Mixing
Find orthogonal projection of $ [n,0,0,…,0]^T$ on subspace $V$
$begingroup$
$n>1$ Given is $$V = left{ vec{x} in mathbb R^n : x_1+x_2 + ... + x_n = 0 right} $$
a) Find orthogonal basis of $V^{perp} $
b) Find orthogonal projection $vec{x} = [n,0,0,...,0]^T$ on subspace $V$
If it comes to a)
$$dim V^{perp} = n - dim V = dim V^{perp} = n - n + 1 = 1$$ So $V^{perp} = span$ one_vector_perpendicular_to_v
Put $[1,1,1,...,1,1]^T$ - it is perpendicular to $V$
Let's start Gram–Schmidt process - but we have $1$ vector so $u_1 = [1,1,1,...,1,1]^T$ = orthogonal basis of $V^{perp}$
b) It seems to be very interesting and hard. I found basis of $V$:
$$[-1,1,0,0,...,0] = vec{v_1}$$
$$[-1,0,1,0,...,0] = vec{v_2}$$
$$[-1,0,0,1,...,0] = vec{v_3}$$
$$...$$
$$[-1,0,0,0,...,1] = vec{v_{n-1}}$$
Now I start Gram–Schmidt process
$$u_1 = v_1 $$
$$u_2 = v_2 - frac{1}{2} cdot v_1$$
$$u_ 3 = v_3 - frac{1}{4} cdot v_2 + frac{1}{8} cdot v_1 $$
$$u_4 = v_4 - frac{7}{16} cdot v_3 + frac{7}{64} cdot v_2 - frac{7}{64} cdot v_1$$
I don't even know if I don't take mistake.
Moreover the calculations getting harder and harder and I still don't see any regular sequence in it. Can somebody help me with this task?
linear-algebra orthogonality
asked Jan 3 at 14:10
VirtualUserVirtualUser
59212
$endgroup$
add a comment |
$begingroup$
$n>1$ Given is $$V = left{ vec{x} in mathbb R^n : x_1+x_2 + ... + x_n = 0 right} $$
a) Find orthogonal basis of $V^{perp} $
b) Find orthogonal projection $vec{x} = [n,0,0,...,0]^T$ on subspace $V$
If it comes to a)
$$dim V^{perp} = n - dim V = dim V^{perp} = n - n + 1 = 1$$ So $V^{perp} = span$ one_vector_perpendicular_to_v
Put $[1,1,1,...,1,1]^T$ - it is perpendicular to $V$
Let's start Gram–Schmidt process - but we have $1$ vector so $u_1 = [1,1,1,...,1,1]^T$ = orthogonal basis of $V^{perp}$
b) It seems to be very interesting and hard. I found basis of $V$:
$$[-1,1,0,0,...,0] = vec{v_1}$$
$$[-1,0,1,0,...,0] = vec{v_2}$$
$$[-1,0,0,1,...,0] = vec{v_3}$$
$$...$$
$$[-1,0,0,0,...,1] = vec{v_{n-1}}$$
Now I start Gram–Schmidt process
$$u_1 = v_1 $$
$$u_2 = v_2 - frac{1}{2} cdot v_1$$
$$u_ 3 = v_3 - frac{1}{4} cdot v_2 + frac{1}{8} cdot v_1 $$
$$u_4 = v_4 - frac{7}{16} cdot v_3 + frac{7}{64} cdot v_2 - frac{7}{64} cdot v_1$$
I don't even know if I don't take mistake.
Moreover the calculations getting harder and harder and I still don't see any regular sequence in it. Can somebody help me with this task?
linear-algebra orthogonality
asked Jan 3 at 14:10
VirtualUserVirtualUser
59212
$endgroup$
$begingroup$
Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
$endgroup$
– A.Γ.
Jan 3 at 14:15
$begingroup$
I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
$endgroup$
– VirtualUser
Jan 3 at 14:18
$begingroup$
You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
$endgroup$
– amd
Jan 3 at 21:40
add a comment |
$begingroup$
$n>1$ Given is $$V = left{ vec{x} in mathbb R^n : x_1+x_2 + ... + x_n = 0 right} $$
a) Find orthogonal basis of $V^{perp} $
b) Find orthogonal projection $vec{x} = [n,0,0,...,0]^T$ on subspace $V$
If it comes to a)
$$dim V^{perp} = n - dim V = dim V^{perp} = n - n + 1 = 1$$ So $V^{perp} = span$ one_vector_perpendicular_to_v
Put $[1,1,1,...,1,1]^T$ - it is perpendicular to $V$
Let's start Gram–Schmidt process - but we have $1$ vector so $u_1 = [1,1,1,...,1,1]^T$ = orthogonal basis of $V^{perp}$
b) It seems to be very interesting and hard. I found basis of $V$:
$$[-1,1,0,0,...,0] = vec{v_1}$$
$$[-1,0,1,0,...,0] = vec{v_2}$$
$$[-1,0,0,1,...,0] = vec{v_3}$$
$$...$$
$$[-1,0,0,0,...,1] = vec{v_{n-1}}$$
Now I start Gram–Schmidt process
$$u_1 = v_1 $$
$$u_2 = v_2 - frac{1}{2} cdot v_1$$
$$u_ 3 = v_3 - frac{1}{4} cdot v_2 + frac{1}{8} cdot v_1 $$
$$u_4 = v_4 - frac{7}{16} cdot v_3 + frac{7}{64} cdot v_2 - frac{7}{64} cdot v_1$$
I don't even know if I don't take mistake.
Moreover the calculations getting harder and harder and I still don't see any regular sequence in it. Can somebody help me with this task?
linear-algebra orthogonality
asked Jan 3 at 14:10
VirtualUserVirtualUser
59212
$endgroup$
$n>1$ Given is $$V = left{ vec{x} in mathbb R^n : x_1+x_2 + ... + x_n = 0 right} $$
a) Find orthogonal basis of $V^{perp} $
b) Find orthogonal projection $vec{x} = [n,0,0,...,0]^T$ on subspace $V$
If it comes to a)
$$dim V^{perp} = n - dim V = dim V^{perp} = n - n + 1 = 1$$ So $V^{perp} = span$ one_vector_perpendicular_to_v
Put $[1,1,1,...,1,1]^T$ - it is perpendicular to $V$
Let's start Gram–Schmidt process - but we have $1$ vector so $u_1 = [1,1,1,...,1,1]^T$ = orthogonal basis of $V^{perp}$
b) It seems to be very interesting and hard. I found basis of $V$:
$$[-1,1,0,0,...,0] = vec{v_1}$$
$$[-1,0,1,0,...,0] = vec{v_2}$$
$$[-1,0,0,1,...,0] = vec{v_3}$$
$$...$$
$$[-1,0,0,0,...,1] = vec{v_{n-1}}$$
Now I start Gram–Schmidt process
$$u_1 = v_1 $$
$$u_2 = v_2 - frac{1}{2} cdot v_1$$
$$u_ 3 = v_3 - frac{1}{4} cdot v_2 + frac{1}{8} cdot v_1 $$
$$u_4 = v_4 - frac{7}{16} cdot v_3 + frac{7}{64} cdot v_2 - frac{7}{64} cdot v_1$$
I don't even know if I don't take mistake.
Moreover the calculations getting harder and harder and I still don't see any regular sequence in it. Can somebody help me with this task?
linear-algebra orthogonality
linear-algebra orthogonality
asked Jan 3 at 14:10
VirtualUserVirtualUser
59212
asked Jan 3 at 14:10
VirtualUserVirtualUser
59212
asked Jan 3 at 14:10
VirtualUserVirtualUser
59212
asked Jan 3 at 14:10
VirtualUserVirtualUser
59212
asked Jan 3 at 14:10
VirtualUserVirtualUser
59212
59212
$begingroup$
Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
$endgroup$
– A.Γ.
Jan 3 at 14:15
$begingroup$
I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
$endgroup$
– VirtualUser
Jan 3 at 14:18
$begingroup$
You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
$endgroup$
– amd
Jan 3 at 21:40
add a comment |
$begingroup$
Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
$endgroup$
– A.Γ.
Jan 3 at 14:15
$begingroup$
I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
$endgroup$
– VirtualUser
Jan 3 at 14:18
$begingroup$
You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
$endgroup$
– amd
Jan 3 at 21:40
$begingroup$
Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
$endgroup$
– A.Γ.
Jan 3 at 14:15
$begingroup$
Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
$endgroup$
– A.Γ.
Jan 3 at 14:15
$begingroup$
I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
$endgroup$
– VirtualUser
Jan 3 at 14:18
$begingroup$
I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
$endgroup$
– VirtualUser
Jan 3 at 14:18
$begingroup$
You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
$endgroup$
– amd
Jan 3 at 21:40
$begingroup$
You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
$endgroup$
– amd
Jan 3 at 21:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us call $vec u$ the projection of $vec{x} = [n,0,0,...,0]^T$ on $V$.
$vec x - vec u$ is orthogonal to $V$, i.e. $vec x - vec u = a vec v$ with $v = (1, dots, 1)^T$ as you already showed
Therefore $vec x = a vec v + vec u$ with $vec u$ satisfying $sum_i u_i = 0$
Then,
$$ sum_i (x_i - a) = 0 $$
And finally $a = 1$ and
$$vec u = (n-1, -1, dots, -1) ^T $$
answered Jan 3 at 14:33
DamienDamien
58214
$endgroup$
$begingroup$
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
$endgroup$
– VirtualUser
Jan 3 at 14:36
$begingroup$
Great, it is a loooot of simpler way than my idea, thanks
$endgroup$
– VirtualUser
Jan 3 at 14:39
1
$begingroup$
@VirtualUser Corrected. You read my answer before I had time to check it!
$endgroup$
– Damien
Jan 3 at 14:39
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let us call $vec u$ the projection of $vec{x} = [n,0,0,...,0]^T$ on $V$.
$vec x - vec u$ is orthogonal to $V$, i.e. $vec x - vec u = a vec v$ with $v = (1, dots, 1)^T$ as you already showed
Therefore $vec x = a vec v + vec u$ with $vec u$ satisfying $sum_i u_i = 0$
Then,
$$ sum_i (x_i - a) = 0 $$
And finally $a = 1$ and
$$vec u = (n-1, -1, dots, -1) ^T $$
answered Jan 3 at 14:33
DamienDamien
58214
$endgroup$
$begingroup$
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
$endgroup$
– VirtualUser
Jan 3 at 14:36
$begingroup$
Great, it is a loooot of simpler way than my idea, thanks
$endgroup$
– VirtualUser
Jan 3 at 14:39
1
$begingroup$
@VirtualUser Corrected. You read my answer before I had time to check it!
$endgroup$
– Damien
Jan 3 at 14:39
add a comment |
$begingroup$
Let us call $vec u$ the projection of $vec{x} = [n,0,0,...,0]^T$ on $V$.
$vec x - vec u$ is orthogonal to $V$, i.e. $vec x - vec u = a vec v$ with $v = (1, dots, 1)^T$ as you already showed
Therefore $vec x = a vec v + vec u$ with $vec u$ satisfying $sum_i u_i = 0$
Then,
$$ sum_i (x_i - a) = 0 $$
And finally $a = 1$ and
$$vec u = (n-1, -1, dots, -1) ^T $$
answered Jan 3 at 14:33
DamienDamien
58214
$endgroup$
$begingroup$
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
$endgroup$
– VirtualUser
Jan 3 at 14:36
$begingroup$
Great, it is a loooot of simpler way than my idea, thanks
$endgroup$
– VirtualUser
Jan 3 at 14:39
1
$begingroup$
@VirtualUser Corrected. You read my answer before I had time to check it!
$endgroup$
– Damien
Jan 3 at 14:39
add a comment |
$begingroup$
Let us call $vec u$ the projection of $vec{x} = [n,0,0,...,0]^T$ on $V$.
$vec x - vec u$ is orthogonal to $V$, i.e. $vec x - vec u = a vec v$ with $v = (1, dots, 1)^T$ as you already showed
Therefore $vec x = a vec v + vec u$ with $vec u$ satisfying $sum_i u_i = 0$
Then,
$$ sum_i (x_i - a) = 0 $$
And finally $a = 1$ and
$$vec u = (n-1, -1, dots, -1) ^T $$
answered Jan 3 at 14:33
DamienDamien
58214
$endgroup$
Let us call $vec u$ the projection of $vec{x} = [n,0,0,...,0]^T$ on $V$.
$vec x - vec u$ is orthogonal to $V$, i.e. $vec x - vec u = a vec v$ with $v = (1, dots, 1)^T$ as you already showed
Therefore $vec x = a vec v + vec u$ with $vec u$ satisfying $sum_i u_i = 0$
Then,
$$ sum_i (x_i - a) = 0 $$
And finally $a = 1$ and
$$vec u = (n-1, -1, dots, -1) ^T $$
answered Jan 3 at 14:33
DamienDamien
58214
edited Jan 3 at 14:38
answered Jan 3 at 14:33
DamienDamien
58214
answered Jan 3 at 14:33
DamienDamien
58214
answered Jan 3 at 14:33
DamienDamien
58214
58214
$begingroup$
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
$endgroup$
– VirtualUser
Jan 3 at 14:36
$begingroup$
Great, it is a loooot of simpler way than my idea, thanks
$endgroup$
– VirtualUser
Jan 3 at 14:39
1
$begingroup$
@VirtualUser Corrected. You read my answer before I had time to check it!
$endgroup$
– Damien
Jan 3 at 14:39
add a comment |
$begingroup$
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
$endgroup$
– VirtualUser
Jan 3 at 14:36
$begingroup$
Great, it is a loooot of simpler way than my idea, thanks
$endgroup$
– VirtualUser
Jan 3 at 14:39
1
$begingroup$
@VirtualUser Corrected. You read my answer before I had time to check it!
$endgroup$
– Damien
Jan 3 at 14:39
$begingroup$
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
$endgroup$
– VirtualUser
Jan 3 at 14:36
$begingroup$
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
$endgroup$
– VirtualUser
Jan 3 at 14:36
$begingroup$
Great, it is a loooot of simpler way than my idea, thanks
$endgroup$
– VirtualUser
Jan 3 at 14:39
$begingroup$
Great, it is a loooot of simpler way than my idea, thanks
$endgroup$
– VirtualUser
Jan 3 at 14:39
1
1
$begingroup$
@VirtualUser Corrected. You read my answer before I had time to check it!
$endgroup$
– Damien
Jan 3 at 14:39
$begingroup$
@VirtualUser Corrected. You read my answer before I had time to check it!
$endgroup$
– Damien
Jan 3 at 14:39
add a comment |
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$begingroup$
Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
$endgroup$
– A.Γ.
Jan 3 at 14:15
$begingroup$
I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
$endgroup$
– VirtualUser
Jan 3 at 14:18
$begingroup$
You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
$endgroup$
– amd
Jan 3 at 21:40