Problem 5 - Algebra

Question

Let \(M\) be an \(n \times n\) matrix over a field \(k\). Assume that \(\operatorname{tr}(M X)=0\) for all \(n \times n\) matrices \(X\) in \(k\). Show that \(M=O\).

Step-by-Step Solution

Answer

The given condition states that \(\operatorname{tr}(MX) = 0\) for all \(n \times n\) matrices \(X\) over field \(k\). We can use this condition by constructing specific matrices \(X_{ij}\) with only one non-zero entry of 1 at position \((i, j)\) and 0 everywhere else. When we multiply \(M\) with these matrices, we get \(MX_{ij}\) with entries equal to the \(ij\)-th entry of \(M\). Since the trace of \(MX_{ij}\) is also zero, this implies that every entry of \(M\) must be zero. Therefore, \(M = O\), the zero matrix.

Step 1: Construct the matrix \(X_ij\)

First, let's construct the matrix \(X_{ij}\) as a matrix with all zero entries except for a 1 at position \((i, j)\). Explicitly for each \(i\) and \(j\), we have: \[X_{ij} = \begin{pmatrix}0& \cdots& 0\\ \vdots & \ddots & \vdots\\ 0 & \cdots & 1 \end{pmatrix}\] with the entry 1 at position \((i, j)\).

Step 2: Calculate the product \(MX_{ij}\)

Now, let's compute the product \(MX_{ij}\). If entry \(a_{rs}\) is the entry in row \(r\) and column \(s\) for matrix \(M\), then we can calculate the entries of \(MX_{ij}\) as follows: \[(MX_{ij})_{rs} = \sum_{t = 1}^n a_{rt} (X_{ij})_{ts}.\] Since \((X_{ij})_{ts}=0\) for all \(t \ne j\), we only need to consider the case where \(t=j\), which simplifies our calculation to: \[(MX_{ij})_{rs} = a_{rj} (X_{ij})_{js} = a_{rj} \delta_{js}.\] Here, \(\delta_{js}\) is the Kronecker delta, which is equal to 1 if \(j=s\), and 0 otherwise. Thus, \((MX_{ij})_{rs} = a_{rj}\) whenever \(s=j\) and 0 elsewhere. We now have: \[MX_{ij} = \begin{pmatrix}0 & \cdots & a_{1j} & \cdots & 0\\ \vdots & \ddots & \vdots & \ddots & \vdots\\ 0 & \cdots & a_{rj} & \cdots & 0 \end{pmatrix}\] with a non-zero entry \(a_{rj}\) at position \((r, j)\) and 0 everywhere else.

Step 3: Calculate the trace of \(MX_{ij}\)

Now, let's compute the trace of \(MX_{ij}\). The trace is the sum of the diagonal entries: \[\operatorname{tr}(MX_{ij}) = \sum_{r = 1}^n (MX_{ij})_{rr} = a_{ij}\] where \(a_{ij}\) is the \(i,j\) entry of matrix \(M\).

Step 4: Use the given condition and conclude

Since we know that the trace of any \(MX\) must be 0, we have: \[\operatorname{tr}(MX_{ij}) = a_{ij} = 0\] for all \(i, j\). This means that every entry of the matrix \(M\) must be 0, so \(M\) is the zero matrix: \[M = O.\]

Key Concepts

Trace of a Matrix

The trace of a matrix is a straightforward concept but central to many areas in matrix theory. It refers to the sum of the elements on the main diagonal of a square matrix. For an \(n \times n\) matrix \(A\), denoted \(A_{ii}\), the trace \(\operatorname{tr}(A)\) is calculated as follows:\[\operatorname{tr}(A) = \sum_{i=1}^n A_{ii}.\]

• The trace allows us to gather important information about the matrix, such as checking matrix similarity.
• It is a basis for defining several concepts and theorems within linear algebra.

By examining the trace of the product of matrices and using properties of specific matrix operations, we can derive certain results, like ensuring a matrix must be zero when its trace equals zero for any arbitrary matrix \(X\). This unique feature is crucial as demonstrated in problems like this, where the trace assists in revealing the nature of \(M\).

Kronecker Delta

The Kronecker delta, often denoted as \(\delta_{ij}\), plays a fundamental role in simplifying calculations involving indices. This symbol equals 1 when the indices are equal, and 0 otherwise:\[\delta_{ij} = \begin{cases}1 & \text{if } i = j,\ 0 & \text{if } i eq j.\end{cases}\]

• Think of it as a mathematical identity switch; it's like saying "yes" (1) when the positions match and "no" (0) otherwise.
• It often appears in expressions to isolate or simplify calculations, particularly in coordinate transformation and tensor analysis.

In the context of our problem, the Kronecker delta reduces the complexity of matrix multiplication by focusing attention on when row and column indices align. This simplification helps in proving that matrices \(MX_{ij}\) have non-diagonal elements remaining zero, simplifying further calculations.

Zero Matrix

The zero matrix is essentially the null element in the set of all matrices under addition. It is a matrix where every element is zero, often denoted by \(O\). For any \(m \times n\) matrix dimensions, the zero matrix is expressed as:\[O = \begin{pmatrix}0 & 0 & \cdots & 0 \0 & 0 & \cdots & 0 \\vdots & \vdots & \ddots & \vdots \0 & 0 & \cdots & 0\end{pmatrix}.\]

• When added to any other matrix of the same dimension, the result is the original matrix (\(A + O = A\)).
• It also acts as a neutral element in matrix multiplication; when any matrix is multiplied by a zero matrix, the result is a zero matrix.

In this problem, the condition \(\operatorname{tr}(MX) = 0\) for all matrices \(X\) implies that no matter what matrix \(X\) is chosen, \(M\)'s effect nullifies it completely, indicating \(M\) must be a zero matrix itself.

Matrix Multiplication

Matrix multiplication is a fundamental operation where two matrices are multiplied to produce a third matrix. For multiplication to be possible, the number of columns in the first matrix must match the number of rows in the second. If \(A\) is an \(m \times n\) matrix and \(B\) is an \(n \times p\) matrix, their product is an \(m \times p\) matrix \(C\) with entries given by:\[C_{ij} = \sum_{k=1}^n A_{ik}B_{kj}.\]

• Notice how each element in the resulting matrix is the sum of products of corresponding elements from the row of the first matrix and the column of the second matrix.
• This operation is not commutative, meaning \(AB eq BA\) in most cases.

In our exercise, matrix multiplication helped redefine and break down the effect \(M\) had on a specifically constructed matrix \(X_{ij}\). This use of multiplication with constructed matrices helped simplify and visualize the contribution of each element in \(M\) on the final result, underpinning the discovery that \(M\) must be a zero matrix.