Find the orthogonal bases of the space $ V $ and $ V^{perp}$
$begingroup$
In space $mathbb R^3 $ f Find the orthogonal bases of the space $ V $ and $ V^{perp}$ where $$V = left{ vec{x} in mathbb R^3 : x_1 - 3x_2 + x_3 = 0 right} $$
On the begining, I know that may I ask you for basics of topic, but I truly have some doubts...
Ok, I found basis of $V$: $ [3,1,0]^T,[-1,0,1]^T$
and I have used Gram–Schmidt process to find orthogonal basis of $V$:
$[3,1,0]^T,[-1/10,3/10,1]^T$
Ok, now I should find basis of $ V^{perp} $ and on the same algorithm find orthogonal basis of $V^{perp}$ - but there is my question - if in basis of $V^{perp}$ I have all vectors which are perpendicular to vectors from basis $V$, the basis of $ V^{perp} $ will be just orthogonal basis of $V$? And orthogonal basis of $V^{perp}$ will be just basis of $V$? Have I right or there is something other to do?
linear-algebra orthogonality
asked Jan 3 at 12:07
VirtualUserVirtualUser
69414
$endgroup$
add a comment |
$begingroup$
In space $mathbb R^3 $ f Find the orthogonal bases of the space $ V $ and $ V^{perp}$ where $$V = left{ vec{x} in mathbb R^3 : x_1 - 3x_2 + x_3 = 0 right} $$
On the begining, I know that may I ask you for basics of topic, but I truly have some doubts...
Ok, I found basis of $V$: $ [3,1,0]^T,[-1,0,1]^T$
and I have used Gram–Schmidt process to find orthogonal basis of $V$:
$[3,1,0]^T,[-1/10,3/10,1]^T$
Ok, now I should find basis of $ V^{perp} $ and on the same algorithm find orthogonal basis of $V^{perp}$ - but there is my question - if in basis of $V^{perp}$ I have all vectors which are perpendicular to vectors from basis $V$, the basis of $ V^{perp} $ will be just orthogonal basis of $V$? And orthogonal basis of $V^{perp}$ will be just basis of $V$? Have I right or there is something other to do?
linear-algebra orthogonality
asked Jan 3 at 12:07
VirtualUserVirtualUser
69414
$endgroup$
$begingroup$
I have asked this again with fixed symbols
$endgroup$
– VirtualUser
Jan 3 at 12:07
$begingroup$
You could simply edit the old question fixing those symbols (see the edit button). It is easier than retyping the whole thing as a new question.
$endgroup$
– A.Γ.
Jan 3 at 12:25
add a comment |
$begingroup$
In space $mathbb R^3 $ f Find the orthogonal bases of the space $ V $ and $ V^{perp}$ where $$V = left{ vec{x} in mathbb R^3 : x_1 - 3x_2 + x_3 = 0 right} $$
On the begining, I know that may I ask you for basics of topic, but I truly have some doubts...
Ok, I found basis of $V$: $ [3,1,0]^T,[-1,0,1]^T$
and I have used Gram–Schmidt process to find orthogonal basis of $V$:
$[3,1,0]^T,[-1/10,3/10,1]^T$
Ok, now I should find basis of $ V^{perp} $ and on the same algorithm find orthogonal basis of $V^{perp}$ - but there is my question - if in basis of $V^{perp}$ I have all vectors which are perpendicular to vectors from basis $V$, the basis of $ V^{perp} $ will be just orthogonal basis of $V$? And orthogonal basis of $V^{perp}$ will be just basis of $V$? Have I right or there is something other to do?
linear-algebra orthogonality
asked Jan 3 at 12:07
VirtualUserVirtualUser
69414
$endgroup$
In space $mathbb R^3 $ f Find the orthogonal bases of the space $ V $ and $ V^{perp}$ where $$V = left{ vec{x} in mathbb R^3 : x_1 - 3x_2 + x_3 = 0 right} $$
On the begining, I know that may I ask you for basics of topic, but I truly have some doubts...
Ok, I found basis of $V$: $ [3,1,0]^T,[-1,0,1]^T$
and I have used Gram–Schmidt process to find orthogonal basis of $V$:
$[3,1,0]^T,[-1/10,3/10,1]^T$
Ok, now I should find basis of $ V^{perp} $ and on the same algorithm find orthogonal basis of $V^{perp}$ - but there is my question - if in basis of $V^{perp}$ I have all vectors which are perpendicular to vectors from basis $V$, the basis of $ V^{perp} $ will be just orthogonal basis of $V$? And orthogonal basis of $V^{perp}$ will be just basis of $V$? Have I right or there is something other to do?
linear-algebra orthogonality
linear-algebra orthogonality
asked Jan 3 at 12:07
VirtualUserVirtualUser
69414
asked Jan 3 at 12:07
VirtualUserVirtualUser
69414
asked Jan 3 at 12:07
VirtualUserVirtualUser
69414
asked Jan 3 at 12:07
VirtualUserVirtualUser
69414
asked Jan 3 at 12:07
VirtualUserVirtualUser
69414
69414
$begingroup$
I have asked this again with fixed symbols
$endgroup$
– VirtualUser
Jan 3 at 12:07
$begingroup$
You could simply edit the old question fixing those symbols (see the edit button). It is easier than retyping the whole thing as a new question.
$endgroup$
– A.Γ.
Jan 3 at 12:25
add a comment |
$begingroup$
I have asked this again with fixed symbols
$endgroup$
– VirtualUser
Jan 3 at 12:07
$begingroup$
You could simply edit the old question fixing those symbols (see the edit button). It is easier than retyping the whole thing as a new question.
$endgroup$
– A.Γ.
Jan 3 at 12:25
$begingroup$
I have asked this again with fixed symbols
$endgroup$
– VirtualUser
Jan 3 at 12:07
$begingroup$
I have asked this again with fixed symbols
$endgroup$
– VirtualUser
Jan 3 at 12:07
$begingroup$
You could simply edit the old question fixing those symbols (see the edit button). It is easier than retyping the whole thing as a new question.
$endgroup$
– A.Γ.
Jan 3 at 12:25
$begingroup$
You could simply edit the old question fixing those symbols (see the edit button). It is easier than retyping the whole thing as a new question.
$endgroup$
– A.Γ.
Jan 3 at 12:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have a mistake in your Gram-Schmidt, it should be:
$$left{frac1{sqrt{10}}begin{bmatrix} 3 \ 1 \ 0end{bmatrix}, frac1{sqrt{110}}begin{bmatrix} -1 \ 3 \ 10end{bmatrix}right}$$
To find the orthonormal basis for $V^perp$, notice that $dim V^perp = dimmathbb{R}^3 - dim V = 3-2 = 1$ so it suffices to find one vector perpendicular to $V$ and normalize it:
$$frac1{sqrt{11}}begin{bmatrix} 1 \ -3 \ 1end{bmatrix}$$
answered Jan 3 at 12:19
mechanodroidmechanodroid
27.4k62447
$endgroup$
$begingroup$
There is no mistake. It says an orthogonal basis, not orthonormal.
$endgroup$
– Shubham Johri
Jan 3 at 12:30
$begingroup$
@ShubhamJohri OP said they used Gram-Schmidt. Gram-Schmidt yields the basis from my answer.
$endgroup$
– mechanodroid
Jan 3 at 12:35
1
$begingroup$
@VirtualUser Since you only want an orthogonal basis, normalization is not necessary, but it won't get you a wrong answer. An orthonormal basis is orthogonal too
$endgroup$
– Shubham Johri
Jan 3 at 12:45
1
$begingroup$
@VirtualUser That is correct. Take $(a,b,c)in V^perp$, then $forall(x_1,x_2,x_3)in V$, we have $(a,b,c)cdot(x_1,x_2,x_3)=0$. Now take $(x_1,x_2,x_3)$ as any two vectors in $V$ to get $(a,b,c)=k(1,-3,1),kinBbb R$.
$endgroup$
– Shubham Johri
Jan 3 at 15:38
1
$begingroup$
Alternatively, it is pretty easy to observe that $V$ is the plane with the equation $x-3y+z=0$, and any vector in $V^perp$ will be parallel to the normal to the plane whose direction ratios are given by the coefficients of $x,y,z$ in the equation of the plane.
$endgroup$
– Shubham Johri
Jan 3 at 15:42
|
show 4 more comments
$begingroup$
No, it's wrong. If you have an orthogonal basis $mathcal{B} = {v_1,cdots, v_n}$ of $V$, then for each $v_i$, it is perpendicular to other $v_j$'s but not to itself. So $v_inotin V^perp$ for all $i$. By the way, you can find an orthonormal basis of $V^perp$ easily from the definition of $V$ as
$$
V = {(1,-3,1)}^perp.
$$ (It's just ${frac{1}{sqrt{11}}(1,-3,1)'}$.)
answered Jan 3 at 12:10
SongSong
11.1k628
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
You have a mistake in your Gram-Schmidt, it should be:
$$left{frac1{sqrt{10}}begin{bmatrix} 3 \ 1 \ 0end{bmatrix}, frac1{sqrt{110}}begin{bmatrix} -1 \ 3 \ 10end{bmatrix}right}$$
To find the orthonormal basis for $V^perp$, notice that $dim V^perp = dimmathbb{R}^3 - dim V = 3-2 = 1$ so it suffices to find one vector perpendicular to $V$ and normalize it:
$$frac1{sqrt{11}}begin{bmatrix} 1 \ -3 \ 1end{bmatrix}$$
answered Jan 3 at 12:19
mechanodroidmechanodroid
27.4k62447
$endgroup$
$begingroup$
There is no mistake. It says an orthogonal basis, not orthonormal.
$endgroup$
– Shubham Johri
Jan 3 at 12:30
$begingroup$
@ShubhamJohri OP said they used Gram-Schmidt. Gram-Schmidt yields the basis from my answer.
$endgroup$
– mechanodroid
Jan 3 at 12:35
1
$begingroup$
@VirtualUser Since you only want an orthogonal basis, normalization is not necessary, but it won't get you a wrong answer. An orthonormal basis is orthogonal too
$endgroup$
– Shubham Johri
Jan 3 at 12:45
1
$begingroup$
@VirtualUser That is correct. Take $(a,b,c)in V^perp$, then $forall(x_1,x_2,x_3)in V$, we have $(a,b,c)cdot(x_1,x_2,x_3)=0$. Now take $(x_1,x_2,x_3)$ as any two vectors in $V$ to get $(a,b,c)=k(1,-3,1),kinBbb R$.
$endgroup$
– Shubham Johri
Jan 3 at 15:38
1
$begingroup$
Alternatively, it is pretty easy to observe that $V$ is the plane with the equation $x-3y+z=0$, and any vector in $V^perp$ will be parallel to the normal to the plane whose direction ratios are given by the coefficients of $x,y,z$ in the equation of the plane.
$endgroup$
– Shubham Johri
Jan 3 at 15:42
|
show 4 more comments
$begingroup$
You have a mistake in your Gram-Schmidt, it should be:
$$left{frac1{sqrt{10}}begin{bmatrix} 3 \ 1 \ 0end{bmatrix}, frac1{sqrt{110}}begin{bmatrix} -1 \ 3 \ 10end{bmatrix}right}$$
To find the orthonormal basis for $V^perp$, notice that $dim V^perp = dimmathbb{R}^3 - dim V = 3-2 = 1$ so it suffices to find one vector perpendicular to $V$ and normalize it:
$$frac1{sqrt{11}}begin{bmatrix} 1 \ -3 \ 1end{bmatrix}$$
answered Jan 3 at 12:19
mechanodroidmechanodroid
27.4k62447
$endgroup$
$begingroup$
There is no mistake. It says an orthogonal basis, not orthonormal.
$endgroup$
– Shubham Johri
Jan 3 at 12:30
$begingroup$
@ShubhamJohri OP said they used Gram-Schmidt. Gram-Schmidt yields the basis from my answer.
$endgroup$
– mechanodroid
Jan 3 at 12:35
1
$begingroup$
@VirtualUser Since you only want an orthogonal basis, normalization is not necessary, but it won't get you a wrong answer. An orthonormal basis is orthogonal too
$endgroup$
– Shubham Johri
Jan 3 at 12:45
1
$begingroup$
@VirtualUser That is correct. Take $(a,b,c)in V^perp$, then $forall(x_1,x_2,x_3)in V$, we have $(a,b,c)cdot(x_1,x_2,x_3)=0$. Now take $(x_1,x_2,x_3)$ as any two vectors in $V$ to get $(a,b,c)=k(1,-3,1),kinBbb R$.
$endgroup$
– Shubham Johri
Jan 3 at 15:38
1
$begingroup$
Alternatively, it is pretty easy to observe that $V$ is the plane with the equation $x-3y+z=0$, and any vector in $V^perp$ will be parallel to the normal to the plane whose direction ratios are given by the coefficients of $x,y,z$ in the equation of the plane.
$endgroup$
– Shubham Johri
Jan 3 at 15:42
|
show 4 more comments
$begingroup$
You have a mistake in your Gram-Schmidt, it should be:
$$left{frac1{sqrt{10}}begin{bmatrix} 3 \ 1 \ 0end{bmatrix}, frac1{sqrt{110}}begin{bmatrix} -1 \ 3 \ 10end{bmatrix}right}$$
To find the orthonormal basis for $V^perp$, notice that $dim V^perp = dimmathbb{R}^3 - dim V = 3-2 = 1$ so it suffices to find one vector perpendicular to $V$ and normalize it:
$$frac1{sqrt{11}}begin{bmatrix} 1 \ -3 \ 1end{bmatrix}$$
answered Jan 3 at 12:19
mechanodroidmechanodroid
27.4k62447
$endgroup$
You have a mistake in your Gram-Schmidt, it should be:
$$left{frac1{sqrt{10}}begin{bmatrix} 3 \ 1 \ 0end{bmatrix}, frac1{sqrt{110}}begin{bmatrix} -1 \ 3 \ 10end{bmatrix}right}$$
To find the orthonormal basis for $V^perp$, notice that $dim V^perp = dimmathbb{R}^3 - dim V = 3-2 = 1$ so it suffices to find one vector perpendicular to $V$ and normalize it:
$$frac1{sqrt{11}}begin{bmatrix} 1 \ -3 \ 1end{bmatrix}$$
answered Jan 3 at 12:19
mechanodroidmechanodroid
27.4k62447
answered Jan 3 at 12:19
mechanodroidmechanodroid
27.4k62447
answered Jan 3 at 12:19
mechanodroidmechanodroid
27.4k62447
answered Jan 3 at 12:19
mechanodroidmechanodroid
27.4k62447
27.4k62447
$begingroup$
There is no mistake. It says an orthogonal basis, not orthonormal.
$endgroup$
– Shubham Johri
Jan 3 at 12:30
$begingroup$
@ShubhamJohri OP said they used Gram-Schmidt. Gram-Schmidt yields the basis from my answer.
$endgroup$
– mechanodroid
Jan 3 at 12:35
1
$begingroup$
@VirtualUser Since you only want an orthogonal basis, normalization is not necessary, but it won't get you a wrong answer. An orthonormal basis is orthogonal too
$endgroup$
– Shubham Johri
Jan 3 at 12:45
1
$begingroup$
@VirtualUser That is correct. Take $(a,b,c)in V^perp$, then $forall(x_1,x_2,x_3)in V$, we have $(a,b,c)cdot(x_1,x_2,x_3)=0$. Now take $(x_1,x_2,x_3)$ as any two vectors in $V$ to get $(a,b,c)=k(1,-3,1),kinBbb R$.
$endgroup$
– Shubham Johri
Jan 3 at 15:38
1
$begingroup$
Alternatively, it is pretty easy to observe that $V$ is the plane with the equation $x-3y+z=0$, and any vector in $V^perp$ will be parallel to the normal to the plane whose direction ratios are given by the coefficients of $x,y,z$ in the equation of the plane.
$endgroup$
– Shubham Johri
Jan 3 at 15:42
|
show 4 more comments
$begingroup$
There is no mistake. It says an orthogonal basis, not orthonormal.
$endgroup$
– Shubham Johri
Jan 3 at 12:30
$begingroup$
@ShubhamJohri OP said they used Gram-Schmidt. Gram-Schmidt yields the basis from my answer.
$endgroup$
– mechanodroid
Jan 3 at 12:35
1
$begingroup$
@VirtualUser Since you only want an orthogonal basis, normalization is not necessary, but it won't get you a wrong answer. An orthonormal basis is orthogonal too
$endgroup$
– Shubham Johri
Jan 3 at 12:45
1
$begingroup$
@VirtualUser That is correct. Take $(a,b,c)in V^perp$, then $forall(x_1,x_2,x_3)in V$, we have $(a,b,c)cdot(x_1,x_2,x_3)=0$. Now take $(x_1,x_2,x_3)$ as any two vectors in $V$ to get $(a,b,c)=k(1,-3,1),kinBbb R$.
$endgroup$
– Shubham Johri
Jan 3 at 15:38
1
$begingroup$
Alternatively, it is pretty easy to observe that $V$ is the plane with the equation $x-3y+z=0$, and any vector in $V^perp$ will be parallel to the normal to the plane whose direction ratios are given by the coefficients of $x,y,z$ in the equation of the plane.
$endgroup$
– Shubham Johri
Jan 3 at 15:42
$begingroup$
There is no mistake. It says an orthogonal basis, not orthonormal.
$endgroup$
– Shubham Johri
Jan 3 at 12:30
$begingroup$
There is no mistake. It says an orthogonal basis, not orthonormal.
$endgroup$
– Shubham Johri
Jan 3 at 12:30
$begingroup$
@ShubhamJohri OP said they used Gram-Schmidt. Gram-Schmidt yields the basis from my answer.
$endgroup$
– mechanodroid
Jan 3 at 12:35
$begingroup$
@ShubhamJohri OP said they used Gram-Schmidt. Gram-Schmidt yields the basis from my answer.
$endgroup$
– mechanodroid
Jan 3 at 12:35
1
1
$begingroup$
@VirtualUser Since you only want an orthogonal basis, normalization is not necessary, but it won't get you a wrong answer. An orthonormal basis is orthogonal too
$endgroup$
– Shubham Johri
Jan 3 at 12:45
$begingroup$
@VirtualUser Since you only want an orthogonal basis, normalization is not necessary, but it won't get you a wrong answer. An orthonormal basis is orthogonal too
$endgroup$
– Shubham Johri
Jan 3 at 12:45
1
1
$begingroup$
@VirtualUser That is correct. Take $(a,b,c)in V^perp$, then $forall(x_1,x_2,x_3)in V$, we have $(a,b,c)cdot(x_1,x_2,x_3)=0$. Now take $(x_1,x_2,x_3)$ as any two vectors in $V$ to get $(a,b,c)=k(1,-3,1),kinBbb R$.
$endgroup$
– Shubham Johri
Jan 3 at 15:38
$begingroup$
@VirtualUser That is correct. Take $(a,b,c)in V^perp$, then $forall(x_1,x_2,x_3)in V$, we have $(a,b,c)cdot(x_1,x_2,x_3)=0$. Now take $(x_1,x_2,x_3)$ as any two vectors in $V$ to get $(a,b,c)=k(1,-3,1),kinBbb R$.
$endgroup$
– Shubham Johri
Jan 3 at 15:38
1
1
$begingroup$
Alternatively, it is pretty easy to observe that $V$ is the plane with the equation $x-3y+z=0$, and any vector in $V^perp$ will be parallel to the normal to the plane whose direction ratios are given by the coefficients of $x,y,z$ in the equation of the plane.
$endgroup$
– Shubham Johri
Jan 3 at 15:42
$begingroup$
Alternatively, it is pretty easy to observe that $V$ is the plane with the equation $x-3y+z=0$, and any vector in $V^perp$ will be parallel to the normal to the plane whose direction ratios are given by the coefficients of $x,y,z$ in the equation of the plane.
$endgroup$
– Shubham Johri
Jan 3 at 15:42
|
show 4 more comments
$begingroup$
No, it's wrong. If you have an orthogonal basis $mathcal{B} = {v_1,cdots, v_n}$ of $V$, then for each $v_i$, it is perpendicular to other $v_j$'s but not to itself. So $v_inotin V^perp$ for all $i$. By the way, you can find an orthonormal basis of $V^perp$ easily from the definition of $V$ as
$$
V = {(1,-3,1)}^perp.
$$ (It's just ${frac{1}{sqrt{11}}(1,-3,1)'}$.)
answered Jan 3 at 12:10
SongSong
11.1k628
$endgroup$
add a comment |
$begingroup$
No, it's wrong. If you have an orthogonal basis $mathcal{B} = {v_1,cdots, v_n}$ of $V$, then for each $v_i$, it is perpendicular to other $v_j$'s but not to itself. So $v_inotin V^perp$ for all $i$. By the way, you can find an orthonormal basis of $V^perp$ easily from the definition of $V$ as
$$
V = {(1,-3,1)}^perp.
$$ (It's just ${frac{1}{sqrt{11}}(1,-3,1)'}$.)
answered Jan 3 at 12:10
SongSong
11.1k628
$endgroup$
add a comment |
$begingroup$
No, it's wrong. If you have an orthogonal basis $mathcal{B} = {v_1,cdots, v_n}$ of $V$, then for each $v_i$, it is perpendicular to other $v_j$'s but not to itself. So $v_inotin V^perp$ for all $i$. By the way, you can find an orthonormal basis of $V^perp$ easily from the definition of $V$ as
$$
V = {(1,-3,1)}^perp.
$$ (It's just ${frac{1}{sqrt{11}}(1,-3,1)'}$.)
answered Jan 3 at 12:10
SongSong
11.1k628
$endgroup$
No, it's wrong. If you have an orthogonal basis $mathcal{B} = {v_1,cdots, v_n}$ of $V$, then for each $v_i$, it is perpendicular to other $v_j$'s but not to itself. So $v_inotin V^perp$ for all $i$. By the way, you can find an orthonormal basis of $V^perp$ easily from the definition of $V$ as
$$
V = {(1,-3,1)}^perp.
$$ (It's just ${frac{1}{sqrt{11}}(1,-3,1)'}$.)
answered Jan 3 at 12:10
SongSong
11.1k628
answered Jan 3 at 12:10
SongSong
11.1k628
answered Jan 3 at 12:10
SongSong
11.1k628
answered Jan 3 at 12:10
SongSong
11.1k628
11.1k628
add a comment |
add a comment |
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$begingroup$
I have asked this again with fixed symbols
$endgroup$
– VirtualUser
Jan 3 at 12:07
$begingroup$
You could simply edit the old question fixing those symbols (see the edit button). It is easier than retyping the whole thing as a new question.
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– A.Γ.
Jan 3 at 12:25