Linear Algebra - Positive-Definiteness in Vectors?
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I was reading up on the inner product over at this Wikipedia page, and I noticed, in the given definition, the use of the term "positive-definiteness". Now, from what I know, this is terminology one would describe a quadratic form with; more specifically, given $Q(vec x) = vec{x}:^TAvec{x}$ and its diagonalised form $Q(vec x) = vec{y}:^TDvec{y}$ with $D$ containing the eigenvalues of $A$ and $vec y$ expressed in the eigenbasis of $A$, one would say that $Q(vec x)$ is positive-definite when the entries on the diagonal of $D$ are strictly positive.
After the use of said term on the linked page, it elaborates that positive-definiteness apparently means $<vec x,vec x>:geq0$, but I have a hard time figuring out what this has to do with quadratic forms, and certainly why the greater-than-or-equal-to sign is used directly after a term describing strict positiveness. It might have something to do with a concept called "bilinear forms", but I haven't been taught about them as of yet.
linear-algebra vector-spaces inner-product-space quadratic-forms positive-definite
asked Jan 9 at 10:55
MewMew
206
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add a comment |
$begingroup$
I was reading up on the inner product over at this Wikipedia page, and I noticed, in the given definition, the use of the term "positive-definiteness". Now, from what I know, this is terminology one would describe a quadratic form with; more specifically, given $Q(vec x) = vec{x}:^TAvec{x}$ and its diagonalised form $Q(vec x) = vec{y}:^TDvec{y}$ with $D$ containing the eigenvalues of $A$ and $vec y$ expressed in the eigenbasis of $A$, one would say that $Q(vec x)$ is positive-definite when the entries on the diagonal of $D$ are strictly positive.
After the use of said term on the linked page, it elaborates that positive-definiteness apparently means $<vec x,vec x>:geq0$, but I have a hard time figuring out what this has to do with quadratic forms, and certainly why the greater-than-or-equal-to sign is used directly after a term describing strict positiveness. It might have something to do with a concept called "bilinear forms", but I haven't been taught about them as of yet.
linear-algebra vector-spaces inner-product-space quadratic-forms positive-definite
asked Jan 9 at 10:55
MewMew
206
$endgroup$
1
$begingroup$
$langle x,xrangle=0$ only for $x=0$
$endgroup$
– Shubham Johri
Jan 9 at 10:58
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an inner product in $Bbb R^n$ can be defined using a positive definite operator $A$ as $$langle x,yrangle_A:=langle Ax,yrangle$$ where the inner product in the RHS is the standard euclidean product. This is used in submanifolds in $Bbb R^n$ to define a local inner product. What is positive definite is the inner product, not the vectors.
$endgroup$
– Masacroso
Jan 9 at 11:17
add a comment |
$begingroup$
I was reading up on the inner product over at this Wikipedia page, and I noticed, in the given definition, the use of the term "positive-definiteness". Now, from what I know, this is terminology one would describe a quadratic form with; more specifically, given $Q(vec x) = vec{x}:^TAvec{x}$ and its diagonalised form $Q(vec x) = vec{y}:^TDvec{y}$ with $D$ containing the eigenvalues of $A$ and $vec y$ expressed in the eigenbasis of $A$, one would say that $Q(vec x)$ is positive-definite when the entries on the diagonal of $D$ are strictly positive.
After the use of said term on the linked page, it elaborates that positive-definiteness apparently means $<vec x,vec x>:geq0$, but I have a hard time figuring out what this has to do with quadratic forms, and certainly why the greater-than-or-equal-to sign is used directly after a term describing strict positiveness. It might have something to do with a concept called "bilinear forms", but I haven't been taught about them as of yet.
linear-algebra vector-spaces inner-product-space quadratic-forms positive-definite
asked Jan 9 at 10:55
MewMew
206
$endgroup$
I was reading up on the inner product over at this Wikipedia page, and I noticed, in the given definition, the use of the term "positive-definiteness". Now, from what I know, this is terminology one would describe a quadratic form with; more specifically, given $Q(vec x) = vec{x}:^TAvec{x}$ and its diagonalised form $Q(vec x) = vec{y}:^TDvec{y}$ with $D$ containing the eigenvalues of $A$ and $vec y$ expressed in the eigenbasis of $A$, one would say that $Q(vec x)$ is positive-definite when the entries on the diagonal of $D$ are strictly positive.
After the use of said term on the linked page, it elaborates that positive-definiteness apparently means $<vec x,vec x>:geq0$, but I have a hard time figuring out what this has to do with quadratic forms, and certainly why the greater-than-or-equal-to sign is used directly after a term describing strict positiveness. It might have something to do with a concept called "bilinear forms", but I haven't been taught about them as of yet.
linear-algebra vector-spaces inner-product-space quadratic-forms positive-definite
linear-algebra vector-spaces inner-product-space quadratic-forms positive-definite
asked Jan 9 at 10:55
MewMew
206
asked Jan 9 at 10:55
MewMew
206
asked Jan 9 at 10:55
MewMew
206
asked Jan 9 at 10:55
MewMew
206
asked Jan 9 at 10:55
MewMew
206
206
1
$begingroup$
$langle x,xrangle=0$ only for $x=0$
$endgroup$
– Shubham Johri
Jan 9 at 10:58
$begingroup$
an inner product in $Bbb R^n$ can be defined using a positive definite operator $A$ as $$langle x,yrangle_A:=langle Ax,yrangle$$ where the inner product in the RHS is the standard euclidean product. This is used in submanifolds in $Bbb R^n$ to define a local inner product. What is positive definite is the inner product, not the vectors.
$endgroup$
– Masacroso
Jan 9 at 11:17
add a comment |
1
$begingroup$
$langle x,xrangle=0$ only for $x=0$
$endgroup$
– Shubham Johri
Jan 9 at 10:58
$begingroup$
an inner product in $Bbb R^n$ can be defined using a positive definite operator $A$ as $$langle x,yrangle_A:=langle Ax,yrangle$$ where the inner product in the RHS is the standard euclidean product. This is used in submanifolds in $Bbb R^n$ to define a local inner product. What is positive definite is the inner product, not the vectors.
$endgroup$
– Masacroso
Jan 9 at 11:17
1
1
$begingroup$
$langle x,xrangle=0$ only for $x=0$
$endgroup$
– Shubham Johri
Jan 9 at 10:58
$begingroup$
$langle x,xrangle=0$ only for $x=0$
$endgroup$
– Shubham Johri
Jan 9 at 10:58
$begingroup$
an inner product in $Bbb R^n$ can be defined using a positive definite operator $A$ as $$langle x,yrangle_A:=langle Ax,yrangle$$ where the inner product in the RHS is the standard euclidean product. This is used in submanifolds in $Bbb R^n$ to define a local inner product. What is positive definite is the inner product, not the vectors.
$endgroup$
– Masacroso
Jan 9 at 11:17
$begingroup$
an inner product in $Bbb R^n$ can be defined using a positive definite operator $A$ as $$langle x,yrangle_A:=langle Ax,yrangle$$ where the inner product in the RHS is the standard euclidean product. This is used in submanifolds in $Bbb R^n$ to define a local inner product. What is positive definite is the inner product, not the vectors.
$endgroup$
– Masacroso
Jan 9 at 11:17
add a comment |
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1
$begingroup$
$langle x,xrangle=0$ only for $x=0$
$endgroup$
– Shubham Johri
Jan 9 at 10:58
$begingroup$
an inner product in $Bbb R^n$ can be defined using a positive definite operator $A$ as $$langle x,yrangle_A:=langle Ax,yrangle$$ where the inner product in the RHS is the standard euclidean product. This is used in submanifolds in $Bbb R^n$ to define a local inner product. What is positive definite is the inner product, not the vectors.
$endgroup$
– Masacroso
Jan 9 at 11:17