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Is eigenvalue of any operator must be real?

Updated: 8/17/2019
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No.

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Q: Is eigenvalue of any operator must be real?
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Is 0 an eigenvalue?

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.


What is meant by the word eigenvalue?

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.


What is a Hermitian operator?

A Hermitian operator is any linear operator for which the following equality property holds: integral from minus infinity to infinity of (f(x)* A^g(x))dx=integral from minus infinity to infinity of (g(x)A*^f(x)*)dx, where A^ is the hermitian operator, * denotes the complex conjugate, and f(x) and g(x) are functions. The eigenvalues of hermitian operators are real and their eigenfunctions are orthonormal.


How do you perform any operation by giving any operator?

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How can you create a new operator through operator overloading?

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How can you can create a new operator through operator overloading?

You cannot create any new operators in C++. You can only overload the existing ones (although some, such as sizeof, new and delete cannot be overloaded). The only way to create a new operator is to implement it as a standard function with a named identifier. For instance, sqrt() is the standard library function that provides the square root operator, for which no real operator exists.


Real square roots of -196?

None. The square of any real number must be non-negative.


Is there is any difference between -algebra and operator algebra?

Only the word "operator"


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