Question

Math

Posted 5 months ago

which one is correct?

`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`

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Answer from Sia

Posted 5 months ago

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.

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