Question 1183383: Let A ∈ Mn×n(F), and let T : Mn×n(F) → Mn×n(F) be the linear transformation X → AXA, for X ∈
Mn×n(F). What is the rank of T? Show that there exists a matrix B such that ABA = A.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's analyze the rank of the linear transformation T and the existence of matrix B.
**1. Rank of T:**
The rank of T depends crucially on the matrix A. There isn't a single, universally applicable rank for T without knowing A. Here's why:
* **Case 1: A is invertible:** If A is invertible (i.e., det(A) ≠ 0), then the rank of T is n². In this case, for any matrix Y in Mₙₓₙ(F), there exists a unique X such that AXA = Y. Specifically, X = A⁻¹YA⁻¹. Thus, T is surjective (onto), and since the domain and codomain have the same dimension (n²), T is also injective (one-to-one). Therefore, T is an isomorphism, and its rank is n².
* **Case 2: A is not invertible:** If A is not invertible, then its rank is less than n. The rank of T will be less than or equal to the rank of A. Let r be the rank of A (r < n). Then the rank of T will be at most r². It might be strictly less than r² because the transformation involves A on both the left and right.
* Example: If A is the zero matrix, then T(X) = AXA = 0 for all X. The rank of T is 0.
* Another example: If A is a projection matrix (A²=A) of rank r, the dimension of the image of T is at most r.
In summary, the rank of T can be anywhere from 0 to n², and it depends directly on the rank of A.
**2. Existence of Matrix B such that ABA = A:**
The statement that there exists a matrix B such that ABA = A is **true**, and it's a consequence of the properties of the Moore-Penrose pseudoinverse, denoted by A⁺. However, we can provide a more elementary argument for this specific case.
* **If A is invertible:** Then we can simply choose B = A⁻¹. Then ABA = AA⁻¹A = IA = A.
* **If A is not invertible:** This is where the argument gets a bit more subtle. We can use the rank factorization of A. Any matrix A of rank r can be factored as A = CR, where C is an n × r matrix of rank r, and R is an r × n matrix of rank r.
We want to find B such that ABA = A, or (CR)B(CR) = CR. If we can find a matrix B' such that RB'C = I (the r × r identity matrix), then we have (CR)B(CR) = C(RB'C)R = CIR = CR = A.
We can always find such a matrix B'. Since C has rank r, its columns are linearly independent, so C has a left inverse L such that LC = I. Since R has rank r, its rows are linearly independent, so R has a right inverse M such that RM = I. Then we can choose B' = MRCL. Then RB'C = RMRCLC = I*I = I.
Therefore, B = B' = MRCL is a matrix such that ABA = A.
In conclusion, such a matrix B always exists, regardless of whether A is invertible. When A is invertible, B is simply A⁻¹. When A is not invertible, a suitable B can be constructed by using the rank factorization of A.
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