Direct sum and intersection of more than two subspaces
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In "Linear Algebra Done Right" Axler points out that if you have more then two subspaces, it is not enough to test that each pair intersects only at 0, in order to say that they have direct sum.
As an example, he asks to consider the following subspaces:
$U_1 = {(x, y, 0)}$
$U_2 = {(0, 0, z)}$
$U_3 = {(0, y, y)}$
where, he says, $U_1 bigcap U_2 = U_1 bigcap U_3 = U_2 bigcap U_3 = {0}$ but they do not have direct sum.
I agree that they do not have direct sum, but if I test pair intersection I see that
$U_1 bigcap U_3 = {(0, y, 0)} ne {0}$
$U_2 bigcap U_3 = {(0, 0, d)} ne {0}$
So, to me it looks like contradiction, and that it is enough to check pair intersection. Where am I wrong?
linear-algebra
asked Jan 18 at 2:19
AlexeyAlexey
132
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add a comment |
$begingroup$
In "Linear Algebra Done Right" Axler points out that if you have more then two subspaces, it is not enough to test that each pair intersects only at 0, in order to say that they have direct sum.
As an example, he asks to consider the following subspaces:
$U_1 = {(x, y, 0)}$
$U_2 = {(0, 0, z)}$
$U_3 = {(0, y, y)}$
where, he says, $U_1 bigcap U_2 = U_1 bigcap U_3 = U_2 bigcap U_3 = {0}$ but they do not have direct sum.
I agree that they do not have direct sum, but if I test pair intersection I see that
$U_1 bigcap U_3 = {(0, y, 0)} ne {0}$
$U_2 bigcap U_3 = {(0, 0, d)} ne {0}$
So, to me it looks like contradiction, and that it is enough to check pair intersection. Where am I wrong?
linear-algebra
asked Jan 18 at 2:19
AlexeyAlexey
132
$endgroup$
add a comment |
$begingroup$
In "Linear Algebra Done Right" Axler points out that if you have more then two subspaces, it is not enough to test that each pair intersects only at 0, in order to say that they have direct sum.
As an example, he asks to consider the following subspaces:
$U_1 = {(x, y, 0)}$
$U_2 = {(0, 0, z)}$
$U_3 = {(0, y, y)}$
where, he says, $U_1 bigcap U_2 = U_1 bigcap U_3 = U_2 bigcap U_3 = {0}$ but they do not have direct sum.
I agree that they do not have direct sum, but if I test pair intersection I see that
$U_1 bigcap U_3 = {(0, y, 0)} ne {0}$
$U_2 bigcap U_3 = {(0, 0, d)} ne {0}$
So, to me it looks like contradiction, and that it is enough to check pair intersection. Where am I wrong?
linear-algebra
asked Jan 18 at 2:19
AlexeyAlexey
132
$endgroup$
In "Linear Algebra Done Right" Axler points out that if you have more then two subspaces, it is not enough to test that each pair intersects only at 0, in order to say that they have direct sum.
As an example, he asks to consider the following subspaces:
$U_1 = {(x, y, 0)}$
$U_2 = {(0, 0, z)}$
$U_3 = {(0, y, y)}$
where, he says, $U_1 bigcap U_2 = U_1 bigcap U_3 = U_2 bigcap U_3 = {0}$ but they do not have direct sum.
I agree that they do not have direct sum, but if I test pair intersection I see that
$U_1 bigcap U_3 = {(0, y, 0)} ne {0}$
$U_2 bigcap U_3 = {(0, 0, d)} ne {0}$
So, to me it looks like contradiction, and that it is enough to check pair intersection. Where am I wrong?
linear-algebra
linear-algebra
asked Jan 18 at 2:19
AlexeyAlexey
132
asked Jan 18 at 2:19
AlexeyAlexey
132
asked Jan 18 at 2:19
AlexeyAlexey
132
asked Jan 18 at 2:19
AlexeyAlexey
132
asked Jan 18 at 2:19
AlexeyAlexey
132
132
add a comment |
add a comment |
1 Answer
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There is no $(0,y,0)$ in $U_3$ with $yneq 0$. The elements of $U_3$ are of the form $(0,y,y)$. So if $y=0$, we will have $(0,0,0)$. Hence we cannot produce $(0,y,0)$ with $yneq 0$.
answered Jan 18 at 2:25
Thomas ShelbyThomas Shelby
4,7832727
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$begingroup$
There is no $(0,y,0)$ in $U_3$ with $yneq 0$. The elements of $U_3$ are of the form $(0,y,y)$. So if $y=0$, we will have $(0,0,0)$. Hence we cannot produce $(0,y,0)$ with $yneq 0$.
answered Jan 18 at 2:25
Thomas ShelbyThomas Shelby
4,7832727
$endgroup$
add a comment |
$begingroup$
There is no $(0,y,0)$ in $U_3$ with $yneq 0$. The elements of $U_3$ are of the form $(0,y,y)$. So if $y=0$, we will have $(0,0,0)$. Hence we cannot produce $(0,y,0)$ with $yneq 0$.
answered Jan 18 at 2:25
Thomas ShelbyThomas Shelby
4,7832727
$endgroup$
add a comment |
$begingroup$
There is no $(0,y,0)$ in $U_3$ with $yneq 0$. The elements of $U_3$ are of the form $(0,y,y)$. So if $y=0$, we will have $(0,0,0)$. Hence we cannot produce $(0,y,0)$ with $yneq 0$.
answered Jan 18 at 2:25
Thomas ShelbyThomas Shelby
4,7832727
$endgroup$
There is no $(0,y,0)$ in $U_3$ with $yneq 0$. The elements of $U_3$ are of the form $(0,y,y)$. So if $y=0$, we will have $(0,0,0)$. Hence we cannot produce $(0,y,0)$ with $yneq 0$.
answered Jan 18 at 2:25
Thomas ShelbyThomas Shelby
4,7832727
answered Jan 18 at 2:25
Thomas ShelbyThomas Shelby
4,7832727
answered Jan 18 at 2:25
Thomas ShelbyThomas Shelby
4,7832727
answered Jan 18 at 2:25
Thomas ShelbyThomas Shelby
4,7832727
4,7832727
add a comment |
add a comment |
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