Find orthogonal projection of $ [n,0,0,…,0]^T$ on subspace $V$
$n>1$ 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 yesterday
VirtualUser
44711
add a comment |
$n>1$ 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 yesterday
VirtualUser
44711
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 $V$.
– A.Γ.
yesterday
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.
– VirtualUser
yesterday
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.
– amd
yesterday
add a comment |
$n>1$ 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 yesterday
VirtualUser
44711
$n>1$ 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 yesterday
VirtualUser
44711
asked yesterday
VirtualUser
44711
asked yesterday
VirtualUser
44711
asked yesterday
VirtualUser
44711
asked yesterday
VirtualUser
44711
44711
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 $V$.
– A.Γ.
yesterday
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.
– VirtualUser
yesterday
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.
– amd
yesterday
add a comment |
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 $V$.
– A.Γ.
yesterday
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.
– VirtualUser
yesterday
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.
– amd
yesterday
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 $V$.
– A.Γ.
yesterday
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 $V$.
– A.Γ.
yesterday
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.
– VirtualUser
yesterday
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.
– VirtualUser
yesterday
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.
– amd
yesterday
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.
– amd
yesterday
add a comment |
1 Answer
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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 yesterday
Damien
50714
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
– VirtualUser
yesterday
Great, it is a loooot of simpler way than my idea, thanks
– VirtualUser
yesterday
1
@VirtualUser Corrected. You read my answer before I had time to check it!
– Damien
yesterday
add a comment |
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1 Answer
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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 yesterday
Damien
50714
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
– VirtualUser
yesterday
Great, it is a loooot of simpler way than my idea, thanks
– VirtualUser
yesterday
1
@VirtualUser Corrected. You read my answer before I had time to check it!
– Damien
yesterday
add a comment |
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 yesterday
Damien
50714
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
– VirtualUser
yesterday
Great, it is a loooot of simpler way than my idea, thanks
– VirtualUser
yesterday
1
@VirtualUser Corrected. You read my answer before I had time to check it!
– Damien
yesterday
add a comment |
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 yesterday
Damien
50714
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 yesterday
Damien
50714
edited yesterday
answered yesterday
Damien
50714
answered yesterday
Damien
50714
answered yesterday
Damien
50714
50714
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
– VirtualUser
yesterday
Great, it is a loooot of simpler way than my idea, thanks
– VirtualUser
yesterday
1
@VirtualUser Corrected. You read my answer before I had time to check it!
– Damien
yesterday
add a comment |
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
– VirtualUser
yesterday
Great, it is a loooot of simpler way than my idea, thanks
– VirtualUser
yesterday
1
@VirtualUser Corrected. You read my answer before I had time to check it!
– Damien
yesterday
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
– VirtualUser
yesterday
the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
– VirtualUser
yesterday
Great, it is a loooot of simpler way than my idea, thanks
– VirtualUser
yesterday
Great, it is a loooot of simpler way than my idea, thanks
– VirtualUser
yesterday
1
1
@VirtualUser Corrected. You read my answer before I had time to check it!
– Damien
yesterday
@VirtualUser Corrected. You read my answer before I had time to check it!
– Damien
yesterday
add a comment |
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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 $V$.
– A.Γ.
yesterday
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.
– VirtualUser
yesterday
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.
– amd
yesterday