Mixing
Proving axiom 1 of inner product
$begingroup$
I am doing one of the problems in my book, but I am uncertain if my approach is correct.
Problem: $(u,v)=-u_1u_2u_3$
Let $u=(u_1,u_2,u_3)$,
$v=(v_1,v_2,v_3)$.
begin{align}
(u,v)&= u_3u_2u_1 \
&= (v,u)
end{align}
Axiom 1 proved. Is this correct?
linear-algebra proof-verification
edited Jan 28 at 17:17
David Kraemer
37516
asked Jan 28 at 17:00
JohnySmith12JohnySmith12
413
$endgroup$
add a comment |
$begingroup$
I am doing one of the problems in my book, but I am uncertain if my approach is correct.
Problem: $(u,v)=-u_1u_2u_3$
Let $u=(u_1,u_2,u_3)$,
$v=(v_1,v_2,v_3)$.
begin{align}
(u,v)&= u_3u_2u_1 \
&= (v,u)
end{align}
Axiom 1 proved. Is this correct?
linear-algebra proof-verification
edited Jan 28 at 17:17
David Kraemer
37516
asked Jan 28 at 17:00
JohnySmith12JohnySmith12
413
$endgroup$
$begingroup$
Can you list the axioms so we know which is "axiom 1"?
$endgroup$
– David Kraemer
Jan 28 at 17:07
add a comment |
$begingroup$
I am doing one of the problems in my book, but I am uncertain if my approach is correct.
Problem: $(u,v)=-u_1u_2u_3$
Let $u=(u_1,u_2,u_3)$,
$v=(v_1,v_2,v_3)$.
begin{align}
(u,v)&= u_3u_2u_1 \
&= (v,u)
end{align}
Axiom 1 proved. Is this correct?
linear-algebra proof-verification
edited Jan 28 at 17:17
David Kraemer
37516
asked Jan 28 at 17:00
JohnySmith12JohnySmith12
413
$endgroup$
I am doing one of the problems in my book, but I am uncertain if my approach is correct.
Problem: $(u,v)=-u_1u_2u_3$
Let $u=(u_1,u_2,u_3)$,
$v=(v_1,v_2,v_3)$.
begin{align}
(u,v)&= u_3u_2u_1 \
&= (v,u)
end{align}
Axiom 1 proved. Is this correct?
linear-algebra proof-verification
linear-algebra proof-verification
edited Jan 28 at 17:17
David Kraemer
37516
asked Jan 28 at 17:00
JohnySmith12JohnySmith12
413
edited Jan 28 at 17:17
David Kraemer
37516
asked Jan 28 at 17:00
JohnySmith12JohnySmith12
413
edited Jan 28 at 17:17
David Kraemer
37516
edited Jan 28 at 17:17
David Kraemer
37516
edited Jan 28 at 17:17
David Kraemer
37516
37516
asked Jan 28 at 17:00
JohnySmith12JohnySmith12
413
asked Jan 28 at 17:00
JohnySmith12JohnySmith12
413
asked Jan 28 at 17:00
JohnySmith12JohnySmith12
413
413
$begingroup$
Can you list the axioms so we know which is "axiom 1"?
$endgroup$
– David Kraemer
Jan 28 at 17:07
add a comment |
$begingroup$
Can you list the axioms so we know which is "axiom 1"?
$endgroup$
– David Kraemer
Jan 28 at 17:07
$begingroup$
Can you list the axioms so we know which is "axiom 1"?
$endgroup$
– David Kraemer
Jan 28 at 17:07
$begingroup$
Can you list the axioms so we know which is "axiom 1"?
$endgroup$
– David Kraemer
Jan 28 at 17:07
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
It's totally incorrect.
It's a function with two arguments that is defined here: $(u, v):=-u_1u_2u_3$.
Putting in $x,y$, we get $(x,y)=-x_1x_2x_3$.
Putting in $v,u$, we get $(v,u)=-v_1v_2v_3$.
Now we can easily show up two specific vectors $u, v$ such that $(u, v) ne (v, u)$.
answered Jan 28 at 17:14
BerciBerci
61.8k23674
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
It's totally incorrect.
It's a function with two arguments that is defined here: $(u, v):=-u_1u_2u_3$.
Putting in $x,y$, we get $(x,y)=-x_1x_2x_3$.
Putting in $v,u$, we get $(v,u)=-v_1v_2v_3$.
Now we can easily show up two specific vectors $u, v$ such that $(u, v) ne (v, u)$.
answered Jan 28 at 17:14
BerciBerci
61.8k23674
$endgroup$
add a comment |
$begingroup$
It's totally incorrect.
It's a function with two arguments that is defined here: $(u, v):=-u_1u_2u_3$.
Putting in $x,y$, we get $(x,y)=-x_1x_2x_3$.
Putting in $v,u$, we get $(v,u)=-v_1v_2v_3$.
Now we can easily show up two specific vectors $u, v$ such that $(u, v) ne (v, u)$.
answered Jan 28 at 17:14
BerciBerci
61.8k23674
$endgroup$
add a comment |
$begingroup$
It's totally incorrect.
It's a function with two arguments that is defined here: $(u, v):=-u_1u_2u_3$.
Putting in $x,y$, we get $(x,y)=-x_1x_2x_3$.
Putting in $v,u$, we get $(v,u)=-v_1v_2v_3$.
Now we can easily show up two specific vectors $u, v$ such that $(u, v) ne (v, u)$.
answered Jan 28 at 17:14
BerciBerci
61.8k23674
$endgroup$
It's totally incorrect.
It's a function with two arguments that is defined here: $(u, v):=-u_1u_2u_3$.
Putting in $x,y$, we get $(x,y)=-x_1x_2x_3$.
Putting in $v,u$, we get $(v,u)=-v_1v_2v_3$.
Now we can easily show up two specific vectors $u, v$ such that $(u, v) ne (v, u)$.
answered Jan 28 at 17:14
BerciBerci
61.8k23674
answered Jan 28 at 17:14
BerciBerci
61.8k23674
answered Jan 28 at 17:14
BerciBerci
61.8k23674
answered Jan 28 at 17:14
BerciBerci
61.8k23674
61.8k23674
add a comment |
add a comment |
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$begingroup$
Can you list the axioms so we know which is "axiom 1"?
$endgroup$
– David Kraemer
Jan 28 at 17:07