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src/sectionIVLT.xml
+<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" />.
+\matrix (m) [matrix of math nodes, row sep=5em, column sep=10em, text height=1.5ex, text depth=0.25ex]
+To understand this diagram, begin in the upperleft 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 upperleft 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 lowerleft 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>