Answer to Question #145001 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

Let us choose standard basis of $\mathbb{R}^4$ and give the linear map $T$ from $\mathbb{R}^4\longrightarrow \mathbb{R}$ as $e_1\mapsto 2e_1, e_2\mapsto 3e_2, e_3\mapsto -e_3, e_4\mapsto -e_4.$ Then clearly, $T(a,b,c,d)=2a+3b-c-d.$ Then A is the kernel of T. Clearly, $T$ being non-zero and image being in $\mathbb{R}$ its rank is 1. Hence by rank-nullity the dimension of A is 4-1=3. Now $2\times 1+3\times 0=2\neq 0+0$ . Hence $e_1 \notin A.$ Similarly $e_2\notin A.$ Let $B_i=\langle e_i\rangle$ for $i=1,2.$ Now $A+B_i$ is a subspace of $\mathbb{R}^4$ . Also $A\cap B_i=\{0\}.$ Hence $A+B_{i}=A\oplus B_i$ . Dimension of $A+B_i=$ dim$A+$ dim $B_i-$ dim $(A\cap B_i)$ =3+1-0=4 . But the only subspace of $\mathbb{R}^4$ of dimension 4 is $\mathbb{R}^4$ itself. Hence $A\oplus B_1=\mathbb{R}^4=A\oplus B_2.$ Now $\lambda e_1=(\lambda,0,0,0)\neq (0,\mu,0,0)=\mu e_2.$ Hence $B_1,B_2$ are distinct.