- #1

- 1,079

- 89

- Homework Statement
- Let ##X## be a vector space and ##Y## a linear subspace. Prove that the Cartesian sum ##Y\oplus X/Y## is isomorphic to ##X##.

- Relevant Equations
- Isomorphic:

Two spaces A and B are isomorphic to each other if there is a one-to-one and onto function f defined from A to B or vice versa.

Quotient space:

##X/Y=\{x\in X:\exists\,x_0 \in X \textrm{ such that } x-x_0=y\textrm{ for some }y\in Y\}##

Cartesian sum:

Let ##A,B## be two sets. Then the Cartesian sum of those sets is denoted as ##A\oplus B## and comprises of elements of the form ##(a,b)## where ##a\in A,b\in B##. Addition and multiplication are defined component-wise.

Theorem:

##\dim Y\oplus {X/Y}=\dim X##

Let ##n=\dim X## and ##m=\dim Y##.

Define a basis for ##X: y_1,...,y_m,z_{m+1},...,z_n##. The first ##m## terms are a basis for ##Y##. The remaining ##n-m## terms are a basis for its complement w.r.t ##X##. Let's call it ##Z##. ##X## is the direct sum of ##Y## and ##Z##; denote it as ##X=Y+Z##. In other words, you can express any ##x\in X## uniquely as ##x=y+z## for some ##y\in Y,z\in Z##.

Define a linear transformation ##T:X\rightarrow (Y\oplus X/Y)## by ##T(x)=(y,z)## where ##x=y+z##. We prove that it is one-to-one and unique.

Suppose for some ##x_1\in X##, ##T(x_1)=T(x)##. Then ##(y_1,z_1)=(y,z)## and in turn, ##(y_1-y,z_1-z)=(0,0)##; ##y_1=y## and ##z_1=z## as a result. It follows that ##x_1=y_1+z_1=y+z=x##.

Let ##(y_0,z_0)\in Y\oplus Z##. Since ##y_0\in Y## and ##z_0\in Z##, then ##y_0+z_0\in Y+Z=X##. Hence, there is a ##x_0\in X## such that ##x_0=y_0+z_0##. ##T(x_0)=(y_0,z_0)## as a result.

Hence, ##Y\oplus X/Y## is isomorphic to ##X##.

Define a basis for ##X: y_1,...,y_m,z_{m+1},...,z_n##. The first ##m## terms are a basis for ##Y##. The remaining ##n-m## terms are a basis for its complement w.r.t ##X##. Let's call it ##Z##. ##X## is the direct sum of ##Y## and ##Z##; denote it as ##X=Y+Z##. In other words, you can express any ##x\in X## uniquely as ##x=y+z## for some ##y\in Y,z\in Z##.

Define a linear transformation ##T:X\rightarrow (Y\oplus X/Y)## by ##T(x)=(y,z)## where ##x=y+z##. We prove that it is one-to-one and unique.

Suppose for some ##x_1\in X##, ##T(x_1)=T(x)##. Then ##(y_1,z_1)=(y,z)## and in turn, ##(y_1-y,z_1-z)=(0,0)##; ##y_1=y## and ##z_1=z## as a result. It follows that ##x_1=y_1+z_1=y+z=x##.

Let ##(y_0,z_0)\in Y\oplus Z##. Since ##y_0\in Y## and ##z_0\in Z##, then ##y_0+z_0\in Y+Z=X##. Hence, there is a ##x_0\in X## such that ##x_0=y_0+z_0##. ##T(x_0)=(y_0,z_0)## as a result.

Hence, ##Y\oplus X/Y## is isomorphic to ##X##.

Last edited: