"In $\mathbb{R}^5$ a subspace $U_1=span\{v_1,v_2,v_3\}$ where
$v_1=(1,1,0,1,1),v_2=(1,3,2,1,3),v_3=(0,6,7,0,4)$
Let V be the matrix $[v_1,v_2,v_3]$. The kernel of the linear transformation $\mathbb{R}^5 \to \mathbb{R}^3$ which has the transformation matrix $V^T$ with respect to standard bases in $\mathbb{R}^5 \text{ and } \mathbb{R}^3$ is a new subspace $U_2$ in $\mathbb{R}^5$
How can one show that the two subspaces $U_1$ and $U_2$ are orthogonal?
Best Answer
Let $u \in U_2$ be any vector then clearly $V^T(u) = 0$ gives us that $v_i^T u=0$ for $ i = 1,2,3$. Now if $v \in U_1$ be any vector, then since $U_1$ is span$\{v_1, v_2, v_3\}$ we get $v = c_1v_1 + c_2v_2 + c_3v_3$ for some $c_1, c_2, c_3 \in \mathbb{R}$. Consider, $$v^Tu = c_1(v_1^Tu) + c_2(v_2^Tu) + c_3(v_3^Tu) = 0$$
Thus $U_1$ and $U_2$ are orthogonal.