In a three-dimensional vector space consider an operator M in 0 iv2 0 0 -iv2 0 2 orthonormal basis {|1), |2), |3)} which is written as M= 3 Question : 1) Is M hermit and unitarity? 2) Calculate the eigenvalues and eigenvectors of M in the normalized expansion of the basis.

Q: 3 9 A -2 -3 1 9 Find an orthonormal basis of the image of A.

A: Orthonormal basis of a vector space V is a basis of a basis for V whose vectors are orthonormal. Two…

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A: For any query with the solution please ask in the comment box.

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A: Dear Bartleby student, according to our guidelines we can answer only three subparts, or first…

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A: first we will find the eigenvalues and eigenvectors of the given matrix A then we will form matrix Q…

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Q: 1. In the {li), j), Ik)} representation, the normalized basis vectors are chosen as li) = (1). 1) =…

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Q: 5. Consider the 4 x 5 matrix A = [u₁|u2|u3|u4| u5], where the columns are U₁= 2 , U2 = 5 Uz = -1 -2…

A: The  4×5  matrix  A= u1 | u2 | u3 | u4 | u5   where  u1=1234 , u2=5102 , u3=   2-1-2   3 , u4=  1…

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Q: 2. Apply the Gramm-Schmidt orthogonalization process to the vectors [1, 3, 2] and [1, 0, 1] in order…

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Q: If b1, b2 form an orthonormal basis for C and x is any vector in C, then O x4- x bị x b2 0 O x+ x bị…

A: We will find out the required value.

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A: We show that A is idempotent matrix   For this we have A2 =A .. And the eigen values of idempotent…

Q: (1) Use Gram-Schmidt process to find an orthonormal basis for the span of the following vectors:

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Q: Consider a physical system whose three-dimensional state space is spanned by the orthonormal basis…

A: Given Data: H^=hω03i0-i30002 ,  B^=b07i1-i-i71+i1+i1-i6,D^ =002α02α 02α0-3α

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A: GRAM-SCHMIDT METHOD The Gram-Schmidt process (or procedure) is a sequence of operations that allow…

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A: v={A in M22 : A is upper triangular}

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Q: {} W = span D. -C If A is an orthonormal basis for VĀ w, then A= a= ,b= B= C= %3D are all positive…

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Q: Consider a physical system whose three-dimensional state space is spanned by the orthonormal basis…

A: Given Data: H^ =hω03i0-i30002 , B^ =b07i1-i-i71+i1+i1-i6D^= 002α02α02α0-3αψ0=e1ψ0e2ψ0e3ψ0=236