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rbeezer committed b874c28

Proof PEEF, second paragraph, insert "product" (Anna Dovzhik)

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 Typo: Theorem BRS, two extra plus signs in proof (Jenna Fontaine)
 Typo: Archetype B as homogeneous system fixed, Example HUSAB (Mike Daven)
 Typo: Solution MM.C20, "-2" should be "-3" (Anna Dovzhik)
+Typo: Proof PEEF, second paragraph, insert "product" (Anna Dovzhik)
 
 
 

src/section-FS.xml

 <proof>
 <p>$J$ is the result of applying a sequence of row operations to $I_m$, and therefore $J$ and $I_m$ are row-equivalent.  $\homosystem{I_m}$ has only the zero solution, since $I_m$ is nonsingular (<acroref type="theorem" acro="NMRRI" />).  Thus, $\homosystem{J}$ also has only the zero solution (<acroref type="theorem" acro="REMES" />, <acroref type="definition" acro="ESYS" />) and $J$ is therefore nonsingular (<acroref type="definition" acro="NSM" />).</p>
 
-<p>To prove the second part of this conclusion, first convince yourself that row operations and the matrix-vector are commutative operations.  By this we mean the following.
+<p>To prove the second part of this conclusion, first convince yourself that row operations and the matrix-vector product are commutative operations.  By this we mean the following.
 Suppose that $F$ is an $m\times n$ matrix that is row-equivalent to the matrix $G$.  Apply to the column vector $F\vect{w}$ the same sequence of row operations that converts $F$ to $G$.  Then the result is $G\vect{w}$.  So we can do row operations on the matrix, then do a matrix-vector product, <em>or</em> do a matrix-vector product and then do row operations on a column vector, and the result will be the same either way.  Since matrix multiplication is defined by a collection of matrix-vector products (<acroref type="definition" acro="MM" />), the matrix product $FH$ will become $GH$ if we apply the same sequence of row operations to $FH$ that convert $F$ to $G$.  (This argument can be made more rigorous using elementary matrices from the upcoming <acroref type="subsection" acro="DM.EM" /> and the associative property of matrix multiplication established in <acroref type="theorem" acro="MMA" />.)  Now apply these observations to $A$.</p>
 
 <p>Write $AI_n=I_mA$ and apply the row operations that convert $M$ to $N$.  $A$ is converted to $B$, while $I_m$ is converted to $J$, so we have $BI_n=JA$.  Simplifying the left side gives the desired conclusion.</p>