Proof verification that if $A_1$ and $A_2$ are affine sets, then $A_1 cap A_2$ is an affine set.
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I have been working through Axler's Linear Algebra Done Right text on my own, and wanted to ensure that the following logic held regarding affine sets:
Let $A_1$ be an affine set in $V$. Then, assuming that $A_1$ is non-empty, we can write $A_1 = {v_1 + u : u in U_1}$ where $U_1$ is a subspace of $V$ and $v_1 in V$. Consider $w, x in A_1$. Then for some $u_1, u_1', u_1'' in U_1$, we have
$$w = v_1 + u_1, quad x = v_1 + u_1' = w + (u_1' - u_1) = w + u_1''$$
which means that $A_1 = {w + u : u in U_1}$. This relationship then helps solve the original problem at hand - show that if $A_1 cap A_2 neq emptyset,$ then $A_1 cap A_2$ is an affine set in $V$. For instance, if $w in A_1 cap A_2$, then $A_1 = w + U_1$ and $A_2 = w + U_2$ (from the reasoning above), so $A_1 cap A_2 = w + U_1 cap U_2$, and since $U_1 cap U_2$ is a subspace in $V$ (we showed that the intersection of subspaces is a subspace earlier in the text), then we can write $A_1 cap A_2 = w + U$ where $U = U_1 cap U_2$ is a subspace of $V$, and thus $A_1 cap A_2$ is an affine set.
I am just getting started in my self-study of higher level math and wanted to ensure that my logic was tight (and also see if you had cleaner/better approaches to this problem)!
linear-algebra proof-verification affine-geometry
asked Nov 30 '18 at 5:26
tucsonman101tucsonman101
134
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add a comment |
$begingroup$
I have been working through Axler's Linear Algebra Done Right text on my own, and wanted to ensure that the following logic held regarding affine sets:
Let $A_1$ be an affine set in $V$. Then, assuming that $A_1$ is non-empty, we can write $A_1 = {v_1 + u : u in U_1}$ where $U_1$ is a subspace of $V$ and $v_1 in V$. Consider $w, x in A_1$. Then for some $u_1, u_1', u_1'' in U_1$, we have
$$w = v_1 + u_1, quad x = v_1 + u_1' = w + (u_1' - u_1) = w + u_1''$$
which means that $A_1 = {w + u : u in U_1}$. This relationship then helps solve the original problem at hand - show that if $A_1 cap A_2 neq emptyset,$ then $A_1 cap A_2$ is an affine set in $V$. For instance, if $w in A_1 cap A_2$, then $A_1 = w + U_1$ and $A_2 = w + U_2$ (from the reasoning above), so $A_1 cap A_2 = w + U_1 cap U_2$, and since $U_1 cap U_2$ is a subspace in $V$ (we showed that the intersection of subspaces is a subspace earlier in the text), then we can write $A_1 cap A_2 = w + U$ where $U = U_1 cap U_2$ is a subspace of $V$, and thus $A_1 cap A_2$ is an affine set.
I am just getting started in my self-study of higher level math and wanted to ensure that my logic was tight (and also see if you had cleaner/better approaches to this problem)!
linear-algebra proof-verification affine-geometry
asked Nov 30 '18 at 5:26
tucsonman101tucsonman101
134
$endgroup$
add a comment |
$begingroup$
I have been working through Axler's Linear Algebra Done Right text on my own, and wanted to ensure that the following logic held regarding affine sets:
Let $A_1$ be an affine set in $V$. Then, assuming that $A_1$ is non-empty, we can write $A_1 = {v_1 + u : u in U_1}$ where $U_1$ is a subspace of $V$ and $v_1 in V$. Consider $w, x in A_1$. Then for some $u_1, u_1', u_1'' in U_1$, we have
$$w = v_1 + u_1, quad x = v_1 + u_1' = w + (u_1' - u_1) = w + u_1''$$
which means that $A_1 = {w + u : u in U_1}$. This relationship then helps solve the original problem at hand - show that if $A_1 cap A_2 neq emptyset,$ then $A_1 cap A_2$ is an affine set in $V$. For instance, if $w in A_1 cap A_2$, then $A_1 = w + U_1$ and $A_2 = w + U_2$ (from the reasoning above), so $A_1 cap A_2 = w + U_1 cap U_2$, and since $U_1 cap U_2$ is a subspace in $V$ (we showed that the intersection of subspaces is a subspace earlier in the text), then we can write $A_1 cap A_2 = w + U$ where $U = U_1 cap U_2$ is a subspace of $V$, and thus $A_1 cap A_2$ is an affine set.
I am just getting started in my self-study of higher level math and wanted to ensure that my logic was tight (and also see if you had cleaner/better approaches to this problem)!
linear-algebra proof-verification affine-geometry
asked Nov 30 '18 at 5:26
tucsonman101tucsonman101
134
$endgroup$
I have been working through Axler's Linear Algebra Done Right text on my own, and wanted to ensure that the following logic held regarding affine sets:
Let $A_1$ be an affine set in $V$. Then, assuming that $A_1$ is non-empty, we can write $A_1 = {v_1 + u : u in U_1}$ where $U_1$ is a subspace of $V$ and $v_1 in V$. Consider $w, x in A_1$. Then for some $u_1, u_1', u_1'' in U_1$, we have
$$w = v_1 + u_1, quad x = v_1 + u_1' = w + (u_1' - u_1) = w + u_1''$$
which means that $A_1 = {w + u : u in U_1}$. This relationship then helps solve the original problem at hand - show that if $A_1 cap A_2 neq emptyset,$ then $A_1 cap A_2$ is an affine set in $V$. For instance, if $w in A_1 cap A_2$, then $A_1 = w + U_1$ and $A_2 = w + U_2$ (from the reasoning above), so $A_1 cap A_2 = w + U_1 cap U_2$, and since $U_1 cap U_2$ is a subspace in $V$ (we showed that the intersection of subspaces is a subspace earlier in the text), then we can write $A_1 cap A_2 = w + U$ where $U = U_1 cap U_2$ is a subspace of $V$, and thus $A_1 cap A_2$ is an affine set.
I am just getting started in my self-study of higher level math and wanted to ensure that my logic was tight (and also see if you had cleaner/better approaches to this problem)!
linear-algebra proof-verification affine-geometry
linear-algebra proof-verification affine-geometry
asked Nov 30 '18 at 5:26
tucsonman101tucsonman101
134
asked Nov 30 '18 at 5:26
tucsonman101tucsonman101
134
asked Nov 30 '18 at 5:26
tucsonman101tucsonman101
134
asked Nov 30 '18 at 5:26
tucsonman101tucsonman101
134
asked Nov 30 '18 at 5:26
tucsonman101tucsonman101
134
134
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