Mixing
Question regarding the algebra of norms
$begingroup$
I am new to linear algebra, as am having some doubts regarding the following question:
True or False
$u,v in R^n$
$leftlVert urightrVert=leftlVert vrightrVert$ if and only if u+v and u-v are orthogonal.- For every $u,v in R^n$ and every $c in R$: $leftlVert cu+vrightrVert^2=c^2leftlVert urightrVert^2+2c(u·v)+leftlVert vrightrVert^2$
I worked out with basic algebra that 1 is true. However I am doubting re 2: algebraically it makes sense, and I have put in two sets of numbers that both came out correct, but I am doubting whether there might be a detail that I am missing (because the question states "for every u,v…"
Thank you!
linear-algebra norm orthogonality
asked Jan 18 at 9:20
daltadalta
1168
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra, as am having some doubts regarding the following question:
True or False
$u,v in R^n$
$leftlVert urightrVert=leftlVert vrightrVert$ if and only if u+v and u-v are orthogonal.- For every $u,v in R^n$ and every $c in R$: $leftlVert cu+vrightrVert^2=c^2leftlVert urightrVert^2+2c(u·v)+leftlVert vrightrVert^2$
I worked out with basic algebra that 1 is true. However I am doubting re 2: algebraically it makes sense, and I have put in two sets of numbers that both came out correct, but I am doubting whether there might be a detail that I am missing (because the question states "for every u,v…"
Thank you!
linear-algebra norm orthogonality
asked Jan 18 at 9:20
daltadalta
1168
$endgroup$
$begingroup$
To state your proof of 2. rigorously: First, write out the single algebraic sentence that "shows" that the LHS of the Equation is equal to the RHS. Second, put brackets around that sentence and precede it with $forall u,vin Bbb R^n,forall cin Bbb R.$ That's all there is to it.
$endgroup$
– DanielWainfleet
Jan 18 at 10:24
add a comment |
$begingroup$
I am new to linear algebra, as am having some doubts regarding the following question:
True or False
$u,v in R^n$
$leftlVert urightrVert=leftlVert vrightrVert$ if and only if u+v and u-v are orthogonal.- For every $u,v in R^n$ and every $c in R$: $leftlVert cu+vrightrVert^2=c^2leftlVert urightrVert^2+2c(u·v)+leftlVert vrightrVert^2$
I worked out with basic algebra that 1 is true. However I am doubting re 2: algebraically it makes sense, and I have put in two sets of numbers that both came out correct, but I am doubting whether there might be a detail that I am missing (because the question states "for every u,v…"
Thank you!
linear-algebra norm orthogonality
asked Jan 18 at 9:20
daltadalta
1168
$endgroup$
I am new to linear algebra, as am having some doubts regarding the following question:
True or False
$u,v in R^n$
$leftlVert urightrVert=leftlVert vrightrVert$ if and only if u+v and u-v are orthogonal.- For every $u,v in R^n$ and every $c in R$: $leftlVert cu+vrightrVert^2=c^2leftlVert urightrVert^2+2c(u·v)+leftlVert vrightrVert^2$
I worked out with basic algebra that 1 is true. However I am doubting re 2: algebraically it makes sense, and I have put in two sets of numbers that both came out correct, but I am doubting whether there might be a detail that I am missing (because the question states "for every u,v…"
Thank you!
linear-algebra norm orthogonality
linear-algebra norm orthogonality
asked Jan 18 at 9:20
daltadalta
1168
asked Jan 18 at 9:20
daltadalta
1168
asked Jan 18 at 9:20
daltadalta
1168
asked Jan 18 at 9:20
daltadalta
1168
asked Jan 18 at 9:20
daltadalta
1168
1168
$begingroup$
To state your proof of 2. rigorously: First, write out the single algebraic sentence that "shows" that the LHS of the Equation is equal to the RHS. Second, put brackets around that sentence and precede it with $forall u,vin Bbb R^n,forall cin Bbb R.$ That's all there is to it.
$endgroup$
– DanielWainfleet
Jan 18 at 10:24
add a comment |
$begingroup$
To state your proof of 2. rigorously: First, write out the single algebraic sentence that "shows" that the LHS of the Equation is equal to the RHS. Second, put brackets around that sentence and precede it with $forall u,vin Bbb R^n,forall cin Bbb R.$ That's all there is to it.
$endgroup$
– DanielWainfleet
Jan 18 at 10:24
$begingroup$
To state your proof of 2. rigorously: First, write out the single algebraic sentence that "shows" that the LHS of the Equation is equal to the RHS. Second, put brackets around that sentence and precede it with $forall u,vin Bbb R^n,forall cin Bbb R.$ That's all there is to it.
$endgroup$
– DanielWainfleet
Jan 18 at 10:24
$begingroup$
To state your proof of 2. rigorously: First, write out the single algebraic sentence that "shows" that the LHS of the Equation is equal to the RHS. Second, put brackets around that sentence and precede it with $forall u,vin Bbb R^n,forall cin Bbb R.$ That's all there is to it.
$endgroup$
– DanielWainfleet
Jan 18 at 10:24
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Just using the definition of the norm in $mathbb{R}^n$: for $u in mathbb{R}^n$ $||u||:=sqrt{(u,u)}$, where $(cdot,cdot)$ represents the standard inner product in $mathbb{R}^n$.
Now $||cu+v||^2={sqrt{(cu+v,cu+v)}}^2=(cu+v,cu+v)=(cu+v,cu)+(cu+v,v)=(cu,cu)+(v,cu)+(cu,v)+(v,v)=||cu||^2+(cu,v)+(cu,v)+||v||^2=c^2||u||^2+2c(u,v)+||v||^2$
using the bilinearity and scalar multiplication of the inner product.
answered Jan 18 at 9:28
user289143user289143
1,002313
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add a comment |
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1 Answer
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active
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1 Answer
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active
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$begingroup$
Just using the definition of the norm in $mathbb{R}^n$: for $u in mathbb{R}^n$ $||u||:=sqrt{(u,u)}$, where $(cdot,cdot)$ represents the standard inner product in $mathbb{R}^n$.
Now $||cu+v||^2={sqrt{(cu+v,cu+v)}}^2=(cu+v,cu+v)=(cu+v,cu)+(cu+v,v)=(cu,cu)+(v,cu)+(cu,v)+(v,v)=||cu||^2+(cu,v)+(cu,v)+||v||^2=c^2||u||^2+2c(u,v)+||v||^2$
using the bilinearity and scalar multiplication of the inner product.
answered Jan 18 at 9:28
user289143user289143
1,002313
$endgroup$
add a comment |
$begingroup$
Just using the definition of the norm in $mathbb{R}^n$: for $u in mathbb{R}^n$ $||u||:=sqrt{(u,u)}$, where $(cdot,cdot)$ represents the standard inner product in $mathbb{R}^n$.
Now $||cu+v||^2={sqrt{(cu+v,cu+v)}}^2=(cu+v,cu+v)=(cu+v,cu)+(cu+v,v)=(cu,cu)+(v,cu)+(cu,v)+(v,v)=||cu||^2+(cu,v)+(cu,v)+||v||^2=c^2||u||^2+2c(u,v)+||v||^2$
using the bilinearity and scalar multiplication of the inner product.
answered Jan 18 at 9:28
user289143user289143
1,002313
$endgroup$
add a comment |
$begingroup$
Just using the definition of the norm in $mathbb{R}^n$: for $u in mathbb{R}^n$ $||u||:=sqrt{(u,u)}$, where $(cdot,cdot)$ represents the standard inner product in $mathbb{R}^n$.
Now $||cu+v||^2={sqrt{(cu+v,cu+v)}}^2=(cu+v,cu+v)=(cu+v,cu)+(cu+v,v)=(cu,cu)+(v,cu)+(cu,v)+(v,v)=||cu||^2+(cu,v)+(cu,v)+||v||^2=c^2||u||^2+2c(u,v)+||v||^2$
using the bilinearity and scalar multiplication of the inner product.
answered Jan 18 at 9:28
user289143user289143
1,002313
$endgroup$
Just using the definition of the norm in $mathbb{R}^n$: for $u in mathbb{R}^n$ $||u||:=sqrt{(u,u)}$, where $(cdot,cdot)$ represents the standard inner product in $mathbb{R}^n$.
Now $||cu+v||^2={sqrt{(cu+v,cu+v)}}^2=(cu+v,cu+v)=(cu+v,cu)+(cu+v,v)=(cu,cu)+(v,cu)+(cu,v)+(v,v)=||cu||^2+(cu,v)+(cu,v)+||v||^2=c^2||u||^2+2c(u,v)+||v||^2$
using the bilinearity and scalar multiplication of the inner product.
answered Jan 18 at 9:28
user289143user289143
1,002313
answered Jan 18 at 9:28
user289143user289143
1,002313
answered Jan 18 at 9:28
user289143user289143
1,002313
answered Jan 18 at 9:28
user289143user289143
1,002313
1,002313
add a comment |
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$begingroup$
To state your proof of 2. rigorously: First, write out the single algebraic sentence that "shows" that the LHS of the Equation is equal to the RHS. Second, put brackets around that sentence and precede it with $forall u,vin Bbb R^n,forall cin Bbb R.$ That's all there is to it.
$endgroup$
– DanielWainfleet
Jan 18 at 10:24