Other inner products for $mathbbR^n$Defining an Inner ProductInner product alternative definitionFinding uniq inner product satisying the following requierments:Prove that there is a unique inner product on $V$Characterisation of inner products preserved by an automorphismThe Redundancy of applying the Gram-Schmidt Procedure on an orthogonal subset of $V$.If complementary subspaces are almost orthogonal, is the same true for their orthogonal complements?Find inner product that meets primary decomposition criteriasan orthogonal map associated with inner productHow to prove Cauchy-Schwarz inequality using multivariable calculus?
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Other inner products for $mathbbR^n$
Defining an Inner ProductInner product alternative definitionFinding uniq inner product satisying the following requierments:Prove that there is a unique inner product on $V$Characterisation of inner products preserved by an automorphismThe Redundancy of applying the Gram-Schmidt Procedure on an orthogonal subset of $V$.If complementary subspaces are almost orthogonal, is the same true for their orthogonal complements?Find inner product that meets primary decomposition criteriasan orthogonal map associated with inner productHow to prove Cauchy-Schwarz inequality using multivariable calculus?
$begingroup$
For $mathbbR^n$, the standard inner product is the dot product. It is defined as $ langle v,,wrangle = sum_i v_i cdot w_i $. I am aware that any scaled version, namely
$ langle v,,wrangle = sum_ilambda_icdot v_i cdot w_i $ will still satisfy the 4 inner product requirements.
Is there any inner product for $mathbbR^n$ that is not just a scaled version of the standard dot product?
I tried for $mathbbR^2$ with $ langle v,,wrangle = v_1 cdot w_2 + v_2 cdot w_1 $ but that is not positive definite.
linear-algebra inner-product-space
$endgroup$
add a comment |
$begingroup$
For $mathbbR^n$, the standard inner product is the dot product. It is defined as $ langle v,,wrangle = sum_i v_i cdot w_i $. I am aware that any scaled version, namely
$ langle v,,wrangle = sum_ilambda_icdot v_i cdot w_i $ will still satisfy the 4 inner product requirements.
Is there any inner product for $mathbbR^n$ that is not just a scaled version of the standard dot product?
I tried for $mathbbR^2$ with $ langle v,,wrangle = v_1 cdot w_2 + v_2 cdot w_1 $ but that is not positive definite.
linear-algebra inner-product-space
$endgroup$
$begingroup$
So, everyone is saying you can also use a matrix, and that works. Is there a proof though that any inner product can be represented by such a matrix type inner product?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
$newcommandRmathbbRnewcommandumathbfe$Yes. Here is a sketch. Suppose that $(cdot, cdot)$ is some inner product on $R^n$. Let $u_1, ldots,u_n$ be the standard basis for $R^n$ (so $u_k$ is a vector whose $k$-th entry is $1$, and all other entries are $0$). Define $A$ to be the $ntimes n$ matrix with $i,j$ entry $colorbluea_ij= (u_i, u_j)$. Then show that this $A$ satisfies $(mathbfx, mathbfy) = mathbfx^T A mathbfy$ for all $mathbfx,mathbfyin R^n$. EDIT: I see Travis has mentioned this in an answer below.
$endgroup$
– Minus One-Twelfth
2 days ago
add a comment |
$begingroup$
For $mathbbR^n$, the standard inner product is the dot product. It is defined as $ langle v,,wrangle = sum_i v_i cdot w_i $. I am aware that any scaled version, namely
$ langle v,,wrangle = sum_ilambda_icdot v_i cdot w_i $ will still satisfy the 4 inner product requirements.
Is there any inner product for $mathbbR^n$ that is not just a scaled version of the standard dot product?
I tried for $mathbbR^2$ with $ langle v,,wrangle = v_1 cdot w_2 + v_2 cdot w_1 $ but that is not positive definite.
linear-algebra inner-product-space
$endgroup$
For $mathbbR^n$, the standard inner product is the dot product. It is defined as $ langle v,,wrangle = sum_i v_i cdot w_i $. I am aware that any scaled version, namely
$ langle v,,wrangle = sum_ilambda_icdot v_i cdot w_i $ will still satisfy the 4 inner product requirements.
Is there any inner product for $mathbbR^n$ that is not just a scaled version of the standard dot product?
I tried for $mathbbR^2$ with $ langle v,,wrangle = v_1 cdot w_2 + v_2 cdot w_1 $ but that is not positive definite.
linear-algebra inner-product-space
linear-algebra inner-product-space
edited 2 days ago
J.G.
29.6k22946
29.6k22946
asked 2 days ago
Ion SmeIon Sme
11510
11510
$begingroup$
So, everyone is saying you can also use a matrix, and that works. Is there a proof though that any inner product can be represented by such a matrix type inner product?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
$newcommandRmathbbRnewcommandumathbfe$Yes. Here is a sketch. Suppose that $(cdot, cdot)$ is some inner product on $R^n$. Let $u_1, ldots,u_n$ be the standard basis for $R^n$ (so $u_k$ is a vector whose $k$-th entry is $1$, and all other entries are $0$). Define $A$ to be the $ntimes n$ matrix with $i,j$ entry $colorbluea_ij= (u_i, u_j)$. Then show that this $A$ satisfies $(mathbfx, mathbfy) = mathbfx^T A mathbfy$ for all $mathbfx,mathbfyin R^n$. EDIT: I see Travis has mentioned this in an answer below.
$endgroup$
– Minus One-Twelfth
2 days ago
add a comment |
$begingroup$
So, everyone is saying you can also use a matrix, and that works. Is there a proof though that any inner product can be represented by such a matrix type inner product?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
$newcommandRmathbbRnewcommandumathbfe$Yes. Here is a sketch. Suppose that $(cdot, cdot)$ is some inner product on $R^n$. Let $u_1, ldots,u_n$ be the standard basis for $R^n$ (so $u_k$ is a vector whose $k$-th entry is $1$, and all other entries are $0$). Define $A$ to be the $ntimes n$ matrix with $i,j$ entry $colorbluea_ij= (u_i, u_j)$. Then show that this $A$ satisfies $(mathbfx, mathbfy) = mathbfx^T A mathbfy$ for all $mathbfx,mathbfyin R^n$. EDIT: I see Travis has mentioned this in an answer below.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
So, everyone is saying you can also use a matrix, and that works. Is there a proof though that any inner product can be represented by such a matrix type inner product?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
So, everyone is saying you can also use a matrix, and that works. Is there a proof though that any inner product can be represented by such a matrix type inner product?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
$newcommandRmathbbRnewcommandumathbfe$Yes. Here is a sketch. Suppose that $(cdot, cdot)$ is some inner product on $R^n$. Let $u_1, ldots,u_n$ be the standard basis for $R^n$ (so $u_k$ is a vector whose $k$-th entry is $1$, and all other entries are $0$). Define $A$ to be the $ntimes n$ matrix with $i,j$ entry $colorbluea_ij= (u_i, u_j)$. Then show that this $A$ satisfies $(mathbfx, mathbfy) = mathbfx^T A mathbfy$ for all $mathbfx,mathbfyin R^n$. EDIT: I see Travis has mentioned this in an answer below.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
$newcommandRmathbbRnewcommandumathbfe$Yes. Here is a sketch. Suppose that $(cdot, cdot)$ is some inner product on $R^n$. Let $u_1, ldots,u_n$ be the standard basis for $R^n$ (so $u_k$ is a vector whose $k$-th entry is $1$, and all other entries are $0$). Define $A$ to be the $ntimes n$ matrix with $i,j$ entry $colorbluea_ij= (u_i, u_j)$. Then show that this $A$ satisfies $(mathbfx, mathbfy) = mathbfx^T A mathbfy$ for all $mathbfx,mathbfyin R^n$. EDIT: I see Travis has mentioned this in an answer below.
$endgroup$
– Minus One-Twelfth
2 days ago
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
Yes. Given any positive definite $n times n$ matrix $A$, $$bf x cdot bf y := bf y^top A bf x$$
defines an inner product on $Bbb R^n$, and for $n > 1$ there are nondiagonal positive definite matrices, for example,
$$pmatrix1&epsilon\epsilon&1\&&1\&&&ddots\&&&&1$$
for any $0 < |epsilon| < 1$.
Conversely all inner products arise this way, as we can recover $A$ by setting $$A_ij = bf e_i cdot bf e_j$$ for the standard basis $(bf e_i)$.
On the other hand, given any inner product on $Bbb R^n$, applying the Gram-Schmidt Process produces an orthonormal basis $(bf f_i)$, so the matrix representation of the inner product with respect to that basis is the identity matrix, $I_n$. In this sense, all inner products on $Bbb R^n$ are equivalent.
$endgroup$
1
$begingroup$
I think that if OP knew what a positive definite symmetric matrix is, there would be no need to pose the question as it is.
$endgroup$
– Marc van Leeuwen
yesterday
$begingroup$
@MarcvanLeeuwen Thanks for the comment. And yes, I agree, but one needs that language in order to state the result precisely. Still, that probably means OP would benefit from a concrete example, which I've added to the answer.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
For any invertible linear transformation $A$ you can define the inner product $langle v,wrangle_A=langle Av,Awrangle$ where $langlecdot,cdotrangle$ denotes the standard inner product. I expect there are no other inner products, which is motivated by the fact that all inner products are known to induce equivalent norms.
$endgroup$
add a comment |
$begingroup$
Technically, you need positive $lambda_i$. Or if we use $sum_ijlambda_ijv_iw_j$, the matrix $lambda$ is without loss of generality equal to $(lambda+lambda^T)/2$, and it has to be positive-definite. (Yes, this matrix property has the same name; it basically means it has only positive eigenvalues.) With an appropriate basis change we can then diagonalize this matrix, which recovers the case you knew about. As for the example you tried, it failed because if you work out the matrix $lambda=left(beginarraycc
0 & 1\
1 & 0
endarrayright)$ (once we make it self-adjoint as explained above), which has $-1$ as an eigenvalue.
$endgroup$
$begingroup$
Where does the condition $lambda = (lambda+lambda^T)/2$ come from?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
@IonSme As I said, it's imposed without loss of generality. If you replace $lambda$ as thus, the resulting inner product won't be changed. (This is obvious in the case $v=w$, but in fact this is exhaustive because norms determine an inner product viz. $langle v,, wrangle = (Vert v+wVert^2-Vert v-wVert^2)/4$.)
$endgroup$
– J.G.
2 days ago
add a comment |
$begingroup$
Inner products $p(x,y)$ on $mathbb R^n$ have the form
$$
p(x,y) = sum_j=1^n sum_k=1^n a_jk x_j y_k
$$
where the matrix $A = [a_jk]$ is positive definite. Choosing a basis of eigenvectors for the matrix $A$, and expanding according to this new basis instead of the original basis, the inner product is then diagonal:
$$
p(x,y) = sum_j=1^n b_j x_j y_j
$$
where $b_j>0$.
$endgroup$
add a comment |
$begingroup$
I agree with SmileyCraft. In finite dimensional vector spaces, bilinear transformations, as linear transformations, can be written in terms of the values that they adopt in a given base:$$left langle x,y right rangle=sum_i,j=1^nx_iy_jleft langle e_i,e_j right rangle.$$
I believe you can arrive in this representation without difficult, proving then you suspicion.
New contributor
$endgroup$
add a comment |
$begingroup$
It is well known (and easy to prove) that any two finite dimensional inner product spaces are isometrically isomorphic. Hence $ langle x, y rangle'$ is an inner product on $mathbb R^n$ iff there is a vector space isomorphism $T: mathbb R^n to mathbb R^n$ such that $langle x, y rangle' =langle Tx, Ty rangle$ for all $x,y$.
$endgroup$
add a comment |
$begingroup$
In general, for any $ntimes n$ matrix $A=(a_i,j)_i,j=1,ldots,n$ the expression $sum_i,j=1^na_i,jx_iy_j$ defines a bilinear form, which will be symmetric if and only $A$ is. Giving a positive definite symmetric bilinear form is a more subtle condition that leads to inequalities for the coefficients of $A$ (and the matrices that satisfy the condition are naturally called positive definite). For $2times2$ symmetric matrices the positive definite condition is $a_1,1>0$, $a_2,2>0$ together with $a_1,1a_2,2-a_1,2^2>0$ (so $det(A)>0$). For a concrete example, the symmetric matrix
$$
A=pmatrix1½\frac12&1
quadtextgives an inner product with
langle v,,wrangle = v_1w_1 +frac12(v_1w_2+v_2w_1) + v_2w_2,.
$$
In higher dimension the condition is more complicated, but in any case one does get many different inner products on $Bbb R^n$ in this way. They do turn out to be all equivalent in the sense that they give rise to the same structure theory, but they are not equal.
$endgroup$
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. Given any positive definite $n times n$ matrix $A$, $$bf x cdot bf y := bf y^top A bf x$$
defines an inner product on $Bbb R^n$, and for $n > 1$ there are nondiagonal positive definite matrices, for example,
$$pmatrix1&epsilon\epsilon&1\&&1\&&&ddots\&&&&1$$
for any $0 < |epsilon| < 1$.
Conversely all inner products arise this way, as we can recover $A$ by setting $$A_ij = bf e_i cdot bf e_j$$ for the standard basis $(bf e_i)$.
On the other hand, given any inner product on $Bbb R^n$, applying the Gram-Schmidt Process produces an orthonormal basis $(bf f_i)$, so the matrix representation of the inner product with respect to that basis is the identity matrix, $I_n$. In this sense, all inner products on $Bbb R^n$ are equivalent.
$endgroup$
1
$begingroup$
I think that if OP knew what a positive definite symmetric matrix is, there would be no need to pose the question as it is.
$endgroup$
– Marc van Leeuwen
yesterday
$begingroup$
@MarcvanLeeuwen Thanks for the comment. And yes, I agree, but one needs that language in order to state the result precisely. Still, that probably means OP would benefit from a concrete example, which I've added to the answer.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
Yes. Given any positive definite $n times n$ matrix $A$, $$bf x cdot bf y := bf y^top A bf x$$
defines an inner product on $Bbb R^n$, and for $n > 1$ there are nondiagonal positive definite matrices, for example,
$$pmatrix1&epsilon\epsilon&1\&&1\&&&ddots\&&&&1$$
for any $0 < |epsilon| < 1$.
Conversely all inner products arise this way, as we can recover $A$ by setting $$A_ij = bf e_i cdot bf e_j$$ for the standard basis $(bf e_i)$.
On the other hand, given any inner product on $Bbb R^n$, applying the Gram-Schmidt Process produces an orthonormal basis $(bf f_i)$, so the matrix representation of the inner product with respect to that basis is the identity matrix, $I_n$. In this sense, all inner products on $Bbb R^n$ are equivalent.
$endgroup$
1
$begingroup$
I think that if OP knew what a positive definite symmetric matrix is, there would be no need to pose the question as it is.
$endgroup$
– Marc van Leeuwen
yesterday
$begingroup$
@MarcvanLeeuwen Thanks for the comment. And yes, I agree, but one needs that language in order to state the result precisely. Still, that probably means OP would benefit from a concrete example, which I've added to the answer.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
Yes. Given any positive definite $n times n$ matrix $A$, $$bf x cdot bf y := bf y^top A bf x$$
defines an inner product on $Bbb R^n$, and for $n > 1$ there are nondiagonal positive definite matrices, for example,
$$pmatrix1&epsilon\epsilon&1\&&1\&&&ddots\&&&&1$$
for any $0 < |epsilon| < 1$.
Conversely all inner products arise this way, as we can recover $A$ by setting $$A_ij = bf e_i cdot bf e_j$$ for the standard basis $(bf e_i)$.
On the other hand, given any inner product on $Bbb R^n$, applying the Gram-Schmidt Process produces an orthonormal basis $(bf f_i)$, so the matrix representation of the inner product with respect to that basis is the identity matrix, $I_n$. In this sense, all inner products on $Bbb R^n$ are equivalent.
$endgroup$
Yes. Given any positive definite $n times n$ matrix $A$, $$bf x cdot bf y := bf y^top A bf x$$
defines an inner product on $Bbb R^n$, and for $n > 1$ there are nondiagonal positive definite matrices, for example,
$$pmatrix1&epsilon\epsilon&1\&&1\&&&ddots\&&&&1$$
for any $0 < |epsilon| < 1$.
Conversely all inner products arise this way, as we can recover $A$ by setting $$A_ij = bf e_i cdot bf e_j$$ for the standard basis $(bf e_i)$.
On the other hand, given any inner product on $Bbb R^n$, applying the Gram-Schmidt Process produces an orthonormal basis $(bf f_i)$, so the matrix representation of the inner product with respect to that basis is the identity matrix, $I_n$. In this sense, all inner products on $Bbb R^n$ are equivalent.
edited yesterday
answered 2 days ago
TravisTravis
63k767150
63k767150
1
$begingroup$
I think that if OP knew what a positive definite symmetric matrix is, there would be no need to pose the question as it is.
$endgroup$
– Marc van Leeuwen
yesterday
$begingroup$
@MarcvanLeeuwen Thanks for the comment. And yes, I agree, but one needs that language in order to state the result precisely. Still, that probably means OP would benefit from a concrete example, which I've added to the answer.
$endgroup$
– Travis
yesterday
add a comment |
1
$begingroup$
I think that if OP knew what a positive definite symmetric matrix is, there would be no need to pose the question as it is.
$endgroup$
– Marc van Leeuwen
yesterday
$begingroup$
@MarcvanLeeuwen Thanks for the comment. And yes, I agree, but one needs that language in order to state the result precisely. Still, that probably means OP would benefit from a concrete example, which I've added to the answer.
$endgroup$
– Travis
yesterday
1
1
$begingroup$
I think that if OP knew what a positive definite symmetric matrix is, there would be no need to pose the question as it is.
$endgroup$
– Marc van Leeuwen
yesterday
$begingroup$
I think that if OP knew what a positive definite symmetric matrix is, there would be no need to pose the question as it is.
$endgroup$
– Marc van Leeuwen
yesterday
$begingroup$
@MarcvanLeeuwen Thanks for the comment. And yes, I agree, but one needs that language in order to state the result precisely. Still, that probably means OP would benefit from a concrete example, which I've added to the answer.
$endgroup$
– Travis
yesterday
$begingroup$
@MarcvanLeeuwen Thanks for the comment. And yes, I agree, but one needs that language in order to state the result precisely. Still, that probably means OP would benefit from a concrete example, which I've added to the answer.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
For any invertible linear transformation $A$ you can define the inner product $langle v,wrangle_A=langle Av,Awrangle$ where $langlecdot,cdotrangle$ denotes the standard inner product. I expect there are no other inner products, which is motivated by the fact that all inner products are known to induce equivalent norms.
$endgroup$
add a comment |
$begingroup$
For any invertible linear transformation $A$ you can define the inner product $langle v,wrangle_A=langle Av,Awrangle$ where $langlecdot,cdotrangle$ denotes the standard inner product. I expect there are no other inner products, which is motivated by the fact that all inner products are known to induce equivalent norms.
$endgroup$
add a comment |
$begingroup$
For any invertible linear transformation $A$ you can define the inner product $langle v,wrangle_A=langle Av,Awrangle$ where $langlecdot,cdotrangle$ denotes the standard inner product. I expect there are no other inner products, which is motivated by the fact that all inner products are known to induce equivalent norms.
$endgroup$
For any invertible linear transformation $A$ you can define the inner product $langle v,wrangle_A=langle Av,Awrangle$ where $langlecdot,cdotrangle$ denotes the standard inner product. I expect there are no other inner products, which is motivated by the fact that all inner products are known to induce equivalent norms.
answered 2 days ago
SmileyCraftSmileyCraft
3,626519
3,626519
add a comment |
add a comment |
$begingroup$
Technically, you need positive $lambda_i$. Or if we use $sum_ijlambda_ijv_iw_j$, the matrix $lambda$ is without loss of generality equal to $(lambda+lambda^T)/2$, and it has to be positive-definite. (Yes, this matrix property has the same name; it basically means it has only positive eigenvalues.) With an appropriate basis change we can then diagonalize this matrix, which recovers the case you knew about. As for the example you tried, it failed because if you work out the matrix $lambda=left(beginarraycc
0 & 1\
1 & 0
endarrayright)$ (once we make it self-adjoint as explained above), which has $-1$ as an eigenvalue.
$endgroup$
$begingroup$
Where does the condition $lambda = (lambda+lambda^T)/2$ come from?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
@IonSme As I said, it's imposed without loss of generality. If you replace $lambda$ as thus, the resulting inner product won't be changed. (This is obvious in the case $v=w$, but in fact this is exhaustive because norms determine an inner product viz. $langle v,, wrangle = (Vert v+wVert^2-Vert v-wVert^2)/4$.)
$endgroup$
– J.G.
2 days ago
add a comment |
$begingroup$
Technically, you need positive $lambda_i$. Or if we use $sum_ijlambda_ijv_iw_j$, the matrix $lambda$ is without loss of generality equal to $(lambda+lambda^T)/2$, and it has to be positive-definite. (Yes, this matrix property has the same name; it basically means it has only positive eigenvalues.) With an appropriate basis change we can then diagonalize this matrix, which recovers the case you knew about. As for the example you tried, it failed because if you work out the matrix $lambda=left(beginarraycc
0 & 1\
1 & 0
endarrayright)$ (once we make it self-adjoint as explained above), which has $-1$ as an eigenvalue.
$endgroup$
$begingroup$
Where does the condition $lambda = (lambda+lambda^T)/2$ come from?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
@IonSme As I said, it's imposed without loss of generality. If you replace $lambda$ as thus, the resulting inner product won't be changed. (This is obvious in the case $v=w$, but in fact this is exhaustive because norms determine an inner product viz. $langle v,, wrangle = (Vert v+wVert^2-Vert v-wVert^2)/4$.)
$endgroup$
– J.G.
2 days ago
add a comment |
$begingroup$
Technically, you need positive $lambda_i$. Or if we use $sum_ijlambda_ijv_iw_j$, the matrix $lambda$ is without loss of generality equal to $(lambda+lambda^T)/2$, and it has to be positive-definite. (Yes, this matrix property has the same name; it basically means it has only positive eigenvalues.) With an appropriate basis change we can then diagonalize this matrix, which recovers the case you knew about. As for the example you tried, it failed because if you work out the matrix $lambda=left(beginarraycc
0 & 1\
1 & 0
endarrayright)$ (once we make it self-adjoint as explained above), which has $-1$ as an eigenvalue.
$endgroup$
Technically, you need positive $lambda_i$. Or if we use $sum_ijlambda_ijv_iw_j$, the matrix $lambda$ is without loss of generality equal to $(lambda+lambda^T)/2$, and it has to be positive-definite. (Yes, this matrix property has the same name; it basically means it has only positive eigenvalues.) With an appropriate basis change we can then diagonalize this matrix, which recovers the case you knew about. As for the example you tried, it failed because if you work out the matrix $lambda=left(beginarraycc
0 & 1\
1 & 0
endarrayright)$ (once we make it self-adjoint as explained above), which has $-1$ as an eigenvalue.
edited yesterday
answered 2 days ago
J.G.J.G.
29.6k22946
29.6k22946
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Where does the condition $lambda = (lambda+lambda^T)/2$ come from?
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– Ion Sme
2 days ago
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@IonSme As I said, it's imposed without loss of generality. If you replace $lambda$ as thus, the resulting inner product won't be changed. (This is obvious in the case $v=w$, but in fact this is exhaustive because norms determine an inner product viz. $langle v,, wrangle = (Vert v+wVert^2-Vert v-wVert^2)/4$.)
$endgroup$
– J.G.
2 days ago
add a comment |
$begingroup$
Where does the condition $lambda = (lambda+lambda^T)/2$ come from?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
@IonSme As I said, it's imposed without loss of generality. If you replace $lambda$ as thus, the resulting inner product won't be changed. (This is obvious in the case $v=w$, but in fact this is exhaustive because norms determine an inner product viz. $langle v,, wrangle = (Vert v+wVert^2-Vert v-wVert^2)/4$.)
$endgroup$
– J.G.
2 days ago
$begingroup$
Where does the condition $lambda = (lambda+lambda^T)/2$ come from?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
Where does the condition $lambda = (lambda+lambda^T)/2$ come from?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
@IonSme As I said, it's imposed without loss of generality. If you replace $lambda$ as thus, the resulting inner product won't be changed. (This is obvious in the case $v=w$, but in fact this is exhaustive because norms determine an inner product viz. $langle v,, wrangle = (Vert v+wVert^2-Vert v-wVert^2)/4$.)
$endgroup$
– J.G.
2 days ago
$begingroup$
@IonSme As I said, it's imposed without loss of generality. If you replace $lambda$ as thus, the resulting inner product won't be changed. (This is obvious in the case $v=w$, but in fact this is exhaustive because norms determine an inner product viz. $langle v,, wrangle = (Vert v+wVert^2-Vert v-wVert^2)/4$.)
$endgroup$
– J.G.
2 days ago
add a comment |
$begingroup$
Inner products $p(x,y)$ on $mathbb R^n$ have the form
$$
p(x,y) = sum_j=1^n sum_k=1^n a_jk x_j y_k
$$
where the matrix $A = [a_jk]$ is positive definite. Choosing a basis of eigenvectors for the matrix $A$, and expanding according to this new basis instead of the original basis, the inner product is then diagonal:
$$
p(x,y) = sum_j=1^n b_j x_j y_j
$$
where $b_j>0$.
$endgroup$
add a comment |
$begingroup$
Inner products $p(x,y)$ on $mathbb R^n$ have the form
$$
p(x,y) = sum_j=1^n sum_k=1^n a_jk x_j y_k
$$
where the matrix $A = [a_jk]$ is positive definite. Choosing a basis of eigenvectors for the matrix $A$, and expanding according to this new basis instead of the original basis, the inner product is then diagonal:
$$
p(x,y) = sum_j=1^n b_j x_j y_j
$$
where $b_j>0$.
$endgroup$
add a comment |
$begingroup$
Inner products $p(x,y)$ on $mathbb R^n$ have the form
$$
p(x,y) = sum_j=1^n sum_k=1^n a_jk x_j y_k
$$
where the matrix $A = [a_jk]$ is positive definite. Choosing a basis of eigenvectors for the matrix $A$, and expanding according to this new basis instead of the original basis, the inner product is then diagonal:
$$
p(x,y) = sum_j=1^n b_j x_j y_j
$$
where $b_j>0$.
$endgroup$
Inner products $p(x,y)$ on $mathbb R^n$ have the form
$$
p(x,y) = sum_j=1^n sum_k=1^n a_jk x_j y_k
$$
where the matrix $A = [a_jk]$ is positive definite. Choosing a basis of eigenvectors for the matrix $A$, and expanding according to this new basis instead of the original basis, the inner product is then diagonal:
$$
p(x,y) = sum_j=1^n b_j x_j y_j
$$
where $b_j>0$.
answered 2 days ago
GEdgarGEdgar
62.9k267171
62.9k267171
add a comment |
add a comment |
$begingroup$
I agree with SmileyCraft. In finite dimensional vector spaces, bilinear transformations, as linear transformations, can be written in terms of the values that they adopt in a given base:$$left langle x,y right rangle=sum_i,j=1^nx_iy_jleft langle e_i,e_j right rangle.$$
I believe you can arrive in this representation without difficult, proving then you suspicion.
New contributor
$endgroup$
add a comment |
$begingroup$
I agree with SmileyCraft. In finite dimensional vector spaces, bilinear transformations, as linear transformations, can be written in terms of the values that they adopt in a given base:$$left langle x,y right rangle=sum_i,j=1^nx_iy_jleft langle e_i,e_j right rangle.$$
I believe you can arrive in this representation without difficult, proving then you suspicion.
New contributor
$endgroup$
add a comment |
$begingroup$
I agree with SmileyCraft. In finite dimensional vector spaces, bilinear transformations, as linear transformations, can be written in terms of the values that they adopt in a given base:$$left langle x,y right rangle=sum_i,j=1^nx_iy_jleft langle e_i,e_j right rangle.$$
I believe you can arrive in this representation without difficult, proving then you suspicion.
New contributor
$endgroup$
I agree with SmileyCraft. In finite dimensional vector spaces, bilinear transformations, as linear transformations, can be written in terms of the values that they adopt in a given base:$$left langle x,y right rangle=sum_i,j=1^nx_iy_jleft langle e_i,e_j right rangle.$$
I believe you can arrive in this representation without difficult, proving then you suspicion.
New contributor
New contributor
answered 2 days ago
Jonathan HonórioJonathan Honório
113
113
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
It is well known (and easy to prove) that any two finite dimensional inner product spaces are isometrically isomorphic. Hence $ langle x, y rangle'$ is an inner product on $mathbb R^n$ iff there is a vector space isomorphism $T: mathbb R^n to mathbb R^n$ such that $langle x, y rangle' =langle Tx, Ty rangle$ for all $x,y$.
$endgroup$
add a comment |
$begingroup$
It is well known (and easy to prove) that any two finite dimensional inner product spaces are isometrically isomorphic. Hence $ langle x, y rangle'$ is an inner product on $mathbb R^n$ iff there is a vector space isomorphism $T: mathbb R^n to mathbb R^n$ such that $langle x, y rangle' =langle Tx, Ty rangle$ for all $x,y$.
$endgroup$
add a comment |
$begingroup$
It is well known (and easy to prove) that any two finite dimensional inner product spaces are isometrically isomorphic. Hence $ langle x, y rangle'$ is an inner product on $mathbb R^n$ iff there is a vector space isomorphism $T: mathbb R^n to mathbb R^n$ such that $langle x, y rangle' =langle Tx, Ty rangle$ for all $x,y$.
$endgroup$
It is well known (and easy to prove) that any two finite dimensional inner product spaces are isometrically isomorphic. Hence $ langle x, y rangle'$ is an inner product on $mathbb R^n$ iff there is a vector space isomorphism $T: mathbb R^n to mathbb R^n$ such that $langle x, y rangle' =langle Tx, Ty rangle$ for all $x,y$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
66.3k42867
add a comment |
add a comment |
$begingroup$
In general, for any $ntimes n$ matrix $A=(a_i,j)_i,j=1,ldots,n$ the expression $sum_i,j=1^na_i,jx_iy_j$ defines a bilinear form, which will be symmetric if and only $A$ is. Giving a positive definite symmetric bilinear form is a more subtle condition that leads to inequalities for the coefficients of $A$ (and the matrices that satisfy the condition are naturally called positive definite). For $2times2$ symmetric matrices the positive definite condition is $a_1,1>0$, $a_2,2>0$ together with $a_1,1a_2,2-a_1,2^2>0$ (so $det(A)>0$). For a concrete example, the symmetric matrix
$$
A=pmatrix1½\frac12&1
quadtextgives an inner product with
langle v,,wrangle = v_1w_1 +frac12(v_1w_2+v_2w_1) + v_2w_2,.
$$
In higher dimension the condition is more complicated, but in any case one does get many different inner products on $Bbb R^n$ in this way. They do turn out to be all equivalent in the sense that they give rise to the same structure theory, but they are not equal.
$endgroup$
add a comment |
$begingroup$
In general, for any $ntimes n$ matrix $A=(a_i,j)_i,j=1,ldots,n$ the expression $sum_i,j=1^na_i,jx_iy_j$ defines a bilinear form, which will be symmetric if and only $A$ is. Giving a positive definite symmetric bilinear form is a more subtle condition that leads to inequalities for the coefficients of $A$ (and the matrices that satisfy the condition are naturally called positive definite). For $2times2$ symmetric matrices the positive definite condition is $a_1,1>0$, $a_2,2>0$ together with $a_1,1a_2,2-a_1,2^2>0$ (so $det(A)>0$). For a concrete example, the symmetric matrix
$$
A=pmatrix1½\frac12&1
quadtextgives an inner product with
langle v,,wrangle = v_1w_1 +frac12(v_1w_2+v_2w_1) + v_2w_2,.
$$
In higher dimension the condition is more complicated, but in any case one does get many different inner products on $Bbb R^n$ in this way. They do turn out to be all equivalent in the sense that they give rise to the same structure theory, but they are not equal.
$endgroup$
add a comment |
$begingroup$
In general, for any $ntimes n$ matrix $A=(a_i,j)_i,j=1,ldots,n$ the expression $sum_i,j=1^na_i,jx_iy_j$ defines a bilinear form, which will be symmetric if and only $A$ is. Giving a positive definite symmetric bilinear form is a more subtle condition that leads to inequalities for the coefficients of $A$ (and the matrices that satisfy the condition are naturally called positive definite). For $2times2$ symmetric matrices the positive definite condition is $a_1,1>0$, $a_2,2>0$ together with $a_1,1a_2,2-a_1,2^2>0$ (so $det(A)>0$). For a concrete example, the symmetric matrix
$$
A=pmatrix1½\frac12&1
quadtextgives an inner product with
langle v,,wrangle = v_1w_1 +frac12(v_1w_2+v_2w_1) + v_2w_2,.
$$
In higher dimension the condition is more complicated, but in any case one does get many different inner products on $Bbb R^n$ in this way. They do turn out to be all equivalent in the sense that they give rise to the same structure theory, but they are not equal.
$endgroup$
In general, for any $ntimes n$ matrix $A=(a_i,j)_i,j=1,ldots,n$ the expression $sum_i,j=1^na_i,jx_iy_j$ defines a bilinear form, which will be symmetric if and only $A$ is. Giving a positive definite symmetric bilinear form is a more subtle condition that leads to inequalities for the coefficients of $A$ (and the matrices that satisfy the condition are naturally called positive definite). For $2times2$ symmetric matrices the positive definite condition is $a_1,1>0$, $a_2,2>0$ together with $a_1,1a_2,2-a_1,2^2>0$ (so $det(A)>0$). For a concrete example, the symmetric matrix
$$
A=pmatrix1½\frac12&1
quadtextgives an inner product with
langle v,,wrangle = v_1w_1 +frac12(v_1w_2+v_2w_1) + v_2w_2,.
$$
In higher dimension the condition is more complicated, but in any case one does get many different inner products on $Bbb R^n$ in this way. They do turn out to be all equivalent in the sense that they give rise to the same structure theory, but they are not equal.
edited 9 hours ago
answered yesterday
Marc van LeeuwenMarc van Leeuwen
88.2k5111228
88.2k5111228
add a comment |
add a comment |
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$begingroup$
So, everyone is saying you can also use a matrix, and that works. Is there a proof though that any inner product can be represented by such a matrix type inner product?
$endgroup$
– Ion Sme
2 days ago
$begingroup$
$newcommandRmathbbRnewcommandumathbfe$Yes. Here is a sketch. Suppose that $(cdot, cdot)$ is some inner product on $R^n$. Let $u_1, ldots,u_n$ be the standard basis for $R^n$ (so $u_k$ is a vector whose $k$-th entry is $1$, and all other entries are $0$). Define $A$ to be the $ntimes n$ matrix with $i,j$ entry $colorbluea_ij= (u_i, u_j)$. Then show that this $A$ satisfies $(mathbfx, mathbfy) = mathbfx^T A mathbfy$ for all $mathbfx,mathbfyin R^n$. EDIT: I see Travis has mentioned this in an answer below.
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– Minus One-Twelfth
2 days ago