A math dictionary (or a mathematics dictionary) is usually a dictionary of the definitions of common mathematical terms, formulae and examples of common diagrams and pictures for the uninitiated or those learning math.
Find values for each of the unknown variables (or at least as many as is possible for the system) that satisfy all the equations.
here are the possible answers:
A) A tridimensional vector
B) A 4D vector
C) A 5D vector
D) An scalar number
E) It is undefined
Answer: When there is no slope stated in the function of x, when y=mx+b It simply means y=0x+b Since "b" is the y intercept, your line would be a horizontal line parallel to the X-axis passing through point (0,b)
Answer: In other cases, you may need to calculate the slope first, from some other information provided. For example, if you are asked to find the equation that passes through two specified points, you can first find the slope between those two points. Then you can use this slope, and one of the points, with the slope-intercept form of the equation.
They are rather quite different: hermitian matrices usually change the norm of vector while unitary ones do not (you can convince yourself by taking the spectral decomposition: eigenvalues of unitary operators are phase factors while an hermitian matrix has real numbers as eigenvalues so they modify the norm of vectors). So unitary matrices are good "maps" whiule hermitian ones are not.
If you think about it a little bit you will be able to demonstrate the following:
for every Hilbert space except C^2 a unitary matrix cannot be hermitian and vice versa.
For the particular case H=C^2 this is not true (e.g. Pauli matrices are hermitian and unitary).
Check related links
A line with no slope is a vertical line. The slope is undefined, and cannot be represented by a real number.
A horizontal line has a slope, but the slope is zero.
Consider the "y = mx + b" form of the straight line equation. For a horizontal line the slope is zero, so y = 0x +b => y = b, which is the equation of a horizontal line. For a vertical line, there is no slope, so you can't substitute for m; the equation can't be written in the form y = mx +b. The equation of a vertical line has the form x = a.
I will give a simple example just to illustrate the idea, but all you have to do is multiply any of the equations by a constant to make them inversely additive values and add the equations.
.2x-.5y = 10
.5x+.3y = 15
.2 = 1/5 and .5 = 1/2 so if I multiply the first equation by -5 and the second by 2, we get the system:
-x+2.5y = -50
x+.6y = 30
(-x+2.5y = -50) + (x+.6y = 30) = (3.1y = -20)
Solving for the variables, we get y = -6.452 and x = 33.871
It is possible to give an example of non-linear, but I have no idea what a non-example is.
Is a non-example of linear a curve. That would be my first thought but not sure
An inconsistent equation (or system of equations) is one that has no possible solutions.
Transposing a matrix is O(n*m) where m and n are the number of rows and columns. For an n-row square matrix, this would be quadratic time-complexity.
The LACSAP fractions are a set of fractions set in a geometric pattern that are part of one of the two portfolio any International Baccalaureate - Diploma student must complete.
1 whole and 3/5
8/5 = 1 3/5 = 1.6
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
First let's be clear on the definitions.
A matrix M is orthogonal if MT=M-1
Or multiply both sides by M and you have
1) M MT=I
Where I is the identity matrix.
So our definition tells us a matrix is orthogonal if its transpose equals its inverse or if the product ( left or right) of the the matrix and its transpose is the identity.
Now we want to show why the inverse of an orthogonal matrix is also orthogonal.
Let A be orthogonal. We are assuming it is square since it has an inverse.
Now we want to show that A-1 is orthogonal.
We need to show that the inverse is equal to the transpose.
Since A is orthogonal, A=AT
Let's multiply both sides by A-1
A-1 A= A-1 AT
Or A-1 AT =I
Compare this to the definition above in 1) (M MT=I)
do you see how A-1 now fits the definition of orthogonal?
Or course we could have multiplied on the left and then we would have arrived at 2) above.
It is true that diagonalizable matrices A and B commute if and only if they are simultaneously diagonalizable. This result can be found in standard texts (e.g. Horn and Johnson, Matrix Analysis, 1999, Theorem 1.3.12.)
One direction of the if and only if proof is straightforward, but the other direction is more technical:
If A and B are diagonalizable matrices of the same order, and have the same eigenvectors, then, without loss of generality, we can write their diagonalizations as A = VDV-1 and B = VLV-1, where V is the matrix composed of the basis eigenvectors of A and B, and D and L are diagonal matrices with the corresponding eigenvalues of A and B as their diagonal elements. Since diagonal matrices commute, DL = LD. So, AB = VDV-1VLV-1 = VDLV-1 = VLDV-1 = VLV-1VDV-1 = BA.
The reverse is harder to prove, but one online proof is given below as a related link. The proof in Horn and Johnson is clear and concise.
Consider the particular case that B is the identity, I. If A = VDV-1 is a diagonalization of A, then I = VIV-1 is a diagonalization of I; i.e., A and I have the same eigenvectors.
When performing the cross product of two vectors (vector A and vector B), one of the properites of the resultant vector C is that it is perpendicular to both vectors A & B. In two dimensional space, this is not possible, because the resultant vector will be perpendicular to the plane that A & B reside in. Using the (i,j,k) unit vector notation, you could add a 0*k to each vector when doing the cross product, and the resultant vector will have zeros for the i & jcomponents, and only have k components.
Two vectors define a plane, and their cross product is always a vector along the normal to that plane, so the three vectors cannot lie in a 2D space which is a plane.
Call your matrix A, the eigenvalues are defined as the numbers e for which a nonzero vector v exists such that Av = ev. This is equivalent to requiring (A-eI)v=0 to have a non zero solution v, where I is the identity matrix of the same dimensions as A. A matrix A-eI with this property is called singular and has a zero determinant. The determinant of A-eI is a polynomial in e, which has the eigenvalues of A as roots. Often setting this polynomial to zero and solving for e is the easiest way to compute the eigenvalues of A.
One common use is in solving simultaneous linear equations.
The indefinite integral of (1/x^2)*dx is -1/x+C.
5 6 0
1 3 2
0 2 1
Graph the line containing the given pair of point and find the slope
If you are ever looking for help on your MAT 116 Quizzes/homework come check out the Essay Assassin. We help all kinds of students with their math classes each and every day. We are always professional and we meet all of your deadlines. We have the best rates in the business too! Adaptive Math, My Math Lab, none of it is a problem!
Use our live chat for instant quotes to help you out with your math!
100x100=10,000 because there are 4 0's left after so they have to be added
such as travelling through an auto rishaw or by any taxi
more examples can be for increase in prises in any field