Given an unknown square matrix $H$ and known vectors $x$ and $y$, is it possible to find $H$ if...
If $Hvec{x}=vec{y}$ can we find $H$ by inverting $vec{x}$? No idea how to invert $vec{x}$ though or if this equation is even solvable.
matrix-equations
edited Nov 21 '18 at 13:10
Vee Hua Zhi
759224
asked Nov 21 '18 at 13:00
PhysicsMan
31
add a comment |
If $Hvec{x}=vec{y}$ can we find $H$ by inverting $vec{x}$? No idea how to invert $vec{x}$ though or if this equation is even solvable.
matrix-equations
edited Nov 21 '18 at 13:10
Vee Hua Zhi
759224
asked Nov 21 '18 at 13:00
PhysicsMan
31
2
No.${}{}{}{}{}{}$
– Saucy O'Path
Nov 21 '18 at 13:04
Given you are in a flight (unknown) can you say it will take you from your current location (known) to Tokyo? Your question can be considered equivalent to this one in some sense.
– P Vanchinathan
Nov 21 '18 at 13:07
add a comment |
If $Hvec{x}=vec{y}$ can we find $H$ by inverting $vec{x}$? No idea how to invert $vec{x}$ though or if this equation is even solvable.
matrix-equations
edited Nov 21 '18 at 13:10
Vee Hua Zhi
759224
asked Nov 21 '18 at 13:00
PhysicsMan
31
If $Hvec{x}=vec{y}$ can we find $H$ by inverting $vec{x}$? No idea how to invert $vec{x}$ though or if this equation is even solvable.
matrix-equations
matrix-equations
edited Nov 21 '18 at 13:10
Vee Hua Zhi
759224
asked Nov 21 '18 at 13:00
PhysicsMan
31
edited Nov 21 '18 at 13:10
Vee Hua Zhi
759224
asked Nov 21 '18 at 13:00
PhysicsMan
31
edited Nov 21 '18 at 13:10
Vee Hua Zhi
759224
edited Nov 21 '18 at 13:10
Vee Hua Zhi
759224
edited Nov 21 '18 at 13:10
Vee Hua Zhi
759224
759224
asked Nov 21 '18 at 13:00
PhysicsMan
31
asked Nov 21 '18 at 13:00
PhysicsMan
31
asked Nov 21 '18 at 13:00
PhysicsMan
31
31
2
No.${}{}{}{}{}{}$
– Saucy O'Path
Nov 21 '18 at 13:04
Given you are in a flight (unknown) can you say it will take you from your current location (known) to Tokyo? Your question can be considered equivalent to this one in some sense.
– P Vanchinathan
Nov 21 '18 at 13:07
add a comment |
2
No.${}{}{}{}{}{}$
– Saucy O'Path
Nov 21 '18 at 13:04
Given you are in a flight (unknown) can you say it will take you from your current location (known) to Tokyo? Your question can be considered equivalent to this one in some sense.
– P Vanchinathan
Nov 21 '18 at 13:07
2
2
No.${}{}{}{}{}{}$
– Saucy O'Path
Nov 21 '18 at 13:04
No.${}{}{}{}{}{}$
– Saucy O'Path
Nov 21 '18 at 13:04
Given you are in a flight (unknown) can you say it will take you from your current location (known) to Tokyo? Your question can be considered equivalent to this one in some sense.
– P Vanchinathan
Nov 21 '18 at 13:07
Given you are in a flight (unknown) can you say it will take you from your current location (known) to Tokyo? Your question can be considered equivalent to this one in some sense.
– P Vanchinathan
Nov 21 '18 at 13:07
add a comment |
2 Answers
2
active
oldest
votes
Let $H_{ij}$ denote the elements of the $mtimes n$-matrix $H$. Then you are asking to solutions of the following linear system of equations:
$$sum_{k=1}^n H_{1k}x_k = y_1$$
$$sum_{k=1}^n H_{2k}x_k = y_2$$
and so on, with the final equation being
$$sum_{k=1}^n H_{mk}x_k = y_m$$
Obviously, we have $ntimes m$ unknown elements (the entries of the matrix $H$) and only $m$ equations. This is, in general, not solveable uniquely.
answered Nov 21 '18 at 13:10
Tom
2,761315
add a comment |
For every vector $vec{x}$ exist a non zero vector $vec{z}$ such that $vec x cdotvec z=0$. Then we can build a new matrix $K$ with as many rows as number of $vec y$ coordinates and each row of $K$ coincides with $vec z$. Then
$$
Kvec x=0\
(H+K)vec x=vec y\
H+Kneq H
$$
then your statement is false because for every $vec x, vec y$ there aren't a single $H$ (there're infinitely many of them).
For example
$$
H=left(begin{array}{cc}
1 & 1\
0 & 1
end{array}right)\
vec x=left(begin{array}{c}
1\
2
end{array}right)\
vec y=left(begin{array}{c}
3\
2
end{array}right)\
vec z=left(begin{array}{c}
2\
-1
end{array}right)\
K=left(begin{array}{cc}
2 & -1\
2 & -1
end{array}right)\
H+K=left(begin{array}{cc}
3 & 0\
2 & 0
end{array}right)\
(H+K)vec x=left(begin{array}{c}
3\
2
end{array}right)=vec y
$$
answered Nov 21 '18 at 13:10
P De Donato
4147
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $H_{ij}$ denote the elements of the $mtimes n$-matrix $H$. Then you are asking to solutions of the following linear system of equations:
$$sum_{k=1}^n H_{1k}x_k = y_1$$
$$sum_{k=1}^n H_{2k}x_k = y_2$$
and so on, with the final equation being
$$sum_{k=1}^n H_{mk}x_k = y_m$$
Obviously, we have $ntimes m$ unknown elements (the entries of the matrix $H$) and only $m$ equations. This is, in general, not solveable uniquely.
answered Nov 21 '18 at 13:10
Tom
2,761315
add a comment |
Let $H_{ij}$ denote the elements of the $mtimes n$-matrix $H$. Then you are asking to solutions of the following linear system of equations:
$$sum_{k=1}^n H_{1k}x_k = y_1$$
$$sum_{k=1}^n H_{2k}x_k = y_2$$
and so on, with the final equation being
$$sum_{k=1}^n H_{mk}x_k = y_m$$
Obviously, we have $ntimes m$ unknown elements (the entries of the matrix $H$) and only $m$ equations. This is, in general, not solveable uniquely.
answered Nov 21 '18 at 13:10
Tom
2,761315
add a comment |
Let $H_{ij}$ denote the elements of the $mtimes n$-matrix $H$. Then you are asking to solutions of the following linear system of equations:
$$sum_{k=1}^n H_{1k}x_k = y_1$$
$$sum_{k=1}^n H_{2k}x_k = y_2$$
and so on, with the final equation being
$$sum_{k=1}^n H_{mk}x_k = y_m$$
Obviously, we have $ntimes m$ unknown elements (the entries of the matrix $H$) and only $m$ equations. This is, in general, not solveable uniquely.
answered Nov 21 '18 at 13:10
Tom
2,761315
Let $H_{ij}$ denote the elements of the $mtimes n$-matrix $H$. Then you are asking to solutions of the following linear system of equations:
$$sum_{k=1}^n H_{1k}x_k = y_1$$
$$sum_{k=1}^n H_{2k}x_k = y_2$$
and so on, with the final equation being
$$sum_{k=1}^n H_{mk}x_k = y_m$$
Obviously, we have $ntimes m$ unknown elements (the entries of the matrix $H$) and only $m$ equations. This is, in general, not solveable uniquely.
answered Nov 21 '18 at 13:10
Tom
2,761315
answered Nov 21 '18 at 13:10
Tom
2,761315
answered Nov 21 '18 at 13:10
Tom
2,761315
answered Nov 21 '18 at 13:10
Tom
2,761315
2,761315
add a comment |
add a comment |
For every vector $vec{x}$ exist a non zero vector $vec{z}$ such that $vec x cdotvec z=0$. Then we can build a new matrix $K$ with as many rows as number of $vec y$ coordinates and each row of $K$ coincides with $vec z$. Then
$$
Kvec x=0\
(H+K)vec x=vec y\
H+Kneq H
$$
then your statement is false because for every $vec x, vec y$ there aren't a single $H$ (there're infinitely many of them).
For example
$$
H=left(begin{array}{cc}
1 & 1\
0 & 1
end{array}right)\
vec x=left(begin{array}{c}
1\
2
end{array}right)\
vec y=left(begin{array}{c}
3\
2
end{array}right)\
vec z=left(begin{array}{c}
2\
-1
end{array}right)\
K=left(begin{array}{cc}
2 & -1\
2 & -1
end{array}right)\
H+K=left(begin{array}{cc}
3 & 0\
2 & 0
end{array}right)\
(H+K)vec x=left(begin{array}{c}
3\
2
end{array}right)=vec y
$$
answered Nov 21 '18 at 13:10
P De Donato
4147
add a comment |
For every vector $vec{x}$ exist a non zero vector $vec{z}$ such that $vec x cdotvec z=0$. Then we can build a new matrix $K$ with as many rows as number of $vec y$ coordinates and each row of $K$ coincides with $vec z$. Then
$$
Kvec x=0\
(H+K)vec x=vec y\
H+Kneq H
$$
then your statement is false because for every $vec x, vec y$ there aren't a single $H$ (there're infinitely many of them).
For example
$$
H=left(begin{array}{cc}
1 & 1\
0 & 1
end{array}right)\
vec x=left(begin{array}{c}
1\
2
end{array}right)\
vec y=left(begin{array}{c}
3\
2
end{array}right)\
vec z=left(begin{array}{c}
2\
-1
end{array}right)\
K=left(begin{array}{cc}
2 & -1\
2 & -1
end{array}right)\
H+K=left(begin{array}{cc}
3 & 0\
2 & 0
end{array}right)\
(H+K)vec x=left(begin{array}{c}
3\
2
end{array}right)=vec y
$$
answered Nov 21 '18 at 13:10
P De Donato
4147
add a comment |
For every vector $vec{x}$ exist a non zero vector $vec{z}$ such that $vec x cdotvec z=0$. Then we can build a new matrix $K$ with as many rows as number of $vec y$ coordinates and each row of $K$ coincides with $vec z$. Then
$$
Kvec x=0\
(H+K)vec x=vec y\
H+Kneq H
$$
then your statement is false because for every $vec x, vec y$ there aren't a single $H$ (there're infinitely many of them).
For example
$$
H=left(begin{array}{cc}
1 & 1\
0 & 1
end{array}right)\
vec x=left(begin{array}{c}
1\
2
end{array}right)\
vec y=left(begin{array}{c}
3\
2
end{array}right)\
vec z=left(begin{array}{c}
2\
-1
end{array}right)\
K=left(begin{array}{cc}
2 & -1\
2 & -1
end{array}right)\
H+K=left(begin{array}{cc}
3 & 0\
2 & 0
end{array}right)\
(H+K)vec x=left(begin{array}{c}
3\
2
end{array}right)=vec y
$$
answered Nov 21 '18 at 13:10
P De Donato
4147
For every vector $vec{x}$ exist a non zero vector $vec{z}$ such that $vec x cdotvec z=0$. Then we can build a new matrix $K$ with as many rows as number of $vec y$ coordinates and each row of $K$ coincides with $vec z$. Then
$$
Kvec x=0\
(H+K)vec x=vec y\
H+Kneq H
$$
then your statement is false because for every $vec x, vec y$ there aren't a single $H$ (there're infinitely many of them).
For example
$$
H=left(begin{array}{cc}
1 & 1\
0 & 1
end{array}right)\
vec x=left(begin{array}{c}
1\
2
end{array}right)\
vec y=left(begin{array}{c}
3\
2
end{array}right)\
vec z=left(begin{array}{c}
2\
-1
end{array}right)\
K=left(begin{array}{cc}
2 & -1\
2 & -1
end{array}right)\
H+K=left(begin{array}{cc}
3 & 0\
2 & 0
end{array}right)\
(H+K)vec x=left(begin{array}{c}
3\
2
end{array}right)=vec y
$$
answered Nov 21 '18 at 13:10
P De Donato
4147
answered Nov 21 '18 at 13:10
P De Donato
4147
answered Nov 21 '18 at 13:10
P De Donato
4147
answered Nov 21 '18 at 13:10
P De Donato
4147
4147
add a comment |
add a comment |
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2
No.${}{}{}{}{}{}$
– Saucy O'Path
Nov 21 '18 at 13:04
Given you are in a flight (unknown) can you say it will take you from your current location (known) to Tokyo? Your question can be considered equivalent to this one in some sense.
– P Vanchinathan
Nov 21 '18 at 13:07