Matrix with no negative elements = Positive Semi Definite?
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A matrix $A$ is positive semi-definite IFF $x^TAxgeq 0$ for all non-zero $xinmathbb{R}^d$. If all elements of $A$ are non-negative, does this guarantee that $A$ is positive semi-definite?
positive-semidefinite
asked Oct 31 at 2:18
Iamanon
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A matrix $A$ is positive semi-definite IFF $x^TAxgeq 0$ for all non-zero $xinmathbb{R}^d$. If all elements of $A$ are non-negative, does this guarantee that $A$ is positive semi-definite?
positive-semidefinite
asked Oct 31 at 2:18
Iamanon
1157
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A matrix $A$ is positive semi-definite IFF $x^TAxgeq 0$ for all non-zero $xinmathbb{R}^d$. If all elements of $A$ are non-negative, does this guarantee that $A$ is positive semi-definite?
positive-semidefinite
asked Oct 31 at 2:18
Iamanon
1157
A matrix $A$ is positive semi-definite IFF $x^TAxgeq 0$ for all non-zero $xinmathbb{R}^d$. If all elements of $A$ are non-negative, does this guarantee that $A$ is positive semi-definite?
positive-semidefinite
positive-semidefinite
asked Oct 31 at 2:18
Iamanon
1157
asked Oct 31 at 2:18
Iamanon
1157
asked Oct 31 at 2:18
Iamanon
1157
asked Oct 31 at 2:18
Iamanon
1157
asked Oct 31 at 2:18
Iamanon
1157
1157
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add a comment |
3 Answers
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No. The matrix $A=begin{pmatrix}0&1\1&0end{pmatrix}$ is not psd, as you can check by seeing that $(1,-1)A(1,-1)^T=-2$.
answered Oct 31 at 2:21
kimchi lover
9,54131128
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$$
left(
begin{array}{cc}
1 & 153 \
153 & 1
end{array}
right)
$$
answered Oct 31 at 2:22
Will Jagy
101k598198
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In general no. One way of defining positive definiteness is through the leading principal minors of a matrix. The $k^{th}$ leading minor if found by computing the deturminant of the matrix after deleting the last $n-k$ colomns and rows in an $n times n $ matrix. It is quite common to see a matrix with all positive entries that has a negative deturminnant, this therefore means this matrix would not be positive definite. For example, If you look at the leading principal minors of the following matrix $ A in mathbb{R}^{ntimes n}$:
$$ A = left( begin{array}{cc}
1 & 1 \ 2 & 1
end{array}right),$$
for $A$ to be positive definite $det(1)>0$ and $det(A)>0$. This is clearly not the case as $det(A)=-1$. In fact this particular matrix is indefinite.
answered Nov 22 at 14:29
p32fr4
357
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
No. The matrix $A=begin{pmatrix}0&1\1&0end{pmatrix}$ is not psd, as you can check by seeing that $(1,-1)A(1,-1)^T=-2$.
answered Oct 31 at 2:21
kimchi lover
9,54131128
add a comment |
up vote
1
down vote
No. The matrix $A=begin{pmatrix}0&1\1&0end{pmatrix}$ is not psd, as you can check by seeing that $(1,-1)A(1,-1)^T=-2$.
answered Oct 31 at 2:21
kimchi lover
9,54131128
add a comment |
up vote
1
down vote
up vote
1
down vote
No. The matrix $A=begin{pmatrix}0&1\1&0end{pmatrix}$ is not psd, as you can check by seeing that $(1,-1)A(1,-1)^T=-2$.
answered Oct 31 at 2:21
kimchi lover
9,54131128
No. The matrix $A=begin{pmatrix}0&1\1&0end{pmatrix}$ is not psd, as you can check by seeing that $(1,-1)A(1,-1)^T=-2$.
answered Oct 31 at 2:21
kimchi lover
9,54131128
answered Oct 31 at 2:21
kimchi lover
9,54131128
answered Oct 31 at 2:21
kimchi lover
9,54131128
answered Oct 31 at 2:21
kimchi lover
9,54131128
9,54131128
add a comment |
add a comment |
up vote
0
down vote
$$
left(
begin{array}{cc}
1 & 153 \
153 & 1
end{array}
right)
$$
answered Oct 31 at 2:22
Will Jagy
101k598198
add a comment |
up vote
0
down vote
$$
left(
begin{array}{cc}
1 & 153 \
153 & 1
end{array}
right)
$$
answered Oct 31 at 2:22
Will Jagy
101k598198
add a comment |
up vote
0
down vote
up vote
0
down vote
$$
left(
begin{array}{cc}
1 & 153 \
153 & 1
end{array}
right)
$$
answered Oct 31 at 2:22
Will Jagy
101k598198
$$
left(
begin{array}{cc}
1 & 153 \
153 & 1
end{array}
right)
$$
answered Oct 31 at 2:22
Will Jagy
101k598198
answered Oct 31 at 2:22
Will Jagy
101k598198
answered Oct 31 at 2:22
Will Jagy
101k598198
answered Oct 31 at 2:22
Will Jagy
101k598198
101k598198
add a comment |
add a comment |
up vote
0
down vote
In general no. One way of defining positive definiteness is through the leading principal minors of a matrix. The $k^{th}$ leading minor if found by computing the deturminant of the matrix after deleting the last $n-k$ colomns and rows in an $n times n $ matrix. It is quite common to see a matrix with all positive entries that has a negative deturminnant, this therefore means this matrix would not be positive definite. For example, If you look at the leading principal minors of the following matrix $ A in mathbb{R}^{ntimes n}$:
$$ A = left( begin{array}{cc}
1 & 1 \ 2 & 1
end{array}right),$$
for $A$ to be positive definite $det(1)>0$ and $det(A)>0$. This is clearly not the case as $det(A)=-1$. In fact this particular matrix is indefinite.
answered Nov 22 at 14:29
p32fr4
357
add a comment |
up vote
0
down vote
In general no. One way of defining positive definiteness is through the leading principal minors of a matrix. The $k^{th}$ leading minor if found by computing the deturminant of the matrix after deleting the last $n-k$ colomns and rows in an $n times n $ matrix. It is quite common to see a matrix with all positive entries that has a negative deturminnant, this therefore means this matrix would not be positive definite. For example, If you look at the leading principal minors of the following matrix $ A in mathbb{R}^{ntimes n}$:
$$ A = left( begin{array}{cc}
1 & 1 \ 2 & 1
end{array}right),$$
for $A$ to be positive definite $det(1)>0$ and $det(A)>0$. This is clearly not the case as $det(A)=-1$. In fact this particular matrix is indefinite.
answered Nov 22 at 14:29
p32fr4
357
add a comment |
up vote
0
down vote
up vote
0
down vote
In general no. One way of defining positive definiteness is through the leading principal minors of a matrix. The $k^{th}$ leading minor if found by computing the deturminant of the matrix after deleting the last $n-k$ colomns and rows in an $n times n $ matrix. It is quite common to see a matrix with all positive entries that has a negative deturminnant, this therefore means this matrix would not be positive definite. For example, If you look at the leading principal minors of the following matrix $ A in mathbb{R}^{ntimes n}$:
$$ A = left( begin{array}{cc}
1 & 1 \ 2 & 1
end{array}right),$$
for $A$ to be positive definite $det(1)>0$ and $det(A)>0$. This is clearly not the case as $det(A)=-1$. In fact this particular matrix is indefinite.
answered Nov 22 at 14:29
p32fr4
357
In general no. One way of defining positive definiteness is through the leading principal minors of a matrix. The $k^{th}$ leading minor if found by computing the deturminant of the matrix after deleting the last $n-k$ colomns and rows in an $n times n $ matrix. It is quite common to see a matrix with all positive entries that has a negative deturminnant, this therefore means this matrix would not be positive definite. For example, If you look at the leading principal minors of the following matrix $ A in mathbb{R}^{ntimes n}$:
$$ A = left( begin{array}{cc}
1 & 1 \ 2 & 1
end{array}right),$$
for $A$ to be positive definite $det(1)>0$ and $det(A)>0$. This is clearly not the case as $det(A)=-1$. In fact this particular matrix is indefinite.
answered Nov 22 at 14:29
p32fr4
357
answered Nov 22 at 14:29
p32fr4
357
answered Nov 22 at 14:29
p32fr4
357
answered Nov 22 at 14:29
p32fr4
357
357
add a comment |
add a comment |
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