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Diagram AIVS, Section IVLT

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File changes.txt

` New:  Exercise LISS.T51, converse of Theorem VRRB`
` New:  Exercise LISS.T20, subset of linearly independent set is linearly independent`
` New:  Reading Questions, Section OD`
`+New:  Diagram AIVS, Section IVLT`
` Edit:  Explained  min(m,n)  in Exercise PD.T15`
` Edit:  More on induction in Proof Technique I (Dan Messenger)`
` Edit:  Added multiplicities to Solution EE.C19 (Duncan Bennett)`

File src/section-IVLT.xml

` `
` </example>`
` `
`+<p>In <acroref type="example" acro="IVSAV" /> we avoided a computation in \$P_3\$ by a conversion of the computation to a new vector space, \$M_{22}\$, via an invertible linear transformation (also known as an isomorphism).  Here is a diagram meant to to illustrate the more general situation of two vector spaces, \$U\$ and \$V\$, and an invertible linear transformation, \$T\$.  The diagram is simply about a sum of two vectors from \$U\$, rather than a more involved linear combination.  It should remind you of <acroref type="diagram" acro="DLTA" />.`
`+<diagram acro="AIVS">`
`+<title>Addition in Isomorphic Vector Spaces</title>`
`+<tikz>`
`+\matrix (m) [matrix of math nodes, row sep=5em, column sep=10em, text height=1.5ex, text depth=0.25ex]`
`+<![CDATA[{ \vect{u}_1,\,\vect{u}_2 & T(\vect{u}_1),\,T(\vect{u}_2) \\]]>`
`+<![CDATA[\vect{u}_1+\vect{u}_2 & T(\vect{u}_1+\vect{u}_2)=T(\vect{u}_1)+T(\vect{u}_2)\\};]]>`
`+\path[->]`
`+(m-1-1) edge[thick] node[auto] {\$T\$}             (m-1-2)`
`+(m-1-2) edge[thick] node[auto] {\$+\$}             (m-2-2)`
`+(m-1-1) edge[thick] node[auto] {\$+\$}             (m-2-1)`
`+(m-2-2) edge[thick] node[auto] {\$\ltinverse{T}\$} (m-2-1);`
`+</tikz>`
`+</diagram>`
`+To understand this diagram, begin in the upper-left corner, and by going straight down we can compute the sum of the two vectors using the addition for the vector space \$U\$.  The more circuitous alternative, in the spirit of <acroref type="example" acro="IVSAV" />, is to begin in the upper-left corner and then proceed clockwise around the other three sides of the rectangle.  Notice that the vector addition is accomplished using the addition in the vector space \$V\$.  Then, because \$T\$ is a linear transformation, we can say that the result of \$\lt{T}{\vect{u}_1}+\lt{T}{\vect{u}_2}\$ is  equal to \$\lt{T}{\vect{u}_1+\vect{u}_2}\$.  Then the key feature is to recognize that applying \$\ltinverse{T}\$ obviously converts the second version of this result into the sum in the lower-left corner.  So there are two routes to the sum \$\vect{u}_1+\vect{u}_2\$, each employing an addition from a different vector space, but one is <q>direct</q> and trhe other is <q>roundabout</q>.  You might try designing a similar diagram for the case of scalar multiplication (see <acroref type="diagram" acro="DLTM" />) or for a full linear combination.</p>`
`+`
` <p>Checking the dimensions of two vector spaces can be a quick way to establish that they are not isomorphic.  Here's the theorem.</p>`
` `
` <theorem acro="IVSED" index="isomorphic vector spaces!dimension">`