Suppose V is a real vector space of finite dimension n. Call the set of matrices from V into itself M(V). Let T be in M(V). Consider the two subspaces U = {X ∈ M(V) : TX = XT} and W = {TX − XT : X ∈ M(V)}. Which of the following must be TRUE? I. If V has a basis containing only eigenvectors of T then U = M(V). II. dim (U) + dim (W) = n^2 III. dim (U)

Question

Suppose V is a real vector space of finite dimension n. Call the set of matrices from V into itself M(V). Let T be in M(V). Consider the two subspaces U = {X ∈ M(V) : TX = XT} and W = {TX − XT : X ∈ M(V)}. Which of the following must be TRUE? I. If V has a basis containing only eigenvectors of T then U = M(V). II. dim (U) + dim (W) = n^2 III. dim (U) < n

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To address the first statement, we consider the case where V has a basis consisting of eigenvectors of T. ‖ ‖ step 2 ⋮ If every vector in the basis is an eigenvector of T, then every matrix that commutes with T must be a diagonal matrix with respect to this basis. ‖ ‖ step 3 ⋮ Since all diagonal matrices commute with T, U contains all such matrices, and thus U = M(V). ‖ ∻Answer∻ ⚹ Statement I is TRUE. ⚹ ∻Key Concept∻ ⚹Commutant of a linear operator with a diagonalizable matrix⚹ ∻Explanation∻ ⚹If a linear operator T has a basis of eigenvectors, then every diagonal matrix with respect to this basis commutes with T, making the commutant the entire space of matrices.⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To address the second statement, we use the fact that the space of matrices M(V) has dimension n^2. ‖ ‖ step 2 ⋮ The dimension of the sum of two subspaces U and W is given by dim(U) + dim(W) - dim(U ∩ W). ‖ ‖ step 3 ⋮ Since U and W are defined such that U ∩ W = {0}, the intersection contains only the zero matrix. ‖ ‖ step 4 ⋮ Therefore, dim(U ∩ W) = 0, and dim(U) + dim(W) = n^2. ‖ ∻Answer∻ ⚹ Statement II is TRUE. ⚹ ∻Key Concept∻ ⚹Dimension of a sum of subspaces⚹ ∻Explanation∻ ⚹The dimension of the sum of two subspaces is the sum of their dimensions minus the dimension of their intersection. Since U and W intersect only at the zero matrix, their dimensions add up to n^2.⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To address the third statement, we consider the dimension of U. ‖ ‖ step 2 ⋮ Since U is a subspace of M(V), its dimension cannot exceed that of M(V). ‖ ‖ step 3 ⋮ The dimension of M(V) is n^2, which is greater than n for all n > 1. ‖ ‖ step 4 ⋮ Therefore, dim(U) ≤ n^2, and since n^2 > n, it follows that dim(U) cannot be less than n unless n = 1. ‖ ∻Answer∻ ⚹ Statement III is FALSE. ⚹ ∻Key Concept∻ ⚹Dimension of subspaces⚹ ∻Explanation∻ ⚹The dimension of a subspace U of M(V) cannot be less than n unless n = 1, because the dimension of M(V) is n^2, which is always greater than n for n > 1.⚹◊What is the dimension of the subspace $U$ in terms of the operator $T$ in $M(V)$?⍭ Generate me a similar question◊

Sia