Is there a scalar product s.t. the following list is orthogonal?

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Let $A_1=beginpmatrix1&0\0&0endpmatrixquad A_2=beginpmatrix1&1\0&0endpmatrix,quad A_3=beginpmatrix1&1\1&0endpmatrix,quad A_4=beginpmatrix1&1\1&1endpmatrix$.

Is there a scalar product s.t. $|A_k|=k$ for $k=1,2,3,4$ and $A_iperp A_j$ for $ineq j$ ?

---

With $langle A, B rangle = mboxTr(A^top B)$ we have that $|A_i|=i$, but unfortunately it's not orthogonal. So, how can we conclude?

linear-algebra

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edited Jul 29 at 11:11

Rodrigo de Azevedo

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up vote
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Let $A_1=beginpmatrix1&0\0&0endpmatrixquad A_2=beginpmatrix1&1\0&0endpmatrix,quad A_3=beginpmatrix1&1\1&0endpmatrix,quad A_4=beginpmatrix1&1\1&1endpmatrix$.

Is there a scalar product s.t. $|A_k|=k$ for $k=1,2,3,4$ and $A_iperp A_j$ for $ineq j$ ?

---

With $langle A, B rangle = mboxTr(A^top B)$ we have that $|A_i|=i$, but unfortunately it's not orthogonal. So, how can we conclude?

linear-algebra

share|cite|improve this question

edited Jul 29 at 11:11

Rodrigo de Azevedo

12.6k41751

asked Jul 29 at 10:44

MSE

1,471315

add a comment | 

up vote
7
down vote

favorite

3

up vote
7
down vote

favorite

3

3

Let $A_1=beginpmatrix1&0\0&0endpmatrixquad A_2=beginpmatrix1&1\0&0endpmatrix,quad A_3=beginpmatrix1&1\1&0endpmatrix,quad A_4=beginpmatrix1&1\1&1endpmatrix$.

Is there a scalar product s.t. $|A_k|=k$ for $k=1,2,3,4$ and $A_iperp A_j$ for $ineq j$ ?

---

With $langle A, B rangle = mboxTr(A^top B)$ we have that $|A_i|=i$, but unfortunately it's not orthogonal. So, how can we conclude?

linear-algebra

share|cite|improve this question

edited Jul 29 at 11:11

Rodrigo de Azevedo

12.6k41751

asked Jul 29 at 10:44

MSE

1,471315

Let $A_1=beginpmatrix1&0\0&0endpmatrixquad A_2=beginpmatrix1&1\0&0endpmatrix,quad A_3=beginpmatrix1&1\1&0endpmatrix,quad A_4=beginpmatrix1&1\1&1endpmatrix$.

Is there a scalar product s.t. $|A_k|=k$ for $k=1,2,3,4$ and $A_iperp A_j$ for $ineq j$ ?

---

With $langle A, B rangle = mboxTr(A^top B)$ we have that $|A_i|=i$, but unfortunately it's not orthogonal. So, how can we conclude?

linear-algebra

share|cite|improve this question

edited Jul 29 at 11:11

Rodrigo de Azevedo

12.6k41751

asked Jul 29 at 10:44

MSE

1,471315

share|cite|improve this question

share|cite|improve this question

edited Jul 29 at 11:11

Rodrigo de Azevedo

12.6k41751

edited Jul 29 at 11:11

Rodrigo de Azevedo

12.6k41751

edited Jul 29 at 11:11

Rodrigo de Azevedo

12.6k41751

12.6k41751

asked Jul 29 at 10:44

MSE

1,471315

asked Jul 29 at 10:44

MSE

1,471315

asked Jul 29 at 10:44

MSE

1,471315

1,471315

add a comment | 

add a comment | 

2 Answers
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accepted

Lemma. Let $V$ be a real vector space and $v_1,...,v_n$ a basis. Then there exists an inner product on $V$ that makes $(v_i)_i$ an orthonormal basis.

Proof. Define $langle v_j,v_krangle = delta_jk$ and extend it linearly.

---

For your Problem: Check that the $A_j$ form a basis of $mathbbR^2times 2$. Then also $(A_j/j)_j$ is a basis and the inner product from the lemma does the job.

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answered Jul 29 at 11:09

Jan Bohr

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up vote
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Let $E_1:=beginbmatrix1&0\0&0endbmatrix$, $E_1:=beginbmatrix0&1\0&0endbmatrix$, $E_3:=beginbmatrix0&0\1&0endbmatrix$, and $E_4:=beginbmatrix0&0\0&1endbmatrix$. Then, the matrix of transformation sending $(E_1,E_2,E_3,E_4)$ to $(A_1,A_2,A_3,A_4)$ is
$$T:=beginbmatrix1&1&1&1\0&1&1&1\0&0&1&1\0&0&0&1endbmatrix,.$$
The required bilinear form in the basis $(A_1,A_2,A_3,A_4)$ is given by the matrix
$$B:=beginbmatrix1&0&0&0\0&4&0&0\0&0&9&0\0&0&0&16endbmatrix,.$$
Therefore, in the basis $(E_1,E_2,E_3,E_4)$, the bilinear form is given by the matrix
$$left(T^-1right)^top,B,T^-1=beginbmatrix1&-1&0&0\-1&5&-4&0\0&-4&13&-9\0&0&-9&25endbmatrix,.$$
In other words, this bilinear form is given by
$$leftlangle beginbmatrixa&b\c&dendbmatrix,beginbmatrixx&y\z&wendbmatrixrightrangle=ax-ay-bx+5by-4bz-4cy+13cz-9cw-9dz+25dw$$
for $a,b,c,d,x,y,z,winmathbbR$.
If the base field is $mathbbC$, then the bilinear form (or rather, the sesquilinear form) is given by
$$leftlangle beginbmatrixa&b\c&dendbmatrix,beginbmatrixx&y\z&wendbmatrixrightrangle=abar x-abar y-bbar x+5bbar y-4bbar z-4cbar y+13cbar z-9cbar w-9dbar z+25dbar w$$
for $a,b,c,d,x,y,z,winmathbbC$.

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edited Jul 29 at 11:21

answered Jul 29 at 11:12

Batominovski

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
23
down vote



accepted

Lemma. Let $V$ be a real vector space and $v_1,...,v_n$ a basis. Then there exists an inner product on $V$ that makes $(v_i)_i$ an orthonormal basis.

Proof. Define $langle v_j,v_krangle = delta_jk$ and extend it linearly.

---

For your Problem: Check that the $A_j$ form a basis of $mathbbR^2times 2$. Then also $(A_j/j)_j$ is a basis and the inner product from the lemma does the job.

share|cite|improve this answer

answered Jul 29 at 11:09

Jan Bohr

3,1241419

add a comment | 

up vote
23
down vote



accepted

Lemma. Let $V$ be a real vector space and $v_1,...,v_n$ a basis. Then there exists an inner product on $V$ that makes $(v_i)_i$ an orthonormal basis.

Proof. Define $langle v_j,v_krangle = delta_jk$ and extend it linearly.

---

For your Problem: Check that the $A_j$ form a basis of $mathbbR^2times 2$. Then also $(A_j/j)_j$ is a basis and the inner product from the lemma does the job.

share|cite|improve this answer

answered Jul 29 at 11:09

Jan Bohr

3,1241419

add a comment | 

up vote
23
down vote



accepted

up vote
23
down vote



accepted

Lemma. Let $V$ be a real vector space and $v_1,...,v_n$ a basis. Then there exists an inner product on $V$ that makes $(v_i)_i$ an orthonormal basis.

Proof. Define $langle v_j,v_krangle = delta_jk$ and extend it linearly.

---

For your Problem: Check that the $A_j$ form a basis of $mathbbR^2times 2$. Then also $(A_j/j)_j$ is a basis and the inner product from the lemma does the job.

share|cite|improve this answer

answered Jul 29 at 11:09

Jan Bohr

3,1241419

Lemma. Let $V$ be a real vector space and $v_1,...,v_n$ a basis. Then there exists an inner product on $V$ that makes $(v_i)_i$ an orthonormal basis.

Proof. Define $langle v_j,v_krangle = delta_jk$ and extend it linearly.

---

For your Problem: Check that the $A_j$ form a basis of $mathbbR^2times 2$. Then also $(A_j/j)_j$ is a basis and the inner product from the lemma does the job.

share|cite|improve this answer

answered Jul 29 at 11:09

Jan Bohr

3,1241419

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 11:09

Jan Bohr

3,1241419

answered Jul 29 at 11:09

Jan Bohr

3,1241419

answered Jul 29 at 11:09

Jan Bohr

3,1241419

3,1241419

add a comment | 

add a comment | 

up vote
4
down vote

Let $E_1:=beginbmatrix1&0\0&0endbmatrix$, $E_1:=beginbmatrix0&1\0&0endbmatrix$, $E_3:=beginbmatrix0&0\1&0endbmatrix$, and $E_4:=beginbmatrix0&0\0&1endbmatrix$. Then, the matrix of transformation sending $(E_1,E_2,E_3,E_4)$ to $(A_1,A_2,A_3,A_4)$ is
$$T:=beginbmatrix1&1&1&1\0&1&1&1\0&0&1&1\0&0&0&1endbmatrix,.$$
The required bilinear form in the basis $(A_1,A_2,A_3,A_4)$ is given by the matrix
$$B:=beginbmatrix1&0&0&0\0&4&0&0\0&0&9&0\0&0&0&16endbmatrix,.$$
Therefore, in the basis $(E_1,E_2,E_3,E_4)$, the bilinear form is given by the matrix
$$left(T^-1right)^top,B,T^-1=beginbmatrix1&-1&0&0\-1&5&-4&0\0&-4&13&-9\0&0&-9&25endbmatrix,.$$
In other words, this bilinear form is given by
$$leftlangle beginbmatrixa&b\c&dendbmatrix,beginbmatrixx&y\z&wendbmatrixrightrangle=ax-ay-bx+5by-4bz-4cy+13cz-9cw-9dz+25dw$$
for $a,b,c,d,x,y,z,winmathbbR$.
If the base field is $mathbbC$, then the bilinear form (or rather, the sesquilinear form) is given by
$$leftlangle beginbmatrixa&b\c&dendbmatrix,beginbmatrixx&y\z&wendbmatrixrightrangle=abar x-abar y-bbar x+5bbar y-4bbar z-4cbar y+13cbar z-9cbar w-9dbar z+25dbar w$$
for $a,b,c,d,x,y,z,winmathbbC$.

share|cite|improve this answer

edited Jul 29 at 11:21

answered Jul 29 at 11:12

Batominovski

22.9k22777

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up vote
4
down vote

Let $E_1:=beginbmatrix1&0\0&0endbmatrix$, $E_1:=beginbmatrix0&1\0&0endbmatrix$, $E_3:=beginbmatrix0&0\1&0endbmatrix$, and $E_4:=beginbmatrix0&0\0&1endbmatrix$. Then, the matrix of transformation sending $(E_1,E_2,E_3,E_4)$ to $(A_1,A_2,A_3,A_4)$ is
$$T:=beginbmatrix1&1&1&1\0&1&1&1\0&0&1&1\0&0&0&1endbmatrix,.$$
The required bilinear form in the basis $(A_1,A_2,A_3,A_4)$ is given by the matrix
$$B:=beginbmatrix1&0&0&0\0&4&0&0\0&0&9&0\0&0&0&16endbmatrix,.$$
Therefore, in the basis $(E_1,E_2,E_3,E_4)$, the bilinear form is given by the matrix
$$left(T^-1right)^top,B,T^-1=beginbmatrix1&-1&0&0\-1&5&-4&0\0&-4&13&-9\0&0&-9&25endbmatrix,.$$
In other words, this bilinear form is given by
$$leftlangle beginbmatrixa&b\c&dendbmatrix,beginbmatrixx&y\z&wendbmatrixrightrangle=ax-ay-bx+5by-4bz-4cy+13cz-9cw-9dz+25dw$$
for $a,b,c,d,x,y,z,winmathbbR$.
If the base field is $mathbbC$, then the bilinear form (or rather, the sesquilinear form) is given by
$$leftlangle beginbmatrixa&b\c&dendbmatrix,beginbmatrixx&y\z&wendbmatrixrightrangle=abar x-abar y-bbar x+5bbar y-4bbar z-4cbar y+13cbar z-9cbar w-9dbar z+25dbar w$$
for $a,b,c,d,x,y,z,winmathbbC$.

share|cite|improve this answer

edited Jul 29 at 11:21

answered Jul 29 at 11:12

Batominovski

22.9k22777

add a comment | 

up vote
4
down vote

up vote
4
down vote

Let $E_1:=beginbmatrix1&0\0&0endbmatrix$, $E_1:=beginbmatrix0&1\0&0endbmatrix$, $E_3:=beginbmatrix0&0\1&0endbmatrix$, and $E_4:=beginbmatrix0&0\0&1endbmatrix$. Then, the matrix of transformation sending $(E_1,E_2,E_3,E_4)$ to $(A_1,A_2,A_3,A_4)$ is
$$T:=beginbmatrix1&1&1&1\0&1&1&1\0&0&1&1\0&0&0&1endbmatrix,.$$
The required bilinear form in the basis $(A_1,A_2,A_3,A_4)$ is given by the matrix
$$B:=beginbmatrix1&0&0&0\0&4&0&0\0&0&9&0\0&0&0&16endbmatrix,.$$
Therefore, in the basis $(E_1,E_2,E_3,E_4)$, the bilinear form is given by the matrix
$$left(T^-1right)^top,B,T^-1=beginbmatrix1&-1&0&0\-1&5&-4&0\0&-4&13&-9\0&0&-9&25endbmatrix,.$$
In other words, this bilinear form is given by
$$leftlangle beginbmatrixa&b\c&dendbmatrix,beginbmatrixx&y\z&wendbmatrixrightrangle=ax-ay-bx+5by-4bz-4cy+13cz-9cw-9dz+25dw$$
for $a,b,c,d,x,y,z,winmathbbR$.
If the base field is $mathbbC$, then the bilinear form (or rather, the sesquilinear form) is given by
$$leftlangle beginbmatrixa&b\c&dendbmatrix,beginbmatrixx&y\z&wendbmatrixrightrangle=abar x-abar y-bbar x+5bbar y-4bbar z-4cbar y+13cbar z-9cbar w-9dbar z+25dbar w$$
for $a,b,c,d,x,y,z,winmathbbC$.

share|cite|improve this answer

edited Jul 29 at 11:21

answered Jul 29 at 11:12

Batominovski

22.9k22777

Let $E_1:=beginbmatrix1&0\0&0endbmatrix$, $E_1:=beginbmatrix0&1\0&0endbmatrix$, $E_3:=beginbmatrix0&0\1&0endbmatrix$, and $E_4:=beginbmatrix0&0\0&1endbmatrix$. Then, the matrix of transformation sending $(E_1,E_2,E_3,E_4)$ to $(A_1,A_2,A_3,A_4)$ is
$$T:=beginbmatrix1&1&1&1\0&1&1&1\0&0&1&1\0&0&0&1endbmatrix,.$$
The required bilinear form in the basis $(A_1,A_2,A_3,A_4)$ is given by the matrix
$$B:=beginbmatrix1&0&0&0\0&4&0&0\0&0&9&0\0&0&0&16endbmatrix,.$$
Therefore, in the basis $(E_1,E_2,E_3,E_4)$, the bilinear form is given by the matrix
$$left(T^-1right)^top,B,T^-1=beginbmatrix1&-1&0&0\-1&5&-4&0\0&-4&13&-9\0&0&-9&25endbmatrix,.$$
In other words, this bilinear form is given by
$$leftlangle beginbmatrixa&b\c&dendbmatrix,beginbmatrixx&y\z&wendbmatrixrightrangle=ax-ay-bx+5by-4bz-4cy+13cz-9cw-9dz+25dw$$
for $a,b,c,d,x,y,z,winmathbbR$.
If the base field is $mathbbC$, then the bilinear form (or rather, the sesquilinear form) is given by
$$leftlangle beginbmatrixa&b\c&dendbmatrix,beginbmatrixx&y\z&wendbmatrixrightrangle=abar x-abar y-bbar x+5bbar y-4bbar z-4cbar y+13cbar z-9cbar w-9dbar z+25dbar w$$
for $a,b,c,d,x,y,z,winmathbbC$.

share|cite|improve this answer

edited Jul 29 at 11:21

answered Jul 29 at 11:12

Batominovski

22.9k22777

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 11:21

edited Jul 29 at 11:21

edited Jul 29 at 11:21

answered Jul 29 at 11:12

Batominovski

22.9k22777

answered Jul 29 at 11:12

Batominovski

22.9k22777

answered Jul 29 at 11:12

Batominovski

22.9k22777

22.9k22777

add a comment | 

add a comment | 

 

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