Answer to Question #138165 in Linear Algebra for Sourav Mondal

Question

Consider the real vector space
A = {(a, b, c, d) I a, b, c, d belongs to R, 2a + 3b = c + d}.
Find dim (A). Also find two distinct subspaces
B1. and B2 of R⁴ such that
A direct sum B1= R⁴=A direct sum B2

Answer 1

The given vector space is

$A=\{(a,b,c,d ) : a,b,c,d \in \R ,2a+3b=c+d\}$

$=\{ (a,b,c,2a+3b-c):a,b,c \in\R\}...........(1)$

$=\{ (a,b,2a+3b-d,d):a,b,d\in \R\}..............(2)$

Since , $A$ is depending on three real number . Therefore $d(A)=3$

Consider the subset $B_1 \ and \ B_2 \ of \ \R^4$ defined as follow,

$B_1=\{(0,0,0,d):d\in \R \}$

$B_2=\{ (0,0,c,0):c\in \R \}$

Clearly, $B_1$ and $B_2$ are subspace of $\R^4$

If $A$ in the form (1) then Our claim is

$\R^4=A\bigoplus B_1$

Let $(a,b,c,d)\in \R^4$ be any arbitrary element.

Then $(a,b,c,d)=(a,b,c,2a+3b-c)+(0,0,0,d-2a-3b+c)$

Clearly , $(a,b,c,2a+3b-c) \in A$

and $(0,0,0,d-2a-3b+c)\in B_1$

Hence , $\R^4=A+ B_1$

Now , My next claim is

$A\cap B_1=\{0\}$

Let $(x,y,z,w)\in A\cap B_1$ be any arbitrary element then ,

$(x,y,z ,w)\in A$ and $(x,y,z,w) \in B_1$

Therefore, $2x+3y=z+w \ and \ x=y=z=0$

$\implies w=0$

$\implies(x,y,z,w)=(0,0,0,0)$

Therefore , $A\cap B_1 =\{0\}$

Hence , $A\bigoplus B_1=\R^4$

Similarly , If $A$ is written in the form (2) then $A\bigoplus B_2=\R^4$