpha, n c_n, delta = sumlimits_n=1^gamma left(sumlimits_p=1^beta a_alpha,p b_p,n right) c_n, delta = sumlimits_n=1^gamma sumlimits_p=1^beta left(a_alpha,p b_p,nright) c_n, delta.
endalign*
For the RHS:
beginalign*
& left(BCright)_beta, delta = sumlimits_n=1^gamma b_beta, n c_n, delta \
& left(Aleft(BCright)right)_alpha,delta = sumlimits_p=1^beta a_alpha,p (BC)_p, delta = sumlimits_p=1^beta a_alpha,p left(sumlimits_n=1^gamma b_p, n c_n, delta right) = sumlimits_p=1^beta sumlimits_n=1^gamma a_alpha,p left(b_p, n c_n, delta right).
endalign*
Assuming I have written these correctly, we can make two observations: first, the summands are equivalent, as multiplication is associative. Second, the order of the summations doesn't matter when we're summing a finite number of entries. Thus, $(AB)C = A(BC)$.
How does this look?
linear-algebra proof-verification
asked Sep 11 at 4:30
Matt.P
1,015414
I'm trying to prove that matrix multiplication is associative, but seem to be making mistakes in each of my past write-ups, so hopefully someone can check over my work.
Theorem. Let $A$ be $alpha times beta$, $B$ be $beta times gamma$, and $C$ be $gamma times delta$. Prove that $(AB)C = A(BC)$.
Proof. Define general entries of the matrices $A$, $B$, and $C$ by $a_g,h$, $b_i,j$, and $c_k,m$, respectively. Then, for the LHS:
beginalign*
& (AB)_alpha, gamma = sumlimits_p=1^beta a_alpha,p b_p,gamma \
& left((AB)Cright)_alpha, delta = sumlimits_n=1^gamma left(ABright)_alpha, n c_n, delta = sumlimits_n=1^gamma left(sumlimits_p=1^beta a_alpha,p b_p,n right) c_n, delta = sumlimits_n=1^gamma sumlimits_p=1^beta left(a_alpha,p b_p,nright) c_n, delta.
endalign*
For the RHS:
beginalign*
& left(BCright)_beta, delta = sumlimits_n=1^gamma b_beta, n c_n, delta \
& left(Aleft(BCright)right)_alpha,delta = sumlimits_p=1^beta a_alpha,p (BC)_p, delta = sumlimits_p=1^beta a_alpha,p left(sumlimits_n=1^gamma b_p, n c_n, delta right) = sumlimits_p=1^beta sumlimits_n=1^gamma a_alpha,p left(b_p, n c_n, delta right).
endalign*
Assuming I have written these correctly, we can make two observations: first, the summands are equivalent, as multiplication is associative. Second, the order of the summations doesn't matter when we're summing a finite number of entries. Thus, $(AB)C = A(BC)$.
How does this look?
linear-algebra proof-verification
linear-algebra proof-verification
asked Sep 11 at 4:30
Matt.P
1,015414
asked Sep 11 at 4:30
Matt.P
1,015414
asked Sep 11 at 4:30
Matt.P
1,015414
asked Sep 11 at 4:30
Matt.P
1,015414
asked Sep 11 at 4:30
Matt.P
1,015414
1,015414
2
$checkmark,!$
– David
Sep 11 at 4:37
1
Just a point on notation: you are using $alpha$ as both the number of rows of $A$ and as a general row-index; and same for$beta$, $gamma$ and $delta$. You should clean that up for a completely correct proof.
– ancientmathematician
Sep 11 at 6:56
This is very true and I hadn't realized it. Thank you. A general entry of $(AB)C$ and $A(BC)$ should have the indices $g,m$, I believe.
– Matt.P
Sep 11 at 14:23
add a comment |
2
$checkmark,!$
– David
Sep 11 at 4:37
1
Just a point on notation: you are using $alpha$ as both the number of rows of $A$ and as a general row-index; and same for$beta$, $gamma$ and $delta$. You should clean that up for a completely correct proof.
– ancientmathematician
Sep 11 at 6:56
This is very true and I hadn't realized it. Thank you. A general entry of $(AB)C$ and $A(BC)$ should have the indices $g,m$, I believe.
– Matt.P
Sep 11 at 14:23
2
2
$checkmark,!$
– David
Sep 11 at 4:37
$checkmark,!$
– David
Sep 11 at 4:37
1
1
Just a point on notation: you are using $alpha$ as both the number of rows of $A$ and as a general row-index; and same for$beta$, $gamma$ and $delta$. You should clean that up for a completely correct proof.
– ancientmathematician
Sep 11 at 6:56
Just a point on notation: you are using $alpha$ as both the number of rows of $A$ and as a general row-index; and same for$beta$, $gamma$ and $delta$. You should clean that up for a completely correct proof.
– ancientmathematician
Sep 11 at 6:56
This is very true and I hadn't realized it. Thank you. A general entry of $(AB)C$ and $A(BC)$ should have the indices $g,m$, I believe.
– Matt.P
Sep 11 at 14:23
This is very true and I hadn't realized it. Thank you. A general entry of $(AB)C$ and $A(BC)$ should have the indices $g,m$, I believe.
– Matt.P
Sep 11 at 14:23
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
Your proof is fine.
We can change the order of summation as the sum is finite. When we mention multiplication is associative, we might want to mention multiplicative of which object, such as multiplicative of real numbers or complex number.
answered Sep 11 at 4:36
Siong Thye Goh
92k1460113
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Your proof is fine.
We can change the order of summation as the sum is finite. When we mention multiplication is associative, we might want to mention multiplicative of which object, such as multiplicative of real numbers or complex number.
answered Sep 11 at 4:36
Siong Thye Goh
92k1460113
add a comment |
up vote
2
down vote
Your proof is fine.
We can change the order of summation as the sum is finite. When we mention multiplication is associative, we might want to mention multiplicative of which object, such as multiplicative of real numbers or complex number.
answered Sep 11 at 4:36
Siong Thye Goh
92k1460113
add a comment |
up vote
2
down vote
up vote
2
down vote
Your proof is fine.
We can change the order of summation as the sum is finite. When we mention multiplication is associative, we might want to mention multiplicative of which object, such as multiplicative of real numbers or complex number.
answered Sep 11 at 4:36
Siong Thye Goh
92k1460113
Your proof is fine.
We can change the order of summation as the sum is finite. When we mention multiplication is associative, we might want to mention multiplicative of which object, such as multiplicative of real numbers or complex number.
answered Sep 11 at 4:36
Siong Thye Goh
92k1460113
answered Sep 11 at 4:36
Siong Thye Goh
92k1460113
answered Sep 11 at 4:36
Siong Thye Goh
92k1460113
answered Sep 11 at 4:36
Siong Thye Goh
92k1460113
92k1460113
add a comment |
add a comment |
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2
$checkmark,!$
– David
Sep 11 at 4:37
1
Just a point on notation: you are using $alpha$ as both the number of rows of $A$ and as a general row-index; and same for$beta$, $gamma$ and $delta$. You should clean that up for a completely correct proof.
– ancientmathematician
Sep 11 at 6:56
This is very true and I hadn't realized it. Thank you. A general entry of $(AB)C$ and $A(BC)$ should have the indices $g,m$, I believe.
– Matt.P
Sep 11 at 14:23