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An idempotent is a matrix whose square is itself. Specifically, A^{2}=A. For example the 2x2 matrix
A= 1 1
0 0
is idempotent.
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An idempotent is a matrix whose square is itself. Specifically, A^{2}=A. For example the 2x2 matrix
A= 1 1
0 0
is idempotent.
1 answer
A square matrix A is idempotent if A^2 = A. It's really simple
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yes,the histogram equalization operation is idempotent
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An idempotent matrix is a matrix which gives the same matrix if we multiply with the same.
in simple words,square of the matrix is equal to the same matrix.
if M is our matrix,then
MM=M.
then M is a idempotent matrix.
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A square matrix K is said to be idempotent if K2=K.
So yes K is a square matrix
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The assertion is true.
Let A be an idempotent matrix. Then we have A.A=A. Since A is invertible, multiplying A-1 to both sides of the equality, we get A = I.
Q. E. D
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Idempotence refers to several definitions involving mathematical operations:
- A unary operation is idempotent if applying it twice gives the same result as applying it once. For example, multiplication by 1 is idempotent as a x 1 = a x 1 x 1 = a.
- Another definition of unary idempotence is that when the operation is applied twice, it returns the original number. An example of this is the use of binary encryption in onetime pads - adding 1 to a binary digit twice (and ignoring any other digits; i.e. modulo 2) returns the original digit.
- A binary operation is idempotent if for both of the operands, the result is the same, e.g. the maximum of the set (x, x) is x.
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A singular matrix is a matrix that is not invertible. If a matrix is not invertible, then:
• The determinant of the matrix is 0.
• Any matrix multiplied by that matrix doesn't give the identity matrix.
There are a lot of examples in which a singular matrix is an idempotent matrix. For instance:
M =
[1 1]
[0 0]
Take the product of two M's to get the same M, the given!
M x M = M
So yes, SOME singular matrices are idempotent matrices! How? Let's take a 2 by 2 identity matrix for instance.
I =
[1 0]
[0 1]
I x I = I obviously.
Then, that nonsingular matrix is also idempotent!
Hope this helps!
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The same way you prove anything else. You need to be clear on what you have and what you want. You can prove it directly, by contradiction, or by induction. If you have an object which is idempotent (x = xx), you will need to use whatever definitions and theorems which apply to that object, according to what set it belongs to.
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The idempotent matrix is also called square root of a matrix. i.e.)A2=A
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Idempotent Matrix:
An idempotent matrix, A, is the specific periodic matrix (see note) where k=1, thus having the property A2=A (we can also say A.A=A).
Inverse Matrix:
Given a square matrix, A, its inverse is B if AB=BA.
Note:
A periodic matrix, A, has the property Ak+1=A where k is a positive integer. If k is the least positive integer for which Ak+1=A, then A is said to be of period k.
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An operation which when performed multiple times, has no further effect on
Its subject after the first time it is performed. - Nacolepsy
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The phrase "idempotent matrix" is an algebraic term. It is defined as a matrix that equals itself when multiplied by itself.
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Maslov. has written:
'Idempotent Analysis/Advances in Soviet Math 1051 8037 (Advances in Soviet Mathematics,)'
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Generally speaking, in mathematics, a projection is a mapping of a set (or of a mathematical structure) which is idempotent, which means that a projection is equal to its composition with itself. A projection may also refer to a mapping which has a left inverse.
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Stateless, by default. HTML's parent protocol, HTTP, is a idempotent, stateless protocol.
However, we have means using Javascript, PHP, Perl, Ruby, ASP, etc. to add a state an application. But, using only HTML, it's not possible.
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An anholonomic space, more commonly referred to as a nonholonomic space, is simply a path-dependent space.
For example, if I went to the kitchen to get a snack, I know that, regardless of what path I take to get back to my room, I will get back to my room. I could have gone outside, on the roof, to a liquor store, or wherever, but the ultimate result from adding up all those paths is that I'll be back in my room. That is because I'm in a holonomic space, or a path-independentspace. Now, if after traveling to all those locations I came back to what I thought should be my room, but instead found myself at, say, the beach, I would be in an anholonomic space, where my destination changes depending on my path taken, ie. my destination is path-dependent.
An idempotent vector doesn't really have any meaning since the concept of idempotence applies to operations. The term idempotence basically just means something that can be applied to something else over and over again without changing it, like adding zero to a real number or multiplying that number by one. That's why a vector, in and of itself, can't be idempotent. However, multiplyinga unit basis vector, ie. one that wouldn't change the magnitude or direction of another vector, to another vector would be an idemtopic operation in a vector space.
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(2+2)/(2+2)=1
(2/2)+(2/2)=2
(2*2)-(2/2)=3
(2+2+2)-2=4
(2+2)+(2/2)=5
(2*2*2)-2=6
((2-.2)/.2)+2=7
2+2+2+2=8
(22/2)+2=9
((2+2)*2)+2=10
((2-.2)/.2)+2=11
(2+2+2)*2+=12
(22/2)+2=13
(2/.2)+(2+2)=14
2+(2/2)/.2=15
(2+2)*(2+2)=16
...??????????=17
((2+2)/.2)-2=18
((2+2)-.2)/.2=19
22- √(2+2)=20
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230.
2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2 = 1073741824
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- 2,222,222
- 2 + 2 + 2 + 2 + 2 + 2 + 2 = 7(2) = 14
- 2 - 2 - 2 - 2 - 2 - 2 - 2 = 2 - 6(2) = -10
- 2*2*2*2*2*2*2 = 27 = 128
- 2/2/2/2/2/2/2 = 2/26 = 1/32
Otherwise, it's just a bunch of 2's.
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256
2, 128
2, 2, 64
2, 2, 2, 32
2, 2, 2, 2, 16
2, 2, 2, 2, 2, 8
2, 2, 2, 2, 2, 2, 4
2, 2, 2, 2, 2, 2, 2, 2
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2+2+2+2+2+2+2+2+2+2+2+2+(2x0)
=2+2+2+2+2+2+2+2+2+2+2+2+0
=24
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2 * 2 * 2 * 2 * 2 * 2 * 2, or 27.
Broken down as follows.....
128
2, 64
2, 2, 32
2, 2, 2, 16
2, 2, 2, 2, 8
2, 2, 2, 2, 2, 4
2, 2, 2, 2, 2, 2, 2
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128
2*64
2*2*32
2*2*2*16
2*2*2*2*8
2*2*2*2*2*4
2*2*2*2*2*2*2
2^7
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128
2, 64
2, 2, 32
2, 2, 2, 16
2, 2, 2, 2, 8
2, 2, 2, 2, 2, 4
2, 2, 2, 2, 2, 2, 2
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2 + 23 = 25
3 + 3 + 19 = 25
3 + 5 + 17 = 25
3 + 11 + 11 = 25
5 + 7 + 13 = 25
7 + 7 + 11 = 25
2 + 2 + 2 + 19 = 25
2 + 2 + 2 + 2 + 17 = 25
2 + 2 + 2 + 2 + 2 + 2 + 13 = 25
2 + 2 + 2 + 2 + 2 + 2 + 2 + 11 = 25
2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 7 = 25
2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 5 = 25
2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 3 = 25
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480
2, 240
2, 2, 120
2, 2, 2, 60
2, 2, 2, 2, 30
2, 2, 2, 2, 2, 15
2, 2, 2, 2, 2, 3, 5
1 answer
800
2, 400
2, 2, 200
2, 2, 2, 100
2, 2, 2, 2, 50
2, 2, 2, 2, 2, 25
2, 2, 2, 2, 2, 5, 5
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Using the distributive property of multiplication over addition, 2*[2 - 2] = 2*[2 + (-2)] = 2*2 + 2*(-2)LHS = 2*0 = 0
RHS = 4 + 2*(-2)
Therefore 2*(-2) = -4 and, by commutativity, -2*2 = -4
Next
-2*[2 - 2] = -2*[2 + (-2)] = -2*2 + -2*(-2)
LHS = -2*0 = 0
RHS = -4 + -2*(-2)
Therefore -2 times -2 = +4
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-1+(2x9)=17
1 and 9 are not twos.
How about this: (2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2) /2 = 17
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They are opposite parallel lines with points of (-2, 2) (2, 2) and (-2, -2) (2, -2)
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LCM factoring
512=2*2*2*2*2*2*2*2*2
648=2*2*2*2*3*3*3*3
lcm=2*2*2*2*2*2*2*2*2*3*3*3*3=41,472
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The answer is 0 because at the end it is multiplied by 0
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