1154: Non-Degenerate Hermitian Matrix

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definition of non-degenerate Hermitian matrix

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Topics

About: matrices space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of Hermitian matrix.
• The reader knows a definition of Hermitian conjugate of complex matrix.
• The reader knows a definition of covectors (dual) space of vectors space.

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Target Context

• The reader will have a definition of non-degenerate Hermitian matrix.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$\ast M$: $\in \{\text{ the }n\times n\text{ Hermitian matrices }\}$
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Conditions:
$\mathrm{\forall }v\in \{\text{ the }n\times 1\text{ complex matrices }\}$
(
$\mathrm{\forall }w\in \{\text{ the }n\times 1\text{ complex matrices }\}(w^{\ast }Mv=0)⟹v=0$
)
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The condition, $\mathrm{\forall }v\in \{\text{ the }n\times 1\text{ complex matrices }\}\text{ such that }v\ne 0(\mathrm{\exists }w\in \{\text{ the }n\times 1\text{ complex matrices }\}(w^{\ast }Mv\ne 0))$, is an equivalent condition (called "the 1st alternative condition"), as will be seen in Note.

The condition, $f:{\mathbb{C}}^n\to {{\mathbb{C}}^n}^{\ast },v\mapsto (w\mapsto v^{\ast }Mw)$, where ${{\mathbb{C}}^n}^{\ast }$ is the covectors space of ${\mathbb{C}}^n$, $L({\mathbb{C}}^n:\mathbb{C})$, is a 'vectors spaces - linear morphisms' isomorphism, is an equivalent condition (called "the 2nd alternative condition"), as will be seen in Note.

The condition, $M$ can be diagonalized with all nonzero diagonal components with a unitary matrix, is an equivalent condition (called "the 3nd alternative condition"), as will be seen in Note.

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2: Note

Let us see that the 1st alternative condition is an equivalent condition.

Let us suppose the condition of this definition.

Let $v\ne 0$ be any.

If there was no $w$ such that $w^{\ast }Mv\ne 0$, for each $w$, $w^{\ast }Mv=0$, which would imply $v=0$, a contradiction.

So, there is a $w$.

Let us suppose that 1st alternative condition.

Let $v$ be any.

Let us suppose that for each $w$, $w^{\ast }Mv=0$.

If $v\ne 0$, there would be a $w$ such that $w^{\ast }Mv\ne 0$, a contradiction against that $w^{\ast }Mv=0$ for each $w$.

So, $v=0$.

Let us see that the 2nd alternative condition is an equivalent condition.

1st, let us see that $f(v)$ is indeed in ${{\mathbb{C}}^n}^{\ast }$.

For each $w\in {\mathbb{C}}^n$, $f(v)(w)=v^{\ast }Mw\in \mathbb{C}$. For each $w,w^{\prime }\in {\mathbb{C}}^n$ and each $r,r^{\prime }\in \mathbb{C}$, $f(v)(rw+r^{\prime }{w}^{\prime })=v^{\ast }M(rw+r^{\prime }{w}^{\prime })=r{v}^{\ast }Mw+r^{\prime }{v}^{\ast }M{w}^{\prime }=rf(v)(w)+r^{\prime }f(v)(w^{\prime })$, which means that $f(v)$ is a multilinear map.

So, $f(v)\in {{\mathbb{C}}^n}^{\ast }$.

Let us suppose the condition of this definition.

Let us see that $f$ is linear.

$f(rv+r^{\prime }{v}^{\prime })(w)=(rv+r^{\prime }{v}^{\prime }{)}^{\ast }Mw=r{v}^{\ast }Mw+r^{\prime }{v}^{\prime \ast }Mw=rf(v)(w)+r^{\prime }f(v^{\prime })(w)=(rf(v)+r^{\prime }f(v^{\prime }))(w)$, which implies that $f(rv+r^{\prime }{v}^{\prime })=rf(v)+r^{\prime }f(v^{\prime })$.

Let us see that $f$ is injective.

Let $v,v^{\prime }\in {\mathbb{C}}^n$ be any such that $v\ne v^{\prime }$.

Let us suppose that $f(v)=f(v^{\prime })$. For each $w\in {\mathbb{C}}^n$, $f(v)(w)-f(v^{\prime })(w)=0$, $f(v-v^{\prime })(w)=(v-v^{\prime }{)}^{\ast }Mw=0$, so, $((v-v^{\prime }{)}^{\ast }Mw{)}^{\ast }=0^{\ast }=0$, but the left hand side is $w^{\ast }{M}^{\ast }(v-v^{\prime })=w^{\ast }M(v-v^{\prime })$, so, $w^{\ast }M(v-v^{\prime })=0$, which would imply that $v-v^{\prime }=0$, so, $v=v^{\prime }$, a contradiction.

So, $f$ is injective.

So, $f$ is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any linear injection between any same-finite-dimensional vectors spaces is a 'vectors spaces - linear morphisms' isomorphism: any covectors space is known to be same-dimensional with the original space.

Let us suppose the 2nd alternative condition.

Let $v\in {\mathbb{C}}^n$ be any such that $v\ne 0$.

There is a $w\in {\mathbb{C}}^n$ such that $v^{\ast }Mw\ne 0$, because as $f$ is injective, $f(v)\ne f(0)$, which means that there is a $w$ such that $f(v)(w)\ne 0$ (otherwise, $f(v)=f(0)$), and $f(v)(w)=v^{\ast }Mw\ne 0$, so, $(v^{\ast }Mw{)}^{\ast }\ne 0^{\ast }=0$, but the left hand side is $w^{\ast }{M}^{\ast }v=w^{\ast }Mv$, so, $w^{\ast }Mv\ne 0$.

So, the 1st alternative condition is satisfied, which implies that the condition of this definition is satisfied.

Let us see that the 3rd alternative condition is an equivalent condition.

Note that it is known that any Hermitian matrix, $M$, can be diagonalized by a unitary matrix, $U$, as $U^{\ast }MU$, with all real diagonal components. The issues is whether there is no zero diagonal component.

Let us suppose the condition of this definition.

$w^{\ast }Mv=w^{\ast }{U^{\ast }}^{-1}{U}^{\ast }MU{U}^{-1}v=w^{\ast }{U^{-1}}^{\ast }(U^{\ast }MU)U^{-1}v$, by the proposition that for any invertible complex matrix, the inverse of the Hermitian conjugate of the matrix is the Hermitian conjugate of the inverse of the matrix, $=(U^{-1}w{)}^{\ast }(U^{\ast }MU)U^{-1}v$, by the proposition that the Hermitian conjugate of the product of any complex matrices is the product of the Hermitian conjugates of the constituents in the reverse order.

If $U^{\ast }MU$ had a zero component, a nonzero $v^{\prime }:=U^{-1}v$ can be chosen such that $U^{\ast }MU{U}^{-1}v=0$ (let only the corresponding component of $U^{-1}v$ be nonzero), and $(U^{-1}w{)}^{\ast }(U^{\ast }MU)U^{-1}v=0$ for each $w^{\prime }:=U^{-1}w$; but $v=U{v}^{\prime }$ would be nonzero because $U$ was unitary and $w^{\ast }Mv=(U^{-1}w{)}^{\ast }(U^{\ast }MU)U^{-1}v$ would be $0$ for each $w$, a contradiction.

So, $U^{\ast }MU$ cannot have any zero component.

Let us suppose the 3rd alternative condition.

As before, $w^{\ast }Mv=(U^{-1}w{)}^{\ast }(U^{\ast }MU)U^{-1}v$.

Let us suppose that for each $w\in {\mathbb{C}}^n$, $w^{\ast }Mv=0$.

$w^{\ast }Mv=(U^{-1}w{)}^{\ast }(U^{\ast }MU)U^{-1}v=0$, and $U^{-1}w$ can be chosen such that only the $j$-th component of $(U^{-1}w{)}^{\ast }(U^{\ast }MU)$ is nonzero (let only the corresponding component of $U^{-1}w$ be nonzero), which implies that the $j$-th component of $U^{-1}v$ is $0$, so, $U^{-1}v=0$, so, $v=0$.

So, the condition of this definition is satisfied.

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References

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