Matrix Decomposition Fundamentals

Understanding matrix decomposition is a key concept in linear algebra. Any square matrix \(A\) can be uniquely expressed as the sum of a symmetric matrix \(B\) and a skew-symmetric matrix \(C\). This is often represented as \(A = B + C\).

• A matrix \(B\) is considered symmetric if its transpose \(B^T\) is equal to the matrix itself, i.e., \(B = B^T\).
• A matrix \(C\) is considered skew-symmetric if its transpose \(C^T\) is equal to the negative of the matrix itself, i.e., \(C = -C^T\).

The formulas to find the symmetric and skew-symmetric parts are derived as follows:

• For the symmetric part \(B\): \(B = \frac{1}{2}(A + A^T)\)
• For the skew-symmetric part \(C\): \(C = \frac{1}{2}(A - A^T)\)

Given Matrix A and Its Transpose

We are given the matrix \(A\):

\(A = \left[ {\begin{array}{c} 0&1&{ - 1} \\ 4&{ - 3}&4 \\ 3&{ - 3}&4 \end{array}} \right]\)

First, we need to find the transpose of matrix \(A\), denoted as \(A^T\). The transpose of a matrix is obtained by interchanging its rows and columns.

\(A^T = \left[ {\begin{array}{c} 0&4&3 \\ 1&{ - 3}&{ - 3} \\ { - 1}&4&4 \end{array}} \right]\)

Calculating the Sum \(A + A^T\)

Next, we add matrix \(A\) and its transpose \(A^T\). To add matrices, we simply add their corresponding elements.

\(A + A^T = \left[ {\begin{array}{c} 0&1&{ - 1} \\ 4&{ - 3}&4 \\ 3&{ - 3}&4 \end{array}} \right] + \left[ {\begin{array}{c} 0&4&3 \\ 1&{ - 3}&{ - 3} \\ { - 1}&4&4 \end{array}} \right]\)

\(A + A^T = \left[ {\begin{array}{c} 0+0&1+4&{ - 1}+3 \\ 4+1&{ - 3}+( { - 3})&4+( { - 3}) \\ 3+( { - 1})&{ - 3}+4&4+4 \end{array}} \right]\)

\(A + A^T = \left[ {\begin{array}{c} 0&5&2 \\ 5&{ - 6}&1 \\ 2&1&8 \end{array}} \right]\)

Determining Matrix B (Symmetric Part)

Now, we use the formula for the symmetric part \(B\): \(B = \frac{1}{2}(A + A^T)\). We multiply the resulting sum by \(\frac{1}{2}\).

\(B = \frac{1}{2}\left[ {\begin{array}{c} 0&5&2 \\ 5&{ - 6}&1 \\ 2&1&8 \end{array}} \right]\)

This result matches one of the given options. Let's verify this step with the options provided.

Verifying the Solution

The calculated matrix \(B\) is:

\(B = \frac{1}{2}\left[ {\begin{array}{c} 0&5&2 \\ 5&{ - 6}&1 \\ 2&1&8 \end{array}} \right]\)

To confirm that this matrix \(B\) is indeed symmetric, we can check if \(B = B^T\):

\(B^T = \left( \frac{1}{2}\left[ {\begin{array}{c} 0&5&2 \\ 5&{ - 6}&1 \\ 2&1&8 \end{array}} \right] \right)^T = \frac{1}{2}\left[ {\begin{array}{c} 0&5&2 \\ 5&{ - 6}&1 \\ 2&1&8 \end{array}} \right]^T = \frac{1}{2}\left[ {\begin{array}{c} 0&5&2 \\ 5&{ - 6}&1 \\ 2&1&8 \end{array}} \right]\)

Since \(B = B^T\), the matrix \(B\) we found is symmetric, as required by the question.

Comparing our result with the given options, we find that the calculated matrix \(B\) matches Option 2.