Mixing
Finding an orthonormal basis of the subspace
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Find an orthonormal basis of the subspace:
$$V = [x, y, z, w]^T:x+y+z+w=0$$ of $mathbbR^4$
First I found a $4times4$ determinant to verify whether they are non-singular or not.
$$beginvmatrix 1 & 1 & 1 & 1 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 endvmatrix=1ne0$$
Then to orthogonalize the basis I applied Gram-Schmidt process,
$$v_1=x_1=(1,1,1,1)$$
Is my basis $(1,1,1,1)$ correct?
linear-algebra gram-schmidt
asked Jul 20 at 18:01
philip
1228
 |Â
show 10 more comments
up vote
0
down vote
favorite
Find an orthonormal basis of the subspace:
$$V = [x, y, z, w]^T:x+y+z+w=0$$ of $mathbbR^4$
First I found a $4times4$ determinant to verify whether they are non-singular or not.
$$beginvmatrix 1 & 1 & 1 & 1 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 endvmatrix=1ne0$$
Then to orthogonalize the basis I applied Gram-Schmidt process,
$$v_1=x_1=(1,1,1,1)$$
Is my basis $(1,1,1,1)$ correct?
linear-algebra gram-schmidt
asked Jul 20 at 18:01
philip
1228
1
"Is my basis $(1,1,1,1)$ correct?" As a basis for the space $V$, the set of vectors from $Bbb R^4$ such that the sum of the entries is equal to zero? No. $(1,1,1,1)$ does not have sum of entries equal to zero, it has sum of entries equal to four and so is not even an element of the space, much less a basis for it.
– JMoravitz
Jul 20 at 18:04
@JMoravitz I followed Problem $3$ HERE
– philip
Jul 20 at 18:06
@JMoravitz: It's actually possible, in rare cases, to use as bases, vectors, none of which are in the subspace. I don't think that's happening here - you probably have to have infinite dimensions to make it work. In any case, another reason $(1,1,1,1)$ cannot be a basis is that the dimension of $V$ is $3$, and you only have one vector.
– Adrian Keister
Jul 20 at 18:06
@philip Problem $3$ is computing a basis of the orthogonal complement, $V^bot$, of $V$. Is that what you want? If that is what is really wanted, then your answer is correct.
– user577471
Jul 20 at 18:07
1
You should first focus on finding a basis for the space you are actually being asked about. Ignore Gram-Schmidt for now. Finding a basis for the space itself ignoring orthonormality should have been covered in the previous chapter or the one before even that. The condition $x+y+z+w=0$, if you let $y,z,w$ be free variables equal to $r,s,t$ respectively becomes $begincases x=-r-s-t\y=r\z=s\w=tendcases$, can you continue from there?
– JMoravitz
Jul 20 at 18:12
 |Â
show 10 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find an orthonormal basis of the subspace:
$$V = [x, y, z, w]^T:x+y+z+w=0$$ of $mathbbR^4$
First I found a $4times4$ determinant to verify whether they are non-singular or not.
$$beginvmatrix 1 & 1 & 1 & 1 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 endvmatrix=1ne0$$
Then to orthogonalize the basis I applied Gram-Schmidt process,
$$v_1=x_1=(1,1,1,1)$$
Is my basis $(1,1,1,1)$ correct?
linear-algebra gram-schmidt
asked Jul 20 at 18:01
philip
1228
Find an orthonormal basis of the subspace:
$$V = [x, y, z, w]^T:x+y+z+w=0$$ of $mathbbR^4$
First I found a $4times4$ determinant to verify whether they are non-singular or not.
$$beginvmatrix 1 & 1 & 1 & 1 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 endvmatrix=1ne0$$
Then to orthogonalize the basis I applied Gram-Schmidt process,
$$v_1=x_1=(1,1,1,1)$$
Is my basis $(1,1,1,1)$ correct?
linear-algebra gram-schmidt
asked Jul 20 at 18:01
philip
1228
asked Jul 20 at 18:01
philip
1228
asked Jul 20 at 18:01
philip
1228
asked Jul 20 at 18:01
philip
1228
1228
1
"Is my basis $(1,1,1,1)$ correct?" As a basis for the space $V$, the set of vectors from $Bbb R^4$ such that the sum of the entries is equal to zero? No. $(1,1,1,1)$ does not have sum of entries equal to zero, it has sum of entries equal to four and so is not even an element of the space, much less a basis for it.
– JMoravitz
Jul 20 at 18:04
@JMoravitz I followed Problem $3$ HERE
– philip
Jul 20 at 18:06
@JMoravitz: It's actually possible, in rare cases, to use as bases, vectors, none of which are in the subspace. I don't think that's happening here - you probably have to have infinite dimensions to make it work. In any case, another reason $(1,1,1,1)$ cannot be a basis is that the dimension of $V$ is $3$, and you only have one vector.
– Adrian Keister
Jul 20 at 18:06
@philip Problem $3$ is computing a basis of the orthogonal complement, $V^bot$, of $V$. Is that what you want? If that is what is really wanted, then your answer is correct.
– user577471
Jul 20 at 18:07
1
You should first focus on finding a basis for the space you are actually being asked about. Ignore Gram-Schmidt for now. Finding a basis for the space itself ignoring orthonormality should have been covered in the previous chapter or the one before even that. The condition $x+y+z+w=0$, if you let $y,z,w$ be free variables equal to $r,s,t$ respectively becomes $begincases x=-r-s-t\y=r\z=s\w=tendcases$, can you continue from there?
– JMoravitz
Jul 20 at 18:12
 |Â
show 10 more comments
1
"Is my basis $(1,1,1,1)$ correct?" As a basis for the space $V$, the set of vectors from $Bbb R^4$ such that the sum of the entries is equal to zero? No. $(1,1,1,1)$ does not have sum of entries equal to zero, it has sum of entries equal to four and so is not even an element of the space, much less a basis for it.
– JMoravitz
Jul 20 at 18:04
@JMoravitz I followed Problem $3$ HERE
– philip
Jul 20 at 18:06
@JMoravitz: It's actually possible, in rare cases, to use as bases, vectors, none of which are in the subspace. I don't think that's happening here - you probably have to have infinite dimensions to make it work. In any case, another reason $(1,1,1,1)$ cannot be a basis is that the dimension of $V$ is $3$, and you only have one vector.
– Adrian Keister
Jul 20 at 18:06
@philip Problem $3$ is computing a basis of the orthogonal complement, $V^bot$, of $V$. Is that what you want? If that is what is really wanted, then your answer is correct.
– user577471
Jul 20 at 18:07
1
You should first focus on finding a basis for the space you are actually being asked about. Ignore Gram-Schmidt for now. Finding a basis for the space itself ignoring orthonormality should have been covered in the previous chapter or the one before even that. The condition $x+y+z+w=0$, if you let $y,z,w$ be free variables equal to $r,s,t$ respectively becomes $begincases x=-r-s-t\y=r\z=s\w=tendcases$, can you continue from there?
– JMoravitz
Jul 20 at 18:12
1
1
"Is my basis $(1,1,1,1)$ correct?" As a basis for the space $V$, the set of vectors from $Bbb R^4$ such that the sum of the entries is equal to zero? No. $(1,1,1,1)$ does not have sum of entries equal to zero, it has sum of entries equal to four and so is not even an element of the space, much less a basis for it.
– JMoravitz
Jul 20 at 18:04
"Is my basis $(1,1,1,1)$ correct?" As a basis for the space $V$, the set of vectors from $Bbb R^4$ such that the sum of the entries is equal to zero? No. $(1,1,1,1)$ does not have sum of entries equal to zero, it has sum of entries equal to four and so is not even an element of the space, much less a basis for it.
– JMoravitz
Jul 20 at 18:04
@JMoravitz I followed Problem $3$ HERE
– philip
Jul 20 at 18:06
@JMoravitz I followed Problem $3$ HERE
– philip
Jul 20 at 18:06
@JMoravitz: It's actually possible, in rare cases, to use as bases, vectors, none of which are in the subspace. I don't think that's happening here - you probably have to have infinite dimensions to make it work. In any case, another reason $(1,1,1,1)$ cannot be a basis is that the dimension of $V$ is $3$, and you only have one vector.
– Adrian Keister
Jul 20 at 18:06
@JMoravitz: It's actually possible, in rare cases, to use as bases, vectors, none of which are in the subspace. I don't think that's happening here - you probably have to have infinite dimensions to make it work. In any case, another reason $(1,1,1,1)$ cannot be a basis is that the dimension of $V$ is $3$, and you only have one vector.
– Adrian Keister
Jul 20 at 18:06
@philip Problem $3$ is computing a basis of the orthogonal complement, $V^bot$, of $V$. Is that what you want? If that is what is really wanted, then your answer is correct.
– user577471
Jul 20 at 18:07
@philip Problem $3$ is computing a basis of the orthogonal complement, $V^bot$, of $V$. Is that what you want? If that is what is really wanted, then your answer is correct.
– user577471
Jul 20 at 18:07
1
1
You should first focus on finding a basis for the space you are actually being asked about. Ignore Gram-Schmidt for now. Finding a basis for the space itself ignoring orthonormality should have been covered in the previous chapter or the one before even that. The condition $x+y+z+w=0$, if you let $y,z,w$ be free variables equal to $r,s,t$ respectively becomes $begincases x=-r-s-t\y=r\z=s\w=tendcases$, can you continue from there?
– JMoravitz
Jul 20 at 18:12
You should first focus on finding a basis for the space you are actually being asked about. Ignore Gram-Schmidt for now. Finding a basis for the space itself ignoring orthonormality should have been covered in the previous chapter or the one before even that. The condition $x+y+z+w=0$, if you let $y,z,w$ be free variables equal to $r,s,t$ respectively becomes $begincases x=-r-s-t\y=r\z=s\w=tendcases$, can you continue from there?
– JMoravitz
Jul 20 at 18:12
 |Â
show 10 more comments
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Your proposed basis vector is not in your subspace $$x+y+z+w=0$$ because $$1+1+1+1=4 ne 0$$
you need to find some linearly independent vectors in the subspace to form a basis and then apply Gram-Schmidt method to find an orthonormal basis.
For example $$(1,-1,0,0), (0,1,-1,0), (0,0,1,-1)$$ are linearly independent vectors in your subspace.
Can you apply Gram-Schmidt to that set to find an orthonormal basis?
answered Jul 20 at 18:18
Mohammad Riazi-Kermani
27.5k41852
Can I have different basis from what you have?
– philip
Jul 20 at 18:20
1
@philip of course. Any vector space over an infinite field that is not zero dimensional has infinitely many valid bases for it.
– JMoravitz
Jul 20 at 18:23
@JMoravitz Thank you very much! Now I clearly understood.
– philip
Jul 20 at 18:24
add a comment |Â
up vote
0
down vote
How about:
$$beginbmatrix1 \ -1 \ 0 \ 0endbmatrix,beginbmatrix1 \ 0 \ -1 \ 0endbmatrix,beginbmatrix1 \ 0 \ 0 \ -1endbmatrix$$
answered Jul 20 at 18:10
InterstellarProbe
2,207518
That is a fine basis for the space, however it fails to be an orthogonal basis as well as it fails to be a normal basis. Still, this is an important first step for the OP to understand. Now that a basis has been found, what remains to be done is to orthonormalize the basis via Gram-Schmidt.
– JMoravitz
Jul 20 at 18:14
I still did not understand how you got those basis. Can you please explain.
– philip
Jul 20 at 18:16
1
@philip from setting $y,z,w$ as free variables equal to $r,s,t$ respectively, we get $begincasesx=-r-s-t\y=r\z=~~s\w=~~~~tendcases$. Rewritten with vectors: $beginbmatrixx\y\z\wendbmatrix=-rbeginbmatrix1\-1\0\0endbmatrix-sbeginbmatrix1\0\-1\0endbmatrix-tbeginbmatrix1\0\0\-1endbmatrix$
– JMoravitz
Jul 20 at 18:22
add a comment |Â
up vote
0
down vote
Note that ($1,1,1,1)$ is not a basis vector and that we need
to find a basis for the subspace (the dimension is three then we need 3 basis vectors),
apply GS process and finally normalize them.
Note that we can easily find by inspection
- $v_1=(1,0,-1,0)$
- $v_2=(0,1,0,-1)$
which are independent and orthogonal, then we need only a third vector to complete a basis. What about $v_3=(0,1,-1,0)$?
From here we need only to apply GS process to vector $v_3$ since $v_1$ and $v_3$ are given for free.
answered Jul 20 at 18:07
gimusi
65.4k73584
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your proposed basis vector is not in your subspace $$x+y+z+w=0$$ because $$1+1+1+1=4 ne 0$$
you need to find some linearly independent vectors in the subspace to form a basis and then apply Gram-Schmidt method to find an orthonormal basis.
For example $$(1,-1,0,0), (0,1,-1,0), (0,0,1,-1)$$ are linearly independent vectors in your subspace.
Can you apply Gram-Schmidt to that set to find an orthonormal basis?
answered Jul 20 at 18:18
Mohammad Riazi-Kermani
27.5k41852
Can I have different basis from what you have?
– philip
Jul 20 at 18:20
1
@philip of course. Any vector space over an infinite field that is not zero dimensional has infinitely many valid bases for it.
– JMoravitz
Jul 20 at 18:23
@JMoravitz Thank you very much! Now I clearly understood.
– philip
Jul 20 at 18:24
add a comment |Â
up vote
2
down vote
accepted
Your proposed basis vector is not in your subspace $$x+y+z+w=0$$ because $$1+1+1+1=4 ne 0$$
you need to find some linearly independent vectors in the subspace to form a basis and then apply Gram-Schmidt method to find an orthonormal basis.
For example $$(1,-1,0,0), (0,1,-1,0), (0,0,1,-1)$$ are linearly independent vectors in your subspace.
Can you apply Gram-Schmidt to that set to find an orthonormal basis?
answered Jul 20 at 18:18
Mohammad Riazi-Kermani
27.5k41852
Can I have different basis from what you have?
– philip
Jul 20 at 18:20
1
@philip of course. Any vector space over an infinite field that is not zero dimensional has infinitely many valid bases for it.
– JMoravitz
Jul 20 at 18:23
@JMoravitz Thank you very much! Now I clearly understood.
– philip
Jul 20 at 18:24
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your proposed basis vector is not in your subspace $$x+y+z+w=0$$ because $$1+1+1+1=4 ne 0$$
you need to find some linearly independent vectors in the subspace to form a basis and then apply Gram-Schmidt method to find an orthonormal basis.
For example $$(1,-1,0,0), (0,1,-1,0), (0,0,1,-1)$$ are linearly independent vectors in your subspace.
Can you apply Gram-Schmidt to that set to find an orthonormal basis?
answered Jul 20 at 18:18
Mohammad Riazi-Kermani
27.5k41852
Your proposed basis vector is not in your subspace $$x+y+z+w=0$$ because $$1+1+1+1=4 ne 0$$
you need to find some linearly independent vectors in the subspace to form a basis and then apply Gram-Schmidt method to find an orthonormal basis.
For example $$(1,-1,0,0), (0,1,-1,0), (0,0,1,-1)$$ are linearly independent vectors in your subspace.
Can you apply Gram-Schmidt to that set to find an orthonormal basis?
answered Jul 20 at 18:18
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 18:18
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 18:18
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 18:18
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
Can I have different basis from what you have?
– philip
Jul 20 at 18:20
1
@philip of course. Any vector space over an infinite field that is not zero dimensional has infinitely many valid bases for it.
– JMoravitz
Jul 20 at 18:23
@JMoravitz Thank you very much! Now I clearly understood.
– philip
Jul 20 at 18:24
add a comment |Â
Can I have different basis from what you have?
– philip
Jul 20 at 18:20
1
@philip of course. Any vector space over an infinite field that is not zero dimensional has infinitely many valid bases for it.
– JMoravitz
Jul 20 at 18:23
@JMoravitz Thank you very much! Now I clearly understood.
– philip
Jul 20 at 18:24
Can I have different basis from what you have?
– philip
Jul 20 at 18:20
Can I have different basis from what you have?
– philip
Jul 20 at 18:20
1
1
@philip of course. Any vector space over an infinite field that is not zero dimensional has infinitely many valid bases for it.
– JMoravitz
Jul 20 at 18:23
@philip of course. Any vector space over an infinite field that is not zero dimensional has infinitely many valid bases for it.
– JMoravitz
Jul 20 at 18:23
@JMoravitz Thank you very much! Now I clearly understood.
– philip
Jul 20 at 18:24
@JMoravitz Thank you very much! Now I clearly understood.
– philip
Jul 20 at 18:24
add a comment |Â
up vote
0
down vote
How about:
$$beginbmatrix1 \ -1 \ 0 \ 0endbmatrix,beginbmatrix1 \ 0 \ -1 \ 0endbmatrix,beginbmatrix1 \ 0 \ 0 \ -1endbmatrix$$
answered Jul 20 at 18:10
InterstellarProbe
2,207518
That is a fine basis for the space, however it fails to be an orthogonal basis as well as it fails to be a normal basis. Still, this is an important first step for the OP to understand. Now that a basis has been found, what remains to be done is to orthonormalize the basis via Gram-Schmidt.
– JMoravitz
Jul 20 at 18:14
I still did not understand how you got those basis. Can you please explain.
– philip
Jul 20 at 18:16
1
@philip from setting $y,z,w$ as free variables equal to $r,s,t$ respectively, we get $begincasesx=-r-s-t\y=r\z=~~s\w=~~~~tendcases$. Rewritten with vectors: $beginbmatrixx\y\z\wendbmatrix=-rbeginbmatrix1\-1\0\0endbmatrix-sbeginbmatrix1\0\-1\0endbmatrix-tbeginbmatrix1\0\0\-1endbmatrix$
– JMoravitz
Jul 20 at 18:22
add a comment |Â
up vote
0
down vote
How about:
$$beginbmatrix1 \ -1 \ 0 \ 0endbmatrix,beginbmatrix1 \ 0 \ -1 \ 0endbmatrix,beginbmatrix1 \ 0 \ 0 \ -1endbmatrix$$
answered Jul 20 at 18:10
InterstellarProbe
2,207518
That is a fine basis for the space, however it fails to be an orthogonal basis as well as it fails to be a normal basis. Still, this is an important first step for the OP to understand. Now that a basis has been found, what remains to be done is to orthonormalize the basis via Gram-Schmidt.
– JMoravitz
Jul 20 at 18:14
I still did not understand how you got those basis. Can you please explain.
– philip
Jul 20 at 18:16
1
@philip from setting $y,z,w$ as free variables equal to $r,s,t$ respectively, we get $begincasesx=-r-s-t\y=r\z=~~s\w=~~~~tendcases$. Rewritten with vectors: $beginbmatrixx\y\z\wendbmatrix=-rbeginbmatrix1\-1\0\0endbmatrix-sbeginbmatrix1\0\-1\0endbmatrix-tbeginbmatrix1\0\0\-1endbmatrix$
– JMoravitz
Jul 20 at 18:22
add a comment |Â
up vote
0
down vote
up vote
0
down vote
How about:
$$beginbmatrix1 \ -1 \ 0 \ 0endbmatrix,beginbmatrix1 \ 0 \ -1 \ 0endbmatrix,beginbmatrix1 \ 0 \ 0 \ -1endbmatrix$$
answered Jul 20 at 18:10
InterstellarProbe
2,207518
How about:
$$beginbmatrix1 \ -1 \ 0 \ 0endbmatrix,beginbmatrix1 \ 0 \ -1 \ 0endbmatrix,beginbmatrix1 \ 0 \ 0 \ -1endbmatrix$$
answered Jul 20 at 18:10
InterstellarProbe
2,207518
answered Jul 20 at 18:10
InterstellarProbe
2,207518
answered Jul 20 at 18:10
InterstellarProbe
2,207518
answered Jul 20 at 18:10
InterstellarProbe
2,207518
2,207518
That is a fine basis for the space, however it fails to be an orthogonal basis as well as it fails to be a normal basis. Still, this is an important first step for the OP to understand. Now that a basis has been found, what remains to be done is to orthonormalize the basis via Gram-Schmidt.
– JMoravitz
Jul 20 at 18:14
I still did not understand how you got those basis. Can you please explain.
– philip
Jul 20 at 18:16
1
@philip from setting $y,z,w$ as free variables equal to $r,s,t$ respectively, we get $begincasesx=-r-s-t\y=r\z=~~s\w=~~~~tendcases$. Rewritten with vectors: $beginbmatrixx\y\z\wendbmatrix=-rbeginbmatrix1\-1\0\0endbmatrix-sbeginbmatrix1\0\-1\0endbmatrix-tbeginbmatrix1\0\0\-1endbmatrix$
– JMoravitz
Jul 20 at 18:22
add a comment |Â
That is a fine basis for the space, however it fails to be an orthogonal basis as well as it fails to be a normal basis. Still, this is an important first step for the OP to understand. Now that a basis has been found, what remains to be done is to orthonormalize the basis via Gram-Schmidt.
– JMoravitz
Jul 20 at 18:14
I still did not understand how you got those basis. Can you please explain.
– philip
Jul 20 at 18:16
1
@philip from setting $y,z,w$ as free variables equal to $r,s,t$ respectively, we get $begincasesx=-r-s-t\y=r\z=~~s\w=~~~~tendcases$. Rewritten with vectors: $beginbmatrixx\y\z\wendbmatrix=-rbeginbmatrix1\-1\0\0endbmatrix-sbeginbmatrix1\0\-1\0endbmatrix-tbeginbmatrix1\0\0\-1endbmatrix$
– JMoravitz
Jul 20 at 18:22
That is a fine basis for the space, however it fails to be an orthogonal basis as well as it fails to be a normal basis. Still, this is an important first step for the OP to understand. Now that a basis has been found, what remains to be done is to orthonormalize the basis via Gram-Schmidt.
– JMoravitz
Jul 20 at 18:14
That is a fine basis for the space, however it fails to be an orthogonal basis as well as it fails to be a normal basis. Still, this is an important first step for the OP to understand. Now that a basis has been found, what remains to be done is to orthonormalize the basis via Gram-Schmidt.
– JMoravitz
Jul 20 at 18:14
I still did not understand how you got those basis. Can you please explain.
– philip
Jul 20 at 18:16
I still did not understand how you got those basis. Can you please explain.
– philip
Jul 20 at 18:16
1
1
@philip from setting $y,z,w$ as free variables equal to $r,s,t$ respectively, we get $begincasesx=-r-s-t\y=r\z=~~s\w=~~~~tendcases$. Rewritten with vectors: $beginbmatrixx\y\z\wendbmatrix=-rbeginbmatrix1\-1\0\0endbmatrix-sbeginbmatrix1\0\-1\0endbmatrix-tbeginbmatrix1\0\0\-1endbmatrix$
– JMoravitz
Jul 20 at 18:22
@philip from setting $y,z,w$ as free variables equal to $r,s,t$ respectively, we get $begincasesx=-r-s-t\y=r\z=~~s\w=~~~~tendcases$. Rewritten with vectors: $beginbmatrixx\y\z\wendbmatrix=-rbeginbmatrix1\-1\0\0endbmatrix-sbeginbmatrix1\0\-1\0endbmatrix-tbeginbmatrix1\0\0\-1endbmatrix$
– JMoravitz
Jul 20 at 18:22
add a comment |Â
up vote
0
down vote
Note that ($1,1,1,1)$ is not a basis vector and that we need
to find a basis for the subspace (the dimension is three then we need 3 basis vectors),
apply GS process and finally normalize them.
Note that we can easily find by inspection
- $v_1=(1,0,-1,0)$
- $v_2=(0,1,0,-1)$
which are independent and orthogonal, then we need only a third vector to complete a basis. What about $v_3=(0,1,-1,0)$?
From here we need only to apply GS process to vector $v_3$ since $v_1$ and $v_3$ are given for free.
answered Jul 20 at 18:07
gimusi
65.4k73584
add a comment |Â
up vote
0
down vote
Note that ($1,1,1,1)$ is not a basis vector and that we need
to find a basis for the subspace (the dimension is three then we need 3 basis vectors),
apply GS process and finally normalize them.
Note that we can easily find by inspection
- $v_1=(1,0,-1,0)$
- $v_2=(0,1,0,-1)$
which are independent and orthogonal, then we need only a third vector to complete a basis. What about $v_3=(0,1,-1,0)$?
From here we need only to apply GS process to vector $v_3$ since $v_1$ and $v_3$ are given for free.
answered Jul 20 at 18:07
gimusi
65.4k73584
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that ($1,1,1,1)$ is not a basis vector and that we need
to find a basis for the subspace (the dimension is three then we need 3 basis vectors),
apply GS process and finally normalize them.
Note that we can easily find by inspection
- $v_1=(1,0,-1,0)$
- $v_2=(0,1,0,-1)$
which are independent and orthogonal, then we need only a third vector to complete a basis. What about $v_3=(0,1,-1,0)$?
From here we need only to apply GS process to vector $v_3$ since $v_1$ and $v_3$ are given for free.
answered Jul 20 at 18:07
gimusi
65.4k73584
Note that ($1,1,1,1)$ is not a basis vector and that we need
to find a basis for the subspace (the dimension is three then we need 3 basis vectors),
apply GS process and finally normalize them.
Note that we can easily find by inspection
- $v_1=(1,0,-1,0)$
- $v_2=(0,1,0,-1)$
which are independent and orthogonal, then we need only a third vector to complete a basis. What about $v_3=(0,1,-1,0)$?
From here we need only to apply GS process to vector $v_3$ since $v_1$ and $v_3$ are given for free.
answered Jul 20 at 18:07
gimusi
65.4k73584
edited Jul 20 at 18:48
answered Jul 20 at 18:07
gimusi
65.4k73584
answered Jul 20 at 18:07
gimusi
65.4k73584
answered Jul 20 at 18:07
gimusi
65.4k73584
65.4k73584
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1
"Is my basis $(1,1,1,1)$ correct?" As a basis for the space $V$, the set of vectors from $Bbb R^4$ such that the sum of the entries is equal to zero? No. $(1,1,1,1)$ does not have sum of entries equal to zero, it has sum of entries equal to four and so is not even an element of the space, much less a basis for it.
– JMoravitz
Jul 20 at 18:04
@JMoravitz I followed Problem $3$ HERE
– philip
Jul 20 at 18:06
@JMoravitz: It's actually possible, in rare cases, to use as bases, vectors, none of which are in the subspace. I don't think that's happening here - you probably have to have infinite dimensions to make it work. In any case, another reason $(1,1,1,1)$ cannot be a basis is that the dimension of $V$ is $3$, and you only have one vector.
– Adrian Keister
Jul 20 at 18:06
@philip Problem $3$ is computing a basis of the orthogonal complement, $V^bot$, of $V$. Is that what you want? If that is what is really wanted, then your answer is correct.
– user577471
Jul 20 at 18:07
1
You should first focus on finding a basis for the space you are actually being asked about. Ignore Gram-Schmidt for now. Finding a basis for the space itself ignoring orthonormality should have been covered in the previous chapter or the one before even that. The condition $x+y+z+w=0$, if you let $y,z,w$ be free variables equal to $r,s,t$ respectively becomes $begincases x=-r-s-t\y=r\z=s\w=tendcases$, can you continue from there?
– JMoravitz
Jul 20 at 18:12