Creating an orthonormal basis with Gram schmidt procedure error.
$begingroup$
I have a question which says the following:
Let $V$ be the span of $v_{1}=(0,1,2)$, $v_{2}=(-1,0,1)$ and $v_{3}=(-1,1,3)$.
Construct an orthonormal basis $B'$ for $V$ (usual dot product).
I know how to do the question. That's easy. I need to apply the gram schmidt procedure thrice and I have my answer. My problem is the 3rd iteration.
$U_{1}=V_{1}=(0,1,2)$
Orthonormalizing, we obtain $V_{1}'=(0,frac{1}{sqrt{5}},frac{2}{sqrt{5}})$
Then $U_2=V_2-frac{V_2 cdot U_1}{||U_1||^2}U_1$
Doing so gives: $U_2=(-1,frac{-2}{5},frac{1}{5})$
Orthonormalizing, we get $V_{2}'=(0,frac{-2}{sqrt{30}},frac{1}{sqrt{30}})$
Finally, $U_3=V_3-frac{V_3 cdot U_1}{||U_1||^2}U_1 - frac{V_3 cdot U_2}{||U_2||^2}U_2$
However this time, if I plug in everything, I get $U_3=(0,0,0)$. While that is indeed orthogonal to both ${v_1}$ and ${v_2}$ there is no way I can orthonormalize $U_3$ since I get division by $0$.
Is the question just badly asked? Because there is simply no way to obtain an orthonormal basis.
linear-algebra orthonormal gram-schmidt
$endgroup$
add a comment |
$begingroup$
I have a question which says the following:
Let $V$ be the span of $v_{1}=(0,1,2)$, $v_{2}=(-1,0,1)$ and $v_{3}=(-1,1,3)$.
Construct an orthonormal basis $B'$ for $V$ (usual dot product).
I know how to do the question. That's easy. I need to apply the gram schmidt procedure thrice and I have my answer. My problem is the 3rd iteration.
$U_{1}=V_{1}=(0,1,2)$
Orthonormalizing, we obtain $V_{1}'=(0,frac{1}{sqrt{5}},frac{2}{sqrt{5}})$
Then $U_2=V_2-frac{V_2 cdot U_1}{||U_1||^2}U_1$
Doing so gives: $U_2=(-1,frac{-2}{5},frac{1}{5})$
Orthonormalizing, we get $V_{2}'=(0,frac{-2}{sqrt{30}},frac{1}{sqrt{30}})$
Finally, $U_3=V_3-frac{V_3 cdot U_1}{||U_1||^2}U_1 - frac{V_3 cdot U_2}{||U_2||^2}U_2$
However this time, if I plug in everything, I get $U_3=(0,0,0)$. While that is indeed orthogonal to both ${v_1}$ and ${v_2}$ there is no way I can orthonormalize $U_3$ since I get division by $0$.
Is the question just badly asked? Because there is simply no way to obtain an orthonormal basis.
linear-algebra orthonormal gram-schmidt
$endgroup$
add a comment |
$begingroup$
I have a question which says the following:
Let $V$ be the span of $v_{1}=(0,1,2)$, $v_{2}=(-1,0,1)$ and $v_{3}=(-1,1,3)$.
Construct an orthonormal basis $B'$ for $V$ (usual dot product).
I know how to do the question. That's easy. I need to apply the gram schmidt procedure thrice and I have my answer. My problem is the 3rd iteration.
$U_{1}=V_{1}=(0,1,2)$
Orthonormalizing, we obtain $V_{1}'=(0,frac{1}{sqrt{5}},frac{2}{sqrt{5}})$
Then $U_2=V_2-frac{V_2 cdot U_1}{||U_1||^2}U_1$
Doing so gives: $U_2=(-1,frac{-2}{5},frac{1}{5})$
Orthonormalizing, we get $V_{2}'=(0,frac{-2}{sqrt{30}},frac{1}{sqrt{30}})$
Finally, $U_3=V_3-frac{V_3 cdot U_1}{||U_1||^2}U_1 - frac{V_3 cdot U_2}{||U_2||^2}U_2$
However this time, if I plug in everything, I get $U_3=(0,0,0)$. While that is indeed orthogonal to both ${v_1}$ and ${v_2}$ there is no way I can orthonormalize $U_3$ since I get division by $0$.
Is the question just badly asked? Because there is simply no way to obtain an orthonormal basis.
linear-algebra orthonormal gram-schmidt
$endgroup$
I have a question which says the following:
Let $V$ be the span of $v_{1}=(0,1,2)$, $v_{2}=(-1,0,1)$ and $v_{3}=(-1,1,3)$.
Construct an orthonormal basis $B'$ for $V$ (usual dot product).
I know how to do the question. That's easy. I need to apply the gram schmidt procedure thrice and I have my answer. My problem is the 3rd iteration.
$U_{1}=V_{1}=(0,1,2)$
Orthonormalizing, we obtain $V_{1}'=(0,frac{1}{sqrt{5}},frac{2}{sqrt{5}})$
Then $U_2=V_2-frac{V_2 cdot U_1}{||U_1||^2}U_1$
Doing so gives: $U_2=(-1,frac{-2}{5},frac{1}{5})$
Orthonormalizing, we get $V_{2}'=(0,frac{-2}{sqrt{30}},frac{1}{sqrt{30}})$
Finally, $U_3=V_3-frac{V_3 cdot U_1}{||U_1||^2}U_1 - frac{V_3 cdot U_2}{||U_2||^2}U_2$
However this time, if I plug in everything, I get $U_3=(0,0,0)$. While that is indeed orthogonal to both ${v_1}$ and ${v_2}$ there is no way I can orthonormalize $U_3$ since I get division by $0$.
Is the question just badly asked? Because there is simply no way to obtain an orthonormal basis.
linear-algebra orthonormal gram-schmidt
linear-algebra orthonormal gram-schmidt
asked Dec 12 '18 at 6:09
Future Math personFuture Math person
972817
972817
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1 Answer
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$begingroup$
Hint: what is the dimension of $V$? This should tell you how big a basis should be.
$endgroup$
$begingroup$
Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
$endgroup$
– Future Math person
Dec 12 '18 at 6:12
$begingroup$
This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
$endgroup$
– platty
Dec 12 '18 at 6:14
$begingroup$
Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
$endgroup$
– PSG
Dec 12 '18 at 6:15
$begingroup$
Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
$endgroup$
– Future Math person
Dec 12 '18 at 6:16
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: what is the dimension of $V$? This should tell you how big a basis should be.
$endgroup$
$begingroup$
Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
$endgroup$
– Future Math person
Dec 12 '18 at 6:12
$begingroup$
This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
$endgroup$
– platty
Dec 12 '18 at 6:14
$begingroup$
Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
$endgroup$
– PSG
Dec 12 '18 at 6:15
$begingroup$
Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
$endgroup$
– Future Math person
Dec 12 '18 at 6:16
add a comment |
$begingroup$
Hint: what is the dimension of $V$? This should tell you how big a basis should be.
$endgroup$
$begingroup$
Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
$endgroup$
– Future Math person
Dec 12 '18 at 6:12
$begingroup$
This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
$endgroup$
– platty
Dec 12 '18 at 6:14
$begingroup$
Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
$endgroup$
– PSG
Dec 12 '18 at 6:15
$begingroup$
Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
$endgroup$
– Future Math person
Dec 12 '18 at 6:16
add a comment |
$begingroup$
Hint: what is the dimension of $V$? This should tell you how big a basis should be.
$endgroup$
Hint: what is the dimension of $V$? This should tell you how big a basis should be.
answered Dec 12 '18 at 6:12
plattyplatty
3,370320
3,370320
$begingroup$
Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
$endgroup$
– Future Math person
Dec 12 '18 at 6:12
$begingroup$
This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
$endgroup$
– platty
Dec 12 '18 at 6:14
$begingroup$
Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
$endgroup$
– PSG
Dec 12 '18 at 6:15
$begingroup$
Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
$endgroup$
– Future Math person
Dec 12 '18 at 6:16
add a comment |
$begingroup$
Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
$endgroup$
– Future Math person
Dec 12 '18 at 6:12
$begingroup$
This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
$endgroup$
– platty
Dec 12 '18 at 6:14
$begingroup$
Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
$endgroup$
– PSG
Dec 12 '18 at 6:15
$begingroup$
Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
$endgroup$
– Future Math person
Dec 12 '18 at 6:16
$begingroup$
Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
$endgroup$
– Future Math person
Dec 12 '18 at 6:12
$begingroup$
Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
$endgroup$
– Future Math person
Dec 12 '18 at 6:12
$begingroup$
This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
$endgroup$
– platty
Dec 12 '18 at 6:14
$begingroup$
This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
$endgroup$
– platty
Dec 12 '18 at 6:14
$begingroup$
Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
$endgroup$
– PSG
Dec 12 '18 at 6:15
$begingroup$
Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
$endgroup$
– PSG
Dec 12 '18 at 6:15
$begingroup$
Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
$endgroup$
– Future Math person
Dec 12 '18 at 6:16
$begingroup$
Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
$endgroup$
– Future Math person
Dec 12 '18 at 6:16
add a comment |
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