Mixing
Need help on understanding linear maps of homogeneous system of linear equations
$begingroup$
In Linear Algebra Done Right, it defines the linear maps of a homogeneous system of linear equations with n variables and m equations as
$$T(x_1,...,x_n) = (sum_{k=1}^{n} A_{1,k}x_k,...,sum_{k=1}^{n} A_{m,k}x_k) = 0$$
My question is why the homogeneous system of linear equations can be expressed as the linear maps from $mathbb{F}^n mapsto mathbb{F}^m$.
In the book it uses this form to prove "A homogeneous system of linear equations with more variables than equations has nonzero solutions".
linear-algebra linear-transformations
edited Jan 24 at 23:18
Bernard
123k741116
asked Jan 24 at 23:04
JOHN JOHN
4279
$endgroup$
add a comment |
$begingroup$
In Linear Algebra Done Right, it defines the linear maps of a homogeneous system of linear equations with n variables and m equations as
$$T(x_1,...,x_n) = (sum_{k=1}^{n} A_{1,k}x_k,...,sum_{k=1}^{n} A_{m,k}x_k) = 0$$
My question is why the homogeneous system of linear equations can be expressed as the linear maps from $mathbb{F}^n mapsto mathbb{F}^m$.
In the book it uses this form to prove "A homogeneous system of linear equations with more variables than equations has nonzero solutions".
linear-algebra linear-transformations
edited Jan 24 at 23:18
Bernard
123k741116
asked Jan 24 at 23:04
JOHN JOHN
4279
$endgroup$
$begingroup$
This because solving the system of linear equations amounts to determining the kernel of the associated linear map.
$endgroup$
– Bernard
Jan 24 at 23:21
1
$begingroup$
You can write the coefficients of the system of linear equations in a matrix $A$. And then define a map $F(x) = Ax$. This map is linear. Solving a homogeneous system of linear equations is then equivalent to finding all vectors $x$ such that $F(x) = Ax = 0$.
$endgroup$
– user635162
Jan 24 at 23:29
add a comment |
$begingroup$
In Linear Algebra Done Right, it defines the linear maps of a homogeneous system of linear equations with n variables and m equations as
$$T(x_1,...,x_n) = (sum_{k=1}^{n} A_{1,k}x_k,...,sum_{k=1}^{n} A_{m,k}x_k) = 0$$
My question is why the homogeneous system of linear equations can be expressed as the linear maps from $mathbb{F}^n mapsto mathbb{F}^m$.
In the book it uses this form to prove "A homogeneous system of linear equations with more variables than equations has nonzero solutions".
linear-algebra linear-transformations
edited Jan 24 at 23:18
Bernard
123k741116
asked Jan 24 at 23:04
JOHN JOHN
4279
$endgroup$
In Linear Algebra Done Right, it defines the linear maps of a homogeneous system of linear equations with n variables and m equations as
$$T(x_1,...,x_n) = (sum_{k=1}^{n} A_{1,k}x_k,...,sum_{k=1}^{n} A_{m,k}x_k) = 0$$
My question is why the homogeneous system of linear equations can be expressed as the linear maps from $mathbb{F}^n mapsto mathbb{F}^m$.
In the book it uses this form to prove "A homogeneous system of linear equations with more variables than equations has nonzero solutions".
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jan 24 at 23:18
Bernard
123k741116
asked Jan 24 at 23:04
JOHN JOHN
4279
edited Jan 24 at 23:18
Bernard
123k741116
asked Jan 24 at 23:04
JOHN JOHN
4279
edited Jan 24 at 23:18
Bernard
123k741116
edited Jan 24 at 23:18
Bernard
123k741116
edited Jan 24 at 23:18
Bernard
123k741116
123k741116
asked Jan 24 at 23:04
JOHN JOHN
4279
asked Jan 24 at 23:04
JOHN JOHN
4279
asked Jan 24 at 23:04
JOHN JOHN
4279
4279
$begingroup$
This because solving the system of linear equations amounts to determining the kernel of the associated linear map.
$endgroup$
– Bernard
Jan 24 at 23:21
1
$begingroup$
You can write the coefficients of the system of linear equations in a matrix $A$. And then define a map $F(x) = Ax$. This map is linear. Solving a homogeneous system of linear equations is then equivalent to finding all vectors $x$ such that $F(x) = Ax = 0$.
$endgroup$
– user635162
Jan 24 at 23:29
add a comment |
$begingroup$
This because solving the system of linear equations amounts to determining the kernel of the associated linear map.
$endgroup$
– Bernard
Jan 24 at 23:21
1
$begingroup$
You can write the coefficients of the system of linear equations in a matrix $A$. And then define a map $F(x) = Ax$. This map is linear. Solving a homogeneous system of linear equations is then equivalent to finding all vectors $x$ such that $F(x) = Ax = 0$.
$endgroup$
– user635162
Jan 24 at 23:29
$begingroup$
This because solving the system of linear equations amounts to determining the kernel of the associated linear map.
$endgroup$
– Bernard
Jan 24 at 23:21
$begingroup$
This because solving the system of linear equations amounts to determining the kernel of the associated linear map.
$endgroup$
– Bernard
Jan 24 at 23:21
1
1
$begingroup$
You can write the coefficients of the system of linear equations in a matrix $A$. And then define a map $F(x) = Ax$. This map is linear. Solving a homogeneous system of linear equations is then equivalent to finding all vectors $x$ such that $F(x) = Ax = 0$.
$endgroup$
– user635162
Jan 24 at 23:29
$begingroup$
You can write the coefficients of the system of linear equations in a matrix $A$. And then define a map $F(x) = Ax$. This map is linear. Solving a homogeneous system of linear equations is then equivalent to finding all vectors $x$ such that $F(x) = Ax = 0$.
$endgroup$
– user635162
Jan 24 at 23:29
add a comment |
1 Answer
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$begingroup$
A homogeneous system of linear equations is, by definition, of the form
$$begin{align} A_{11}x_1+A_{12}x_2+dots+A_{1n}x_n &=0\
&vdots \
A_{m1}x_1+A_{m2}x_2+dots+A_{mn}x_n &=0 end{align}$$
where all $A_{ij}$ are fixed scalars, i.e. are $inBbb F$ (which is typically $Bbb R$ or $Bbb Q$ for the first round).
Now, consider the linear map $T:Bbb F^nmapstoBbb F^m$ defined by the given formula, i.e. a tuple $(x_1,dots,x_n)$ of scalars is mapped to the tuple (column vector) given by the left hand sides of the equations, that is,
$$T := (x_1,dots,x_n)mapsto(sum_kA_{1k}x_k,dots,sum_kA_{mk}x_k)$$
You can readily verify that this indeed defines a linear map, and that the set of equations is exactly its kernel $ker T={xinBbb F^n:T(x)=0}$.
Moreover, and most importantly, if we put the coefficients in a matrix $A$, and regard tuples as column vectors, then we simply have
$$forall xinBbb F^n: T(x)=Acdot x$$
To prove it, check it first for the standard basis elements $x=e_i$, and then use that both $T$ and $xmapsto Acdot x$ are linear.
answered Jan 25 at 0:00
BerciBerci
61.5k23674
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add a comment |
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$begingroup$
A homogeneous system of linear equations is, by definition, of the form
$$begin{align} A_{11}x_1+A_{12}x_2+dots+A_{1n}x_n &=0\
&vdots \
A_{m1}x_1+A_{m2}x_2+dots+A_{mn}x_n &=0 end{align}$$
where all $A_{ij}$ are fixed scalars, i.e. are $inBbb F$ (which is typically $Bbb R$ or $Bbb Q$ for the first round).
Now, consider the linear map $T:Bbb F^nmapstoBbb F^m$ defined by the given formula, i.e. a tuple $(x_1,dots,x_n)$ of scalars is mapped to the tuple (column vector) given by the left hand sides of the equations, that is,
$$T := (x_1,dots,x_n)mapsto(sum_kA_{1k}x_k,dots,sum_kA_{mk}x_k)$$
You can readily verify that this indeed defines a linear map, and that the set of equations is exactly its kernel $ker T={xinBbb F^n:T(x)=0}$.
Moreover, and most importantly, if we put the coefficients in a matrix $A$, and regard tuples as column vectors, then we simply have
$$forall xinBbb F^n: T(x)=Acdot x$$
To prove it, check it first for the standard basis elements $x=e_i$, and then use that both $T$ and $xmapsto Acdot x$ are linear.
answered Jan 25 at 0:00
BerciBerci
61.5k23674
$endgroup$
add a comment |
$begingroup$
A homogeneous system of linear equations is, by definition, of the form
$$begin{align} A_{11}x_1+A_{12}x_2+dots+A_{1n}x_n &=0\
&vdots \
A_{m1}x_1+A_{m2}x_2+dots+A_{mn}x_n &=0 end{align}$$
where all $A_{ij}$ are fixed scalars, i.e. are $inBbb F$ (which is typically $Bbb R$ or $Bbb Q$ for the first round).
Now, consider the linear map $T:Bbb F^nmapstoBbb F^m$ defined by the given formula, i.e. a tuple $(x_1,dots,x_n)$ of scalars is mapped to the tuple (column vector) given by the left hand sides of the equations, that is,
$$T := (x_1,dots,x_n)mapsto(sum_kA_{1k}x_k,dots,sum_kA_{mk}x_k)$$
You can readily verify that this indeed defines a linear map, and that the set of equations is exactly its kernel $ker T={xinBbb F^n:T(x)=0}$.
Moreover, and most importantly, if we put the coefficients in a matrix $A$, and regard tuples as column vectors, then we simply have
$$forall xinBbb F^n: T(x)=Acdot x$$
To prove it, check it first for the standard basis elements $x=e_i$, and then use that both $T$ and $xmapsto Acdot x$ are linear.
answered Jan 25 at 0:00
BerciBerci
61.5k23674
$endgroup$
add a comment |
$begingroup$
A homogeneous system of linear equations is, by definition, of the form
$$begin{align} A_{11}x_1+A_{12}x_2+dots+A_{1n}x_n &=0\
&vdots \
A_{m1}x_1+A_{m2}x_2+dots+A_{mn}x_n &=0 end{align}$$
where all $A_{ij}$ are fixed scalars, i.e. are $inBbb F$ (which is typically $Bbb R$ or $Bbb Q$ for the first round).
Now, consider the linear map $T:Bbb F^nmapstoBbb F^m$ defined by the given formula, i.e. a tuple $(x_1,dots,x_n)$ of scalars is mapped to the tuple (column vector) given by the left hand sides of the equations, that is,
$$T := (x_1,dots,x_n)mapsto(sum_kA_{1k}x_k,dots,sum_kA_{mk}x_k)$$
You can readily verify that this indeed defines a linear map, and that the set of equations is exactly its kernel $ker T={xinBbb F^n:T(x)=0}$.
Moreover, and most importantly, if we put the coefficients in a matrix $A$, and regard tuples as column vectors, then we simply have
$$forall xinBbb F^n: T(x)=Acdot x$$
To prove it, check it first for the standard basis elements $x=e_i$, and then use that both $T$ and $xmapsto Acdot x$ are linear.
answered Jan 25 at 0:00
BerciBerci
61.5k23674
$endgroup$
A homogeneous system of linear equations is, by definition, of the form
$$begin{align} A_{11}x_1+A_{12}x_2+dots+A_{1n}x_n &=0\
&vdots \
A_{m1}x_1+A_{m2}x_2+dots+A_{mn}x_n &=0 end{align}$$
where all $A_{ij}$ are fixed scalars, i.e. are $inBbb F$ (which is typically $Bbb R$ or $Bbb Q$ for the first round).
Now, consider the linear map $T:Bbb F^nmapstoBbb F^m$ defined by the given formula, i.e. a tuple $(x_1,dots,x_n)$ of scalars is mapped to the tuple (column vector) given by the left hand sides of the equations, that is,
$$T := (x_1,dots,x_n)mapsto(sum_kA_{1k}x_k,dots,sum_kA_{mk}x_k)$$
You can readily verify that this indeed defines a linear map, and that the set of equations is exactly its kernel $ker T={xinBbb F^n:T(x)=0}$.
Moreover, and most importantly, if we put the coefficients in a matrix $A$, and regard tuples as column vectors, then we simply have
$$forall xinBbb F^n: T(x)=Acdot x$$
To prove it, check it first for the standard basis elements $x=e_i$, and then use that both $T$ and $xmapsto Acdot x$ are linear.
answered Jan 25 at 0:00
BerciBerci
61.5k23674
answered Jan 25 at 0:00
BerciBerci
61.5k23674
answered Jan 25 at 0:00
BerciBerci
61.5k23674
answered Jan 25 at 0:00
BerciBerci
61.5k23674
61.5k23674
add a comment |
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$begingroup$
This because solving the system of linear equations amounts to determining the kernel of the associated linear map.
$endgroup$
– Bernard
Jan 24 at 23:21
1
$begingroup$
You can write the coefficients of the system of linear equations in a matrix $A$. And then define a map $F(x) = Ax$. This map is linear. Solving a homogeneous system of linear equations is then equivalent to finding all vectors $x$ such that $F(x) = Ax = 0$.
$endgroup$
– user635162
Jan 24 at 23:29