Commits

rbeezer committed 1409888

Theorem HMIP, more Theorem EMMVP motivation (Anna Dovzhik)

  • Participants
  • Parent commits d588dfa

Comments (0)

Files changed (2)

 Edit: Stronger finish to proof of Theorem TT (Anna Dovzhik)
 Edit: Wording in Exercise MM.T52 (Dan Drake)
 Edit: Section VS, added a semi-colon (Gavin Tranter)
+Edit: Theorem HMIP, more Theorem EMMVP motivation (Anna Dovzhik)
 Change: Theorem ISRN demoted to Exercise TSS.T11
 Change: Prefer "pivot columns" over "leading 1", Chapter SLE
 Change: Diagram DTSLS, use Theorem CSRN rather than FVCS (Chris Beaulaurier)

src/section-MM.xml

 </alignmath>
 </p>
 
-<p><implyreverse /> This <q>half</q> will take a bit more work.  Assume that $\innerproduct{A\vect{x}}{\vect{y}}=\innerproduct{\vect{x}}{A\vect{y}}$ for all $\vect{x},\,\vect{y}\in\complex{n}$.   Choose any $\vect{x}\in\complex{n}$.  We want to show that $A=\adjoint{A}$ by establishing that $A\vect{x}=\adjoint{A}\vect{x}$.  With only this much motivation, consider the inner product,
+<p><implyreverse /> This <q>half</q> will take a bit more work.  Assume that $\innerproduct{A\vect{x}}{\vect{y}}=\innerproduct{\vect{x}}{A\vect{y}}$ for all $\vect{x},\,\vect{y}\in\complex{n}$.  We show that $A=\adjoint{A}$ by establishing that $A\vect{x}=\adjoint{A}\vect{x}$ for all $\vect{x}$, so we can then apply <acroref type="theorem" acro="EMMVP" />.  With only this much motivation, consider the inner product for any $\vect{x}\in\complex{n}$.
 <alignmath>
 \innerproduct{A\vect{x}-\adjoint{A}\vect{x}}{A\vect{x}-\adjoint{A}\vect{x}}
 <![CDATA[&=]]>