91Ó°ÊÓ

Let \(B = \{v_1, v_2, \dots, v_n\}\) be an arbitrary basis for the vector space V, and let \(B' = \{v_1', v_2', \dots, v_n'\}\) be another basis for V. Since T has the same matrix representation with respect to any basis, let \(A = [T]_B = [T]_{B'}\). Now, we want to show that T is a scalar multiple of the identity operator.

Since A represents T with respect to basis B, we can write for all 1 ≤ i ≤ n: \[T(v_i) = \sum_{j=1}^n a_{ij} v_j\] Similarly, since A is also the matrix representation of T with respect to B', we have for all 1 ≤ i ≤ n: \[T(v_i') = \sum_{j=1}^n a_{ij} v_j'\]

Notice that for any 1 ≤ i, j ≤ n, if we take the linear combination of the equations for the action of T on the basis vector in the two basis B and B': \[\sum_{k=1}^n a_{ik} (v_k - v_k') = 0\] Since T has the same matrix representation with respect to any basis, these linear combinations of basis vectors must be equal to 0 for all i, j such that 1 ≤ i, j ≤ n.

From the previous step, we have that: \[\sum_{k=1}^n a_{ik} (v_k - v_k') = 0\] This equation implies that for each i, the vector \(\sum_{k=1}^n a_{ik} v_k\) must be orthogonal to the vector \(\sum_{k=1}^n a_{ik} v_k'\). Thus, the only way to satisfy this equation is if the matrices A are diagonal. That is, \(a_{ij} = 0\) for all i ≠ j.

Since A is a diagonal matrix, we can write it as: \[A = \begin{pmatrix} a_{11} & 0 & \dots & 0 \\ 0 & a_{22} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_{nn} \end{pmatrix}\] Now, we need to show that all diagonal elements are equal to each other in order for A to be a scalar multiple of the identity matrix. To prove this, consider \(T(v_1 + v_2)\). We have: \[T(v_1 + v_2) = a_{11}(v_1 + v_2) + a_{21} v_2\] As \(a_{21} = 0\), we must have: \[T(v_1 + v_2) = a_{11}(v_1 + v_2)\] Also, \[T(v_1 + v_2) = T(v_1) + T(v_2) = a_{11} v_1 + a_{22} v_2\] Comparing the two expressions for \(T(v_1 + v_2)\) we can conclude that \(a_{11} = a_{22}\). By using similar arguments, we can show that all diagonal elements are equal. Therefore, A is indeed a scalar multiple of the identity matrix.

Since A is a scalar multiple of the identity matrix, it represents a linear operator that is a scalar multiple of the identity operator. Hence, we can conclude that T is a scalar multiple of the identity operator.