Question

suppose a in l(v) has the same matrix with respect to every basis of v. prove that a is a scalar multiple of the indentity operator.

Answer 1

To prove that an operator A with the same matrix in every basis of V is a scalar multiple of the identity operator, we compare its matrix with respect to two different bases. The consistencies in representation imply A must be a diagonal matrix with equal diagonal entries. Hence, A is a scalar multiple of the identity operator.

Step-by-step explanation:

We will prove that an operator A in L(V), the space of all linear operators on a vector space V, must be a scalar multiple of the identity operator if it has the same matrix with respect to every basis of V.

First, we consider two bases of V, say B and C. The matrix representation of A with respect to both bases B and C is the same. Let's denote this matrix by M. Then by the definition of matrix representation of a linear transformation, we have:

• A(v) = M[v]B = M[v]C for all vectors v in V.

Here, [v]B and [v]C denote the coordinate vectors of v with respect to bases B and C respectively. Since [v]B and [v]C are generally different, the only way for M[v]B to always equal M[v]C for all v is if M is a diagonal matrix where all the diagonal entries are the same.

Therefore, M is of the form aI, where a is a scalar and I is the identity matrix. This means A is a scalar multiple of the identity operator, specifically A = aI.