Yes it is. In fact, every singular operator (read singular matrix) has 0 as an eigenvalue (the converse is also true). To see this, just note that, by definition, for any singular operator A, there exists a nonzero vector x such that Ax = 0. Since 0 = 0x we have Ax = 0x, i.e. 0 is an eigenvalue of A.
Recall that if a matrix is singular, it's determinant is zero. Let our nxn matrix be called A and let k stand for the eigenvalue. To find eigenvalues we solve the equation det(A-kI)=0for k, where I is the nxn identity matrix. (<==) Assume that k=0 is an eigenvalue. Notice that if we plug zero into this equation for k, we just get det(A)=0. This means the matrix is singluar. (==>) Assume that det(A)=0. Then as stated above we need to find solutions of the equation det(A-kI)=0. Notice that k=0 is a solution since det(A-(0)I) = det(A) which we already know is zero. Thus zero is an eigenvalue.
Given some matrix A, an eigenvector of A is a vector that, when acted on by A, will result in a scalar multiple of itself, i.e. Ax=[lambda]x, where lambda is a real scalar multiple, called an eigenvalue, and x is the eigenvector described.To find x you will normally have to find lambda first, which means solving the "characteristic equation": det(A-[lambda]I)=0, where I is the identity matrix.The derivation of the "characteristic equation" is as follows:Rearrange the equation Ax=[lambda]x -> Ax-[lambda]x=0 -> (A-[lambda]I)x=0 and then use the property from linear algebra that says if (A-[lambda]x) has an inverse, then x=0. Since this is trivial, we must instead prove that (A-[lambda]x) does not have an inverse. Because the inverse of a matrix is equal to its transpose divided by its determinant, and because you can't divide by 0, a 0 valued determinant means that the inverse can't exist. This is why we must solve det(A-[lambda]I)=0 for lambda.Once we have found lambda, we can put it in the equation Ax=[lambda]x, and it's then just a simple matter of solving the resulting linear equations.
0 has no factors.
Usually, the identity of addition property is defined to be an axiom (which only specifies the existence of zero, not uniqueness), and the zero property of multiplication is a consequence of existence of zero, existence of an additive inverse, distributivity of multiplication over addition and associativity of addition. Proof of 0 * a = 0: 0 * a = (0 + 0) * a [additive identity] 0 * a = 0 * a + 0 * a [distributivity of multiplication over addition] 0 * a + (-(0 * a)) = (0 * a + 0 * a) + (-(0 * a)) [existence of additive inverse] 0 = (0 * a + 0 * a) + (-(0 * a)) [property of additive inverses] 0 = 0 * a + (0 * a + (-(0 * a))) [associativity of addition] 0 = 0 * a + 0 [property of additive inverses] 0 = 0 * a [additive identity] A similar proof works for a * 0 = 0 (with the other distributive law if commutativity of multiplication is not assumed).
0
Recall that if a matrix is singular, it's determinant is zero. Let our nxn matrix be called A and let k stand for the eigenvalue. To find eigenvalues we solve the equation det(A-kI)=0for k, where I is the nxn identity matrix. (<==) Assume that k=0 is an eigenvalue. Notice that if we plug zero into this equation for k, we just get det(A)=0. This means the matrix is singluar. (==>) Assume that det(A)=0. Then as stated above we need to find solutions of the equation det(A-kI)=0. Notice that k=0 is a solution since det(A-(0)I) = det(A) which we already know is zero. Thus zero is an eigenvalue.
No.
Yes, it is.
define eigen value problem
No. Say your matrix is called A, then a number e is an eigenvalue of A exactly when A-eI is singular, where I is the identity matrix of the same dimensions as A. A-eI is singular exactly when (A-eI)T is singular, but (A-eI)T=AT-(eI)T =AT-eI. Therefore we can conclude that e is an eigenvalue of A exactly when it is an eigenvalue of AT.
how does ahp use eigen values and eigen vectors
If a linear transformation acts on a vector and the result is only a change in the vector's magnitude, not direction, that vector is called an eigenvector of that particular linear transformation, and the magnitude that the vector is changed by is called an eigenvalue of that eigenvector.Formulaically, this statement is expressed as Av=kv, where A is the linear transformation, vis the eigenvector, and k is the eigenvalue. Keep in mind that A is usually a matrix and k is a scalar multiple that must exist in the field of which is over the vector space in question.
The term "eigenvalue" refers to a noun which means each set of values of parameter for which differential equation has a nonzero solution. It can also refers to any number such that given matrix subtracted by the same number and multiply to the identity matrix has a zero determinant.
There's not nearly enough information here to answer. (Among other things, what the heck is "shere" supposed to be?)The general formula of an eigenvalue equation is Of = Ef. (Sorry, I can't do the normal mathematical notation here, but O is supposed to be an operator, and f is a function of some kind... E is, of course, the eigenvalue). If you know how to do differential equations, the rest is easy (assuming you actually know what O and f are). If you don't, you're not going to understand the answer anyway.
Yes, do write. That's what you always have to do when you have got a homework-program.
I'm seeking the answer too. What's the meaning of the principal eigenvector of an MI matrix?
This is the definition of eigenvectors and eigenvalues according to Wikipedia:Specifically, a non-zero column vector v is a (right) eigenvector of a matrix A if (and only if) there exists a number λ such that Av = λv. The number λ is called the eigenvalue corresponding to that vector. The set of all eigenvectors of a matrix, each paired with its corresponding eigenvalue, is called the eigensystemof that matrix