Let's consider an N dimensional vector where each coordinate takes the value 1. For example, for N=5 we have: (1,1,1,1,1). Does this type of vector have a name in the literature? Perhaps "unary?". Also, are there any conventions on how to refer to it in terms of notation? (e.g. 1^T?)

Hana Buck 2022-09-25 Answered
Name of a vector of 1s?
Let's consider an N dimensional vector where each coordinate takes the value 1. For example, for N = 5 we have: ( 1 , 1 , 1 , 1 , 1 )
Does this type of vector have a name in the literature? Perhaps "unary?". Also, are there any conventions on how to refer to it in terms of notation? (e.g. 1 T ?)
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Kellen Blackburn
Answered 2022-09-26 Author has 8 answers
It's not quite common enough to have a standard notation, but a reasonably well-accepted notation would be something like 1 n = ( 1 , 1 , , 1 ) R n , and if you needed a column vector then you'd write 1 n . It may sometimes be called the 1-vector of size n or a size n vector of 1s.
As such, it's the kind of thing that when you use it you would probably be best off defining it explicitly so that it's clear what you're doing with it.
Did you like this example?
Subscribe for all access
Daniella Reyes
Answered 2022-09-27 Author has 3 answers
I'm not aware of any conventional terminology. However, vector of ones is pretty compact and seems to get the job done. Similarly, I'm not aware of any special notation. Were I using it, I'd just define some notation. For example, let a R d denote the vector of ones (ie, a = ( 1 , , 1 )).
Of course, depending on the context, there could be special notation. That could represent the identity in something like the multiplicative group Z × × Z × . Then the standard notation for an identity element would be used.
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2021-05-17
Find the scalar and vector projections of b onto a.
a=(4,7,4),b=(3,1,1)
asked 2021-02-11
Let F be a fixed 3x2 matrix, and let H be the set of all matrices A in M2×4 with the property that FA = 0 (the zero matrix in M3×4). Determine if H is a subspace of M2×4
asked 2021-05-29

Find the vector and parametric equations for the line segment connecting P to Q.
P(0, - 1, 1), Q(1/2, 1/3, 1/4)

asked 2021-05-29
Which of the following expressions are meaningful? Which are meaningless? Explain.
a) (ab)c
(ab)c has ? because it is the dot product of ?.
b) (ab)c
(ab)c has ? because it is a scalar multiple of ?.
c) |a|(bc)
|a|(bc) has ? because it is the product of ?.
d) a(b+c)
a(b+c) has ? because it is the dot product of ?.
e) ab+c
ab+c has ? because it is the sum of ?.
f) |a|(b+c)
|a|(b+c) has ? because it is the dot product of ?.
asked 2022-07-18
Given a vector-valued function defined by
r ( t ) = ( t 3 + 1 t 3 + 1 2 t + 1 )
Let T denote the tangent to the curve at A = ( 2 , 2 , 3 )
Then find the equation of the line L passing through the point u=(1,−1,2),parallel to the plane 2x+y+z=0 which intersects the tangent line T
The equation of the line is in the form:
L = u + v s
Since the line is parallel to the plane,we conclude that the direction of the line is the same as the plane's,let n = ( 2 , 1 , 1 ) denote the normal to the plane,then n × v = ( 1 , 1 , 1 ), which implies:
( v 2 v 3 , 2 v 3 v 1 , v 1 2 v 2 ) = 1
So:
(1) v 2 v 3 = 1
2 v 3 v 1 = 1
(2) v 1 2 v 2 = 1
moreover n v = 0,which implies:
(3) 2 v 1 + v 2 + v 3 = 0
Substituting (1) and (2) into (3) follows:
v 1 = 2 / 3
v 2 = 1 / 6
v 3 = 7 / 6
So the equation of the line is :
L = ( 1 , 1 , 2 ) + ( 2 3 , 1 6 , 7 6 ) s
With the parametric equation :
x = 2 3 s + 1
y = 1 6 s 1
z = 7 6 s + 2
Since the line intersects the tangent line to the curve at a point with coordinate (2,2,3),we see that s=3/2,however substituting this to the y and z we don't get y=2 and z=3 respectively,so where was I wrong?
asked 2022-08-20
L 1 is defined by x + 1 2 = y 3 3 = 1 z
So we can write this as ( ( 2 t 1 ) , ( 3 t + 3 ) , ( 1 t ) ) = r
L 2 passes through (5,4,2) and intersects with L 1 at right angles.
I am asked to determine the point of intersections between those two lines.
I have tried to find the dot product of the directional vector of L1 and another directional vector (x,y,z) So I got this 2x+3y−z=0 and said that x=3,y=2 and z=12
So now L 2 has the following equation r=((5+3t),(4+2t),(2+12t)) However when i equate L 1 and L 2 to find t. The equations aren't satisfied, meaning that t isn't common.
asked 2022-07-20
I want to find the vertical or you could say perpendicular component of a on b
Now I know that it can be found out using a ( a b | b | ) b
However I wanted to know why it cannot be found out using this method I tried. What is the flaw in it?
a × b = | a | | b | sin θ n ^ Now n ^ should be equal to ( a × b ) | a × b |
Using this I can write the expression as a × b = | a | | b | sin θ × ( a × b ) | a × b | which on simplifying gives me
| a × b | | b | = | a | sin θ which I believe should be the perpendicular component
Now I'm probably doing something really stupid but I can't really understand where am I going wrong ?

New questions