Prove that Gramian matrix is Invertible iff $(v_1,…,v_k) $ is linearly independent
$begingroup$
Prove that Gramian matrix is Invertible iff $(v_1,...,v_k) $ is linearly independent
$G=G(v_1,...,v_k) = [leftlangle v_i,v_jrightrangle ]_{i,j=1}^k $
I have great idea to calculate $det G$ and show that if $(v_1,...,v_k) $ is linearly independent then $det G neq0$ and vice versa.
Plan sounds good (?) but how to calculate $det$ of this?
begin{vmatrix} langle v_1,v_1rangle & langle v_1,v_2rangle &dots & langle v_1,v_nrangle\
langle v_2,v_1rangle & langle v_2,v_2rangle &dots & langle v_2,v_nrangle\
vdots&vdots&ddots&vdots\
langle v_n,v_1rangle & langle v_n,v_2rangle &dots & langle v_n,v_nrangleend{vmatrix}.
Probably I should use some scalar product propeties but it isn't clear for me how can I do that.
linear-algebra matrices inner-product-space
asked Jan 2 at 14:12
VirtualUserVirtualUser
69112
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add a comment |
$begingroup$
Prove that Gramian matrix is Invertible iff $(v_1,...,v_k) $ is linearly independent
$G=G(v_1,...,v_k) = [leftlangle v_i,v_jrightrangle ]_{i,j=1}^k $
I have great idea to calculate $det G$ and show that if $(v_1,...,v_k) $ is linearly independent then $det G neq0$ and vice versa.
Plan sounds good (?) but how to calculate $det$ of this?
begin{vmatrix} langle v_1,v_1rangle & langle v_1,v_2rangle &dots & langle v_1,v_nrangle\
langle v_2,v_1rangle & langle v_2,v_2rangle &dots & langle v_2,v_nrangle\
vdots&vdots&ddots&vdots\
langle v_n,v_1rangle & langle v_n,v_2rangle &dots & langle v_n,v_nrangleend{vmatrix}.
Probably I should use some scalar product propeties but it isn't clear for me how can I do that.
linear-algebra matrices inner-product-space
asked Jan 2 at 14:12
VirtualUserVirtualUser
69112
$endgroup$
$begingroup$
Since $G=V^T V$, $det(G)=det(V)^2$ is non-zero iff $det(V)$ is non-zero, i.e. iff the convex hull of $0,v_1,ldots,v_n$ has a non-zero (oriented) volume.
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– Jack D'Aurizio
Jan 2 at 14:25
add a comment |
$begingroup$
Prove that Gramian matrix is Invertible iff $(v_1,...,v_k) $ is linearly independent
$G=G(v_1,...,v_k) = [leftlangle v_i,v_jrightrangle ]_{i,j=1}^k $
I have great idea to calculate $det G$ and show that if $(v_1,...,v_k) $ is linearly independent then $det G neq0$ and vice versa.
Plan sounds good (?) but how to calculate $det$ of this?
begin{vmatrix} langle v_1,v_1rangle & langle v_1,v_2rangle &dots & langle v_1,v_nrangle\
langle v_2,v_1rangle & langle v_2,v_2rangle &dots & langle v_2,v_nrangle\
vdots&vdots&ddots&vdots\
langle v_n,v_1rangle & langle v_n,v_2rangle &dots & langle v_n,v_nrangleend{vmatrix}.
Probably I should use some scalar product propeties but it isn't clear for me how can I do that.
linear-algebra matrices inner-product-space
asked Jan 2 at 14:12
VirtualUserVirtualUser
69112
$endgroup$
Prove that Gramian matrix is Invertible iff $(v_1,...,v_k) $ is linearly independent
$G=G(v_1,...,v_k) = [leftlangle v_i,v_jrightrangle ]_{i,j=1}^k $
I have great idea to calculate $det G$ and show that if $(v_1,...,v_k) $ is linearly independent then $det G neq0$ and vice versa.
Plan sounds good (?) but how to calculate $det$ of this?
begin{vmatrix} langle v_1,v_1rangle & langle v_1,v_2rangle &dots & langle v_1,v_nrangle\
langle v_2,v_1rangle & langle v_2,v_2rangle &dots & langle v_2,v_nrangle\
vdots&vdots&ddots&vdots\
langle v_n,v_1rangle & langle v_n,v_2rangle &dots & langle v_n,v_nrangleend{vmatrix}.
Probably I should use some scalar product propeties but it isn't clear for me how can I do that.
linear-algebra matrices inner-product-space
linear-algebra matrices inner-product-space
asked Jan 2 at 14:12
VirtualUserVirtualUser
69112
asked Jan 2 at 14:12
VirtualUserVirtualUser
69112
asked Jan 2 at 14:12
VirtualUserVirtualUser
69112
asked Jan 2 at 14:12
VirtualUserVirtualUser
69112
asked Jan 2 at 14:12
VirtualUserVirtualUser
69112
69112
$begingroup$
Since $G=V^T V$, $det(G)=det(V)^2$ is non-zero iff $det(V)$ is non-zero, i.e. iff the convex hull of $0,v_1,ldots,v_n$ has a non-zero (oriented) volume.
$endgroup$
– Jack D'Aurizio
Jan 2 at 14:25
add a comment |
$begingroup$
Since $G=V^T V$, $det(G)=det(V)^2$ is non-zero iff $det(V)$ is non-zero, i.e. iff the convex hull of $0,v_1,ldots,v_n$ has a non-zero (oriented) volume.
$endgroup$
– Jack D'Aurizio
Jan 2 at 14:25
$begingroup$
Since $G=V^T V$, $det(G)=det(V)^2$ is non-zero iff $det(V)$ is non-zero, i.e. iff the convex hull of $0,v_1,ldots,v_n$ has a non-zero (oriented) volume.
$endgroup$
– Jack D'Aurizio
Jan 2 at 14:25
$begingroup$
Since $G=V^T V$, $det(G)=det(V)^2$ is non-zero iff $det(V)$ is non-zero, i.e. iff the convex hull of $0,v_1,ldots,v_n$ has a non-zero (oriented) volume.
$endgroup$
– Jack D'Aurizio
Jan 2 at 14:25
add a comment |
1 Answer
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Let $v_i=sum_{k=1}^n a_{ki}e_k$, where ${e_1,dots,e_n}$ is an orthonormal basis. Then
$$
langle v_i,v_jrangle=sum_{k}sum_{l}a_{ki}a_{lj}langle e_k,e_lrangle=
sum_{k}a_{ki}a_{kj}
$$
and therefore, if $A=[a_{ij}]$, we have $G=A^TA$.
If $Gx=0$, then also $x^TA^TAx=(Ax)^T(Ax)=0$, so $Ax=0$.
The matrix $A$ is the matrix (with respect to the basis ${e_1,dots,e_n}$) of the linear map defined by $e_imapsto v_i$, for $i=1,dots,n$. The rank of this matrix is $n$ if and only if ${v_1,dots,v_n}$ is linearly independent.
If the set is linearly independent, then $Ax=0$ implies $x=0$; thus $Gx=0$ implies $x=0$ and $G$ has rank $n$.
If the set is not linearly independent, then there is $xne0$ with $Ax=0$, so also $Gx=0$ and $G$ is not invertible.
Notes. If the inner product is over the complex numbers, then $G=A^HA$ (the Hermitian transpose) and the argument goes through using the Hermitian transpose instead of the transpose. An orthonormal basis exists because of Gram-Schmidt algorithm.
answered Jan 2 at 15:15
egregegreg
180k1485202
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$begingroup$
Thanks for clear solution
$endgroup$
– VirtualUser
Jan 2 at 18:39
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $v_i=sum_{k=1}^n a_{ki}e_k$, where ${e_1,dots,e_n}$ is an orthonormal basis. Then
$$
langle v_i,v_jrangle=sum_{k}sum_{l}a_{ki}a_{lj}langle e_k,e_lrangle=
sum_{k}a_{ki}a_{kj}
$$
and therefore, if $A=[a_{ij}]$, we have $G=A^TA$.
If $Gx=0$, then also $x^TA^TAx=(Ax)^T(Ax)=0$, so $Ax=0$.
The matrix $A$ is the matrix (with respect to the basis ${e_1,dots,e_n}$) of the linear map defined by $e_imapsto v_i$, for $i=1,dots,n$. The rank of this matrix is $n$ if and only if ${v_1,dots,v_n}$ is linearly independent.
If the set is linearly independent, then $Ax=0$ implies $x=0$; thus $Gx=0$ implies $x=0$ and $G$ has rank $n$.
If the set is not linearly independent, then there is $xne0$ with $Ax=0$, so also $Gx=0$ and $G$ is not invertible.
Notes. If the inner product is over the complex numbers, then $G=A^HA$ (the Hermitian transpose) and the argument goes through using the Hermitian transpose instead of the transpose. An orthonormal basis exists because of Gram-Schmidt algorithm.
answered Jan 2 at 15:15
egregegreg
180k1485202
$endgroup$
$begingroup$
Thanks for clear solution
$endgroup$
– VirtualUser
Jan 2 at 18:39
add a comment |
$begingroup$
Let $v_i=sum_{k=1}^n a_{ki}e_k$, where ${e_1,dots,e_n}$ is an orthonormal basis. Then
$$
langle v_i,v_jrangle=sum_{k}sum_{l}a_{ki}a_{lj}langle e_k,e_lrangle=
sum_{k}a_{ki}a_{kj}
$$
and therefore, if $A=[a_{ij}]$, we have $G=A^TA$.
If $Gx=0$, then also $x^TA^TAx=(Ax)^T(Ax)=0$, so $Ax=0$.
The matrix $A$ is the matrix (with respect to the basis ${e_1,dots,e_n}$) of the linear map defined by $e_imapsto v_i$, for $i=1,dots,n$. The rank of this matrix is $n$ if and only if ${v_1,dots,v_n}$ is linearly independent.
If the set is linearly independent, then $Ax=0$ implies $x=0$; thus $Gx=0$ implies $x=0$ and $G$ has rank $n$.
If the set is not linearly independent, then there is $xne0$ with $Ax=0$, so also $Gx=0$ and $G$ is not invertible.
Notes. If the inner product is over the complex numbers, then $G=A^HA$ (the Hermitian transpose) and the argument goes through using the Hermitian transpose instead of the transpose. An orthonormal basis exists because of Gram-Schmidt algorithm.
answered Jan 2 at 15:15
egregegreg
180k1485202
$endgroup$
$begingroup$
Thanks for clear solution
$endgroup$
– VirtualUser
Jan 2 at 18:39
add a comment |
$begingroup$
Let $v_i=sum_{k=1}^n a_{ki}e_k$, where ${e_1,dots,e_n}$ is an orthonormal basis. Then
$$
langle v_i,v_jrangle=sum_{k}sum_{l}a_{ki}a_{lj}langle e_k,e_lrangle=
sum_{k}a_{ki}a_{kj}
$$
and therefore, if $A=[a_{ij}]$, we have $G=A^TA$.
If $Gx=0$, then also $x^TA^TAx=(Ax)^T(Ax)=0$, so $Ax=0$.
The matrix $A$ is the matrix (with respect to the basis ${e_1,dots,e_n}$) of the linear map defined by $e_imapsto v_i$, for $i=1,dots,n$. The rank of this matrix is $n$ if and only if ${v_1,dots,v_n}$ is linearly independent.
If the set is linearly independent, then $Ax=0$ implies $x=0$; thus $Gx=0$ implies $x=0$ and $G$ has rank $n$.
If the set is not linearly independent, then there is $xne0$ with $Ax=0$, so also $Gx=0$ and $G$ is not invertible.
Notes. If the inner product is over the complex numbers, then $G=A^HA$ (the Hermitian transpose) and the argument goes through using the Hermitian transpose instead of the transpose. An orthonormal basis exists because of Gram-Schmidt algorithm.
answered Jan 2 at 15:15
egregegreg
180k1485202
$endgroup$
Let $v_i=sum_{k=1}^n a_{ki}e_k$, where ${e_1,dots,e_n}$ is an orthonormal basis. Then
$$
langle v_i,v_jrangle=sum_{k}sum_{l}a_{ki}a_{lj}langle e_k,e_lrangle=
sum_{k}a_{ki}a_{kj}
$$
and therefore, if $A=[a_{ij}]$, we have $G=A^TA$.
If $Gx=0$, then also $x^TA^TAx=(Ax)^T(Ax)=0$, so $Ax=0$.
The matrix $A$ is the matrix (with respect to the basis ${e_1,dots,e_n}$) of the linear map defined by $e_imapsto v_i$, for $i=1,dots,n$. The rank of this matrix is $n$ if and only if ${v_1,dots,v_n}$ is linearly independent.
If the set is linearly independent, then $Ax=0$ implies $x=0$; thus $Gx=0$ implies $x=0$ and $G$ has rank $n$.
If the set is not linearly independent, then there is $xne0$ with $Ax=0$, so also $Gx=0$ and $G$ is not invertible.
Notes. If the inner product is over the complex numbers, then $G=A^HA$ (the Hermitian transpose) and the argument goes through using the Hermitian transpose instead of the transpose. An orthonormal basis exists because of Gram-Schmidt algorithm.
answered Jan 2 at 15:15
egregegreg
180k1485202
answered Jan 2 at 15:15
egregegreg
180k1485202
answered Jan 2 at 15:15
egregegreg
180k1485202
answered Jan 2 at 15:15
egregegreg
180k1485202
180k1485202
$begingroup$
Thanks for clear solution
$endgroup$
– VirtualUser
Jan 2 at 18:39
add a comment |
$begingroup$
Thanks for clear solution
$endgroup$
– VirtualUser
Jan 2 at 18:39
$begingroup$
Thanks for clear solution
$endgroup$
– VirtualUser
Jan 2 at 18:39
$begingroup$
Thanks for clear solution
$endgroup$
– VirtualUser
Jan 2 at 18:39
add a comment |
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$begingroup$
Since $G=V^T V$, $det(G)=det(V)^2$ is non-zero iff $det(V)$ is non-zero, i.e. iff the convex hull of $0,v_1,ldots,v_n$ has a non-zero (oriented) volume.
$endgroup$
– Jack D'Aurizio
Jan 2 at 14:25