# Date 1352914740 28800
# Node ID 3c65a31d0edea66a9f795c4337942c621b9bc3e6
# Parent 1356d6a59cdf890114b026b50b58b8d35f54a5be
Diagram AIVS, Section IVLT
diff --git a/changes.txt b/changes.txt
--- a/changes.txt
+++ b/changes.txt
@@ -8,6 +8,7 @@
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)
diff --git a/src/section-IVLT.xml b/src/section-IVLT.xml
--- a/src/section-IVLT.xml
+++ b/src/section-IVLT.xml
@@ -697,6 +697,22 @@
+In 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 .
+
+Addition in Isomorphic Vector Spaces
+
+\matrix (m) [matrix of math nodes, row sep=5em, column sep=10em, text height=1.5ex, text depth=0.25ex]
+
+
+\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);
+
+
+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 , 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 direct

and trhe other is roundabout

. You might try designing a similar diagram for the case of scalar multiplication (see ) or for a full linear combination.

+
Checking the dimensions of two vector spaces can be a quick way to establish that they are not isomorphic. Here's the theorem.