Question

1. Show that the norm of a vector u in the vector space (V, F) is the distance from the u to the zero vector 0V. 2. Verify that the usual length of a vector in three-dimensional space satisfies the properties of a norm. 3. Verify that in Rn ||u|| = ?(?_{j=1}^n |u_i|^2) ?u = (u1, ..., un) ? Rn defines a norm.

          1. Show that the norm of a vector u in the vector space (V, F) is the distance from the u to the zero vector 0V.

2. Verify that the usual length of a vector in three-dimensional space satisfies the properties of a norm.

3. Verify that in Rn

||u|| = ?(?_{j=1}^n |u_i|^2) ?u = (u1, ..., un) ? Rn

defines a norm.

Show more…

1. Show that the norm of a vector u in the vector space (V, F) is the distance from the u to the zero vector 0V.

2. Verify that the usual length of a vector in three-dimensional space satisfies the properties of a norm.

3. Verify that in Rn

||u|| = ?(?j=1^n |ui|^2) ?u = (u1, ..., un) ? Rn

defines a norm.

Added by Brittany C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

07/08/2023

Step 1

To show that the norm of a vector $u$ in the vector space $(V,F)$ is the distance from $u$ to the zero vector $O_v$, we need to show that the norm of $u$ is equal to the distance between $u$ and $O_v$. The distance between two vectors $u$ and $v$ in a vector space Show more…

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1. Show that the norm of a vector u in the vector space (V,F) is the distance from the u to the zero vector 0V. 2. Verify that the usual length of a vector in three-dimensional space satisfies the properties of a norm. 3. Verify that in Rn ||u|| = √(∑_{j=1}^n |u_i|^2) ∀u = (u1, ..., un) ∈ Rn defines a norm.

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00:01 So here in this question we have to show that the norm of a vector, let's say vector is given to us that is vector u in a vector span of vf is the distance from u to 0 vector 0v.

00:14 So in this part, so let here we are considering that u mod is equals to the under root of summation j is from 1 to n where u of i raise to the power 2 defines a norm.

00:32 From here we are considering let's say n belongs to n where n square is greater than and equals to 0.

00:37 So u from here is equals to u1, u2 up to un.

00:41 So the value of summation j is from 1 to n of u of j raise to the power 2 is greater than and equals to 0.

00:48 This from here implies that summation j is from 1 to n of u of j raise to the power 2 is greater than and equals to 0.

00:59 If the value of u is equals to 0 which implies that the value of uj2 is equals to 0 for the value of j that is from 1, 2 to n.

01:08 The value of uj square 2 is equals to 0 again for the same values of j that is from 1 to n.

01:17 So the value of summation j is from 1 to n of uj raise to the power 2 is equals to u1 raise to the power 2 plus u2 raise to the power 2 plus up to un raise to the power 2 that from here is equals to 0.

01:33 So the value of summation j is from 1 to n of u of j raise to the power 2 is equals to 0.

01:39 Now conversely we can also say that the mod of u is equals to 0 which implies that summation j is from 1 to n of mod of u of j raise to the power 2 is equals to 0 squaring on all both side.

01:51 So we get the value of summation that is j from 1 to n of u of j raise to the power 2 is equals to 0.

01:57 Let's say this is the equation number 1.

01:59 Now we are considering that the value of summation j is from 1 to n of mod of uj raise to the power 2 is greater than mod of uj raise to the power 2 that is in from here we can say that that summation j is from 1 to n of u of j raise to the power 2 is equals to 0 for all j that is from 1, 2 to n.

02:19 So the value of u of j is equals to 0.

02:23 Now we are considering about the homogeneity of the equation where the value of u is equals to u1, u2 up to un where this is belonging to rn.

02:33 So the term from here become equals to summation j is from 1 to n of a raise to the power 2 of uj raise to the power 2...

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