# HG changeset patch # User Rob Beezer # 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.