Is a square zero matrix positive semidefinite?
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Does the fact that a square zero matrix contains non-negative eigenvalues (zeros) make it proper to say it is positive semidefinite?
linear-algebra matrices positive-semidefinite
asked Mar 18 at 19:11
KayKay
647
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add a comment |
$begingroup$
Does the fact that a square zero matrix contains non-negative eigenvalues (zeros) make it proper to say it is positive semidefinite?
linear-algebra matrices positive-semidefinite
asked Mar 18 at 19:11
KayKay
647
$endgroup$
add a comment |
$begingroup$
Does the fact that a square zero matrix contains non-negative eigenvalues (zeros) make it proper to say it is positive semidefinite?
linear-algebra matrices positive-semidefinite
asked Mar 18 at 19:11
KayKay
647
$endgroup$
Does the fact that a square zero matrix contains non-negative eigenvalues (zeros) make it proper to say it is positive semidefinite?
linear-algebra matrices positive-semidefinite
linear-algebra matrices positive-semidefinite
asked Mar 18 at 19:11
KayKay
647
asked Mar 18 at 19:11
KayKay
647
edited Mar 18 at 19:25
Kay
asked Mar 18 at 19:11
KayKay
647
asked Mar 18 at 19:11
KayKay
647
asked Mar 18 at 19:11
KayKay
647
647
add a comment |
add a comment |
2 Answers
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The $n times n$ zero matrix is positive semidefinite and negative semidefinite.
answered Mar 18 at 19:23
Gary MoonGary Moon
89616
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"When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.
answered Mar 18 at 19:35
user247327user247327
11.6k1516
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2
$begingroup$
Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
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– Misha Lavrov
Mar 19 at 3:56
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@Misha Lavrov Nitpicking the nitpicking. Because both are equivalent, either can be taken as the definition, so some authors could choose the nonnegative eigenvalue criterion as the definition of positive-semidefinite. Anyhow, this is so presuming attention is restricted to symmetric matrices.
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– Mark L. Stone
Mar 24 at 11:41
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
$begingroup$
The $n times n$ zero matrix is positive semidefinite and negative semidefinite.
answered Mar 18 at 19:23
Gary MoonGary Moon
89616
$endgroup$
add a comment |
$begingroup$
The $n times n$ zero matrix is positive semidefinite and negative semidefinite.
answered Mar 18 at 19:23
Gary MoonGary Moon
89616
$endgroup$
add a comment |
$begingroup$
The $n times n$ zero matrix is positive semidefinite and negative semidefinite.
answered Mar 18 at 19:23
Gary MoonGary Moon
89616
$endgroup$
The $n times n$ zero matrix is positive semidefinite and negative semidefinite.
answered Mar 18 at 19:23
Gary MoonGary Moon
89616
answered Mar 18 at 19:23
Gary MoonGary Moon
89616
answered Mar 18 at 19:23
Gary MoonGary Moon
89616
answered Mar 18 at 19:23
Gary MoonGary Moon
89616
89616
add a comment |
add a comment |
$begingroup$
"When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.
answered Mar 18 at 19:35
user247327user247327
11.6k1516
$endgroup$
2
$begingroup$
Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
$endgroup$
– Misha Lavrov
Mar 19 at 3:56
$begingroup$
@Misha Lavrov Nitpicking the nitpicking. Because both are equivalent, either can be taken as the definition, so some authors could choose the nonnegative eigenvalue criterion as the definition of positive-semidefinite. Anyhow, this is so presuming attention is restricted to symmetric matrices.
$endgroup$
– Mark L. Stone
Mar 24 at 11:41
add a comment |
$begingroup$
"When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.
answered Mar 18 at 19:35
user247327user247327
11.6k1516
$endgroup$
2
$begingroup$
Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
$endgroup$
– Misha Lavrov
Mar 19 at 3:56
$begingroup$
@Misha Lavrov Nitpicking the nitpicking. Because both are equivalent, either can be taken as the definition, so some authors could choose the nonnegative eigenvalue criterion as the definition of positive-semidefinite. Anyhow, this is so presuming attention is restricted to symmetric matrices.
$endgroup$
– Mark L. Stone
Mar 24 at 11:41
add a comment |
$begingroup$
"When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.
answered Mar 18 at 19:35
user247327user247327
11.6k1516
$endgroup$
"When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.
answered Mar 18 at 19:35
user247327user247327
11.6k1516
answered Mar 18 at 19:35
user247327user247327
11.6k1516
answered Mar 18 at 19:35
user247327user247327
11.6k1516
answered Mar 18 at 19:35
user247327user247327
11.6k1516
11.6k1516
2
$begingroup$
Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
$endgroup$
– Misha Lavrov
Mar 19 at 3:56
$begingroup$
@Misha Lavrov Nitpicking the nitpicking. Because both are equivalent, either can be taken as the definition, so some authors could choose the nonnegative eigenvalue criterion as the definition of positive-semidefinite. Anyhow, this is so presuming attention is restricted to symmetric matrices.
$endgroup$
– Mark L. Stone
Mar 24 at 11:41
add a comment |
2
$begingroup$
Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
$endgroup$
– Misha Lavrov
Mar 19 at 3:56
$begingroup$
@Misha Lavrov Nitpicking the nitpicking. Because both are equivalent, either can be taken as the definition, so some authors could choose the nonnegative eigenvalue criterion as the definition of positive-semidefinite. Anyhow, this is so presuming attention is restricted to symmetric matrices.
$endgroup$
– Mark L. Stone
Mar 24 at 11:41
2
2
$begingroup$
Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
$endgroup$
– Misha Lavrov
Mar 19 at 3:56
$begingroup$
Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
$endgroup$
– Misha Lavrov
Mar 19 at 3:56
$begingroup$
@Misha Lavrov Nitpicking the nitpicking. Because both are equivalent, either can be taken as the definition, so some authors could choose the nonnegative eigenvalue criterion as the definition of positive-semidefinite. Anyhow, this is so presuming attention is restricted to symmetric matrices.
$endgroup$
– Mark L. Stone
Mar 24 at 11:41
$begingroup$
@Misha Lavrov Nitpicking the nitpicking. Because both are equivalent, either can be taken as the definition, so some authors could choose the nonnegative eigenvalue criterion as the definition of positive-semidefinite. Anyhow, this is so presuming attention is restricted to symmetric matrices.
$endgroup$
– Mark L. Stone
Mar 24 at 11:41
add a comment |
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