Talk:Value of Adjugate of Determinant

29 October 2019
The Theorem statement:


 * Expected in the statement are $n \times n$ matrices $A$ and $\adj {A}$.


 * A determinant value $D^*$ cannot be called an adjugate matrix.


 * Notation det(A) crosses language barriers. Notation D depends on the English sentence defining it.--Gbgustafson (talk) 03:27,29 October 2019 (EST)


 * But this is an English language website, I don't understand what you mean. --prime mover (talk) 04:09, 29 October 2019 (EDT)


 * If $D=0$ (1x1 matrix), then $D^{n-1}$ is $0^0$. What is true: $\det \paren { \adj {A} } \det \paren {A} = \paren{ \det \paren {A} }^n$.

The proof:


 * One tool is the adjugate identity: $\adj {A} \, A = A \, \adj {A} = \det \paren {A} I$, valid for any square matrix $A$.


 * Another tool is the determinant product theorem $\det \paren {EF} = \det \paren {E} \det \paren {F}$.


 * Equalities $\det \paren { c X } = \det \paren { \paren {c I} X} = \det \paren {cI} \det \paren {X} = c^n \det \paren {X}$ apply.

--Gbgustafson (talk) 03:27, 29 October 2019 (EST)

''But this is an English language website, I don't understand what you mean. --prime mover (talk) 04:09, 29 October 2019 (EDT)''


 * Foreign readers of proofWiki have minimal English skills, but they publish a gaggle of research in English.
 * Chinese reading $\det \paren {A}$ understand it. Their understanding declines with English sentences.
 * ProofWiki house style aids the deaf, whose English skills are usually impaired (Think: Oliver Heaviside).
 * House style aids the blind, who use a screen reader to audio app (Think: Bernard Morin).--Gbgustafson (talk) 06:27, 29 October 2019 (EST)

More on the proof in progress--Gbgustafson (talk) 03:27, 3 November 2019 (EST)

Theorem statement $D^* = D^{n-1}$ requires explanation for a $1\times 1$ matrix with $D=0$.

Proof details:

Let $D = \det \paren {A}$ with $D=0$ allowed.


 * Apply the determinant product theorem to adjugate identity $A \, \adj {A} = \adj {A} \, A = \det \paren {A} \, I$:


 * $(1)\quad \det \paren {A} \, \det \paren { \adj {A} } = \det \paren { \det \paren {A} \, I}$


 * On the right side of (1), factor $c = \det \paren {A}$ from each row, then:


 * $(2)\quad \det {A} \, \det \paren { \adj {A} } = c^n { \det \paren {I} }$


 * $(3)\quad D\, D^* = D^n \quad$ Use the theorem statement notation.


 * If $D \neq 0$, then divide by $D$ in (3) to finish the proof.


 * If $D=0$, then by the adjugate identity $A \, \adj {A} = \paren {\det {A} }\, I$:


 * $(4)\quad \adj {A} \, A = \bszero \quad$ Symbol $\bszero$ is the zero matrix.


 * If $A = \bszero$, then $\adj {A} = \bszero$ implies $D^* = \det \paren { \adj {A} } = 0$.


 * If $A \neq \bszero$, then $\adj {A}\, A = \bszero$ implies the nullspace of $\adj {A}$ contains a nonzero vector.


 * Then $D^* = \det \paren { \adj {A} } = 0$ by the invertible matrix theorem.

--Gbgustafson (talk) 03:32, 3 November 2019 (EST)

Detail about the nullspace of adj(A) --Gbgustafson (talk) 14:23, 8 March 2021 (UTC)

Assume $\det(A)=0$. Then $\adj(A) A = |A| I = \bszero$. Choose column $\vec{e}$ of the identity matrix to insure $\vec{x}=A\vec{e} \ne \vec{0}$. Then $\adj{ A }\vec{x}=\bszero \vec{e}=\vec{0}$ shows $\vec{x}$ is a nonzero vector in the nullspace of $\adj {A}$. This proves $\det \paren { {\adj {A} }  }=0$.