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Struggling to prove that the elements of an inverse matrix satisfy a certain equation.
The Next CEO of Stack OverflowHow to find change of basis matricesWhat comes first, a vector base or orthogonality?Show that a vector can be represented in term of its componentsUnderstading wikipedia explanation of tranformation matrixOrthonormal basis for a subspace of a Hilbert spaceSolve linear system with $A_i,j = langle e_i, e_jrangle^2$, edges of a tetrahedronAnti-diagonal matrix symmetric bilinear formDefinition of exterior productConverting Spherical Vector to Cylindrical VectorFind distance from point to line in not standart coordinate system
$begingroup$
We have three vectors $vece_1,vece_2,vece_3$ that are not necessarily orthogonal or normalised, but do form a basis.
We also have a matrix $G$ with elements $G_ij=vece_i . vece_j$, and we are told that $H=G^-1$.
The question asks to show that the vectors $vecf_i=sum_jH_ijvece_j$ are the reciprocal vectors of $vece_i$, which I have done. It then asks to show that $H_ij=vecf_i .vecf_j$. I've been trying to prove this latter statement for about two hours and I've had no luck whatsoever. How do I go about proving this?
linear-algebra matrices
asked Mar 19 at 11:57
Pancake_SenpaiPancake_Senpai
26017
$endgroup$
add a comment |
$begingroup$
We have three vectors $vece_1,vece_2,vece_3$ that are not necessarily orthogonal or normalised, but do form a basis.
We also have a matrix $G$ with elements $G_ij=vece_i . vece_j$, and we are told that $H=G^-1$.
The question asks to show that the vectors $vecf_i=sum_jH_ijvece_j$ are the reciprocal vectors of $vece_i$, which I have done. It then asks to show that $H_ij=vecf_i .vecf_j$. I've been trying to prove this latter statement for about two hours and I've had no luck whatsoever. How do I go about proving this?
linear-algebra matrices
asked Mar 19 at 11:57
Pancake_SenpaiPancake_Senpai
26017
$endgroup$
add a comment |
$begingroup$
We have three vectors $vece_1,vece_2,vece_3$ that are not necessarily orthogonal or normalised, but do form a basis.
We also have a matrix $G$ with elements $G_ij=vece_i . vece_j$, and we are told that $H=G^-1$.
The question asks to show that the vectors $vecf_i=sum_jH_ijvece_j$ are the reciprocal vectors of $vece_i$, which I have done. It then asks to show that $H_ij=vecf_i .vecf_j$. I've been trying to prove this latter statement for about two hours and I've had no luck whatsoever. How do I go about proving this?
linear-algebra matrices
asked Mar 19 at 11:57
Pancake_SenpaiPancake_Senpai
26017
$endgroup$
We have three vectors $vece_1,vece_2,vece_3$ that are not necessarily orthogonal or normalised, but do form a basis.
We also have a matrix $G$ with elements $G_ij=vece_i . vece_j$, and we are told that $H=G^-1$.
The question asks to show that the vectors $vecf_i=sum_jH_ijvece_j$ are the reciprocal vectors of $vece_i$, which I have done. It then asks to show that $H_ij=vecf_i .vecf_j$. I've been trying to prove this latter statement for about two hours and I've had no luck whatsoever. How do I go about proving this?
linear-algebra matrices
linear-algebra matrices
asked Mar 19 at 11:57
Pancake_SenpaiPancake_Senpai
26017
asked Mar 19 at 11:57
Pancake_SenpaiPancake_Senpai
26017
asked Mar 19 at 11:57
Pancake_SenpaiPancake_Senpai
26017
asked Mar 19 at 11:57
Pancake_SenpaiPancake_Senpai
26017
asked Mar 19 at 11:57
Pancake_SenpaiPancake_Senpai
26017
26017
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is not true. Here is a counter-example.
$$
e_1 = pmatrix1\0\0, e_2 = pmatrix1\1\0, e_3 = pmatrix0\0\1,
$$
$$
G=pmatrix1&1&0\1&2&0\0&0&1,
H= pmatrix2&-1&0\-1&1&0\0&0&1
$$
$$
f_1=pmatrix2\-1\0, f_2 = pmatrix1\0\0, f_3=e_3,
$$
$$
(f_icdot f_j) = pmatrix 5&2&0\2&1&0\0&0&1 ne H
$$
answered Mar 19 at 12:51
dawdaw
24.9k1745
$endgroup$
$begingroup$
Interestingly your $f_1$ isn't the reciprocal vector of $e_1$, so something must have been wrong with my first proof also. Thank you :)
$endgroup$
– Pancake_Senpai
Mar 19 at 15:30
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is not true. Here is a counter-example.
$$
e_1 = pmatrix1\0\0, e_2 = pmatrix1\1\0, e_3 = pmatrix0\0\1,
$$
$$
G=pmatrix1&1&0\1&2&0\0&0&1,
H= pmatrix2&-1&0\-1&1&0\0&0&1
$$
$$
f_1=pmatrix2\-1\0, f_2 = pmatrix1\0\0, f_3=e_3,
$$
$$
(f_icdot f_j) = pmatrix 5&2&0\2&1&0\0&0&1 ne H
$$
answered Mar 19 at 12:51
dawdaw
24.9k1745
$endgroup$
$begingroup$
Interestingly your $f_1$ isn't the reciprocal vector of $e_1$, so something must have been wrong with my first proof also. Thank you :)
$endgroup$
– Pancake_Senpai
Mar 19 at 15:30
add a comment |
$begingroup$
It is not true. Here is a counter-example.
$$
e_1 = pmatrix1\0\0, e_2 = pmatrix1\1\0, e_3 = pmatrix0\0\1,
$$
$$
G=pmatrix1&1&0\1&2&0\0&0&1,
H= pmatrix2&-1&0\-1&1&0\0&0&1
$$
$$
f_1=pmatrix2\-1\0, f_2 = pmatrix1\0\0, f_3=e_3,
$$
$$
(f_icdot f_j) = pmatrix 5&2&0\2&1&0\0&0&1 ne H
$$
answered Mar 19 at 12:51
dawdaw
24.9k1745
$endgroup$
$begingroup$
Interestingly your $f_1$ isn't the reciprocal vector of $e_1$, so something must have been wrong with my first proof also. Thank you :)
$endgroup$
– Pancake_Senpai
Mar 19 at 15:30
add a comment |
$begingroup$
It is not true. Here is a counter-example.
$$
e_1 = pmatrix1\0\0, e_2 = pmatrix1\1\0, e_3 = pmatrix0\0\1,
$$
$$
G=pmatrix1&1&0\1&2&0\0&0&1,
H= pmatrix2&-1&0\-1&1&0\0&0&1
$$
$$
f_1=pmatrix2\-1\0, f_2 = pmatrix1\0\0, f_3=e_3,
$$
$$
(f_icdot f_j) = pmatrix 5&2&0\2&1&0\0&0&1 ne H
$$
answered Mar 19 at 12:51
dawdaw
24.9k1745
$endgroup$
It is not true. Here is a counter-example.
$$
e_1 = pmatrix1\0\0, e_2 = pmatrix1\1\0, e_3 = pmatrix0\0\1,
$$
$$
G=pmatrix1&1&0\1&2&0\0&0&1,
H= pmatrix2&-1&0\-1&1&0\0&0&1
$$
$$
f_1=pmatrix2\-1\0, f_2 = pmatrix1\0\0, f_3=e_3,
$$
$$
(f_icdot f_j) = pmatrix 5&2&0\2&1&0\0&0&1 ne H
$$
answered Mar 19 at 12:51
dawdaw
24.9k1745
answered Mar 19 at 12:51
dawdaw
24.9k1745
answered Mar 19 at 12:51
dawdaw
24.9k1745
answered Mar 19 at 12:51
dawdaw
24.9k1745
24.9k1745
$begingroup$
Interestingly your $f_1$ isn't the reciprocal vector of $e_1$, so something must have been wrong with my first proof also. Thank you :)
$endgroup$
– Pancake_Senpai
Mar 19 at 15:30
add a comment |
$begingroup$
Interestingly your $f_1$ isn't the reciprocal vector of $e_1$, so something must have been wrong with my first proof also. Thank you :)
$endgroup$
– Pancake_Senpai
Mar 19 at 15:30
$begingroup$
Interestingly your $f_1$ isn't the reciprocal vector of $e_1$, so something must have been wrong with my first proof also. Thank you :)
$endgroup$
– Pancake_Senpai
Mar 19 at 15:30
$begingroup$
Interestingly your $f_1$ isn't the reciprocal vector of $e_1$, so something must have been wrong with my first proof also. Thank you :)
$endgroup$
– Pancake_Senpai
Mar 19 at 15:30
add a comment |
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