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"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?

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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.

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Let $u_1\in U_1$, $u_2\in U_2$, you need to show that $u_1^Tu_2=0$. What can you derive from "$U_2$ is the kernel of the matrix $V^T$"?

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  • $\begingroup$ Thanks for your answer. The fact that $U_2$ is the kernel with the transformation matrix $V^T$, wouldn't that mean that any vector in $\mathbb{R}^5$ would be transformed into the zerovector in $\mathbb{R}^3$? $\endgroup$ – Alex5207 Dec 2 '17 at 7:16
  • $\begingroup$ It also means $V^Tu_2=0$ which essentially leads to the result. $\endgroup$ – KittyL Dec 2 '17 at 13:02
  • $\begingroup$ Could you elaborate on how that leads to the result? I get that $V^Tu_2=0$ due to the fact the $u_2 \in U_2$, but how does that that $U_1$ and $U_2$ are orthogonal? $\endgroup$ – Alex5207 Dec 2 '17 at 13:37
  • $\begingroup$ Since the product $V^Tu_2$ is obtained by the row vector of $V^T$ multiplied by $u_2$, and the row vectors of $V^T$ are exactly $v_1,v_2,v_3$, which span $U_1$. And that's what you want to show: $u_1^Tu_2=0$. $\endgroup$ – KittyL Dec 2 '17 at 14:16

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