Understanding Matrix Decomposition and Symmetric Matrices

The question asks us to find the symmetric part, denoted by P, of a given matrix A. We are told that matrix A can be expressed as the sum of a symmetric matrix P and a skew-symmetric matrix Q, i.e., A = P + Q.

Any square matrix A can be uniquely decomposed into the sum of a symmetric matrix and a skew-symmetric matrix. The formulas for P and Q are:

• Symmetric part P: \(P = \frac{1}{2}(A + A^T)\)
• Skew-symmetric part Q: \(Q = \frac{1}{2}(A - A^T)\)

Here, \(A^T\) represents the transpose of matrix A.

Finding the Symmetric Part P

The given matrix A is:

\(A = \left(\begin{array}{ccc}0 & \sin^2 \theta & \cos^2 \theta \\ \cos^2 \theta & 0 & \sin^2 \theta \\ \sin^2 \theta & \cos^2 \theta & 0\end{array}\right)\)

First, let's find the transpose of matrix A, \(A^T\). The transpose is obtained by swapping the rows and columns of A.

\(A^T = \left(\begin{array}{ccc}0 & \cos^2 \theta & \sin^2 \theta \\ \sin^2 \theta & 0 & \cos^2 \theta \\ \cos^2 \theta & \sin^2 \theta & 0\end{array}\right)\)

Next, we calculate the sum of A and \(A^T\):

\(A + A^T = \left(\begin{array}{ccc}0 & \sin^2 \theta & \cos^2 \theta \\ \cos^2 \theta & 0 & \sin^2 \theta \\ \sin^2 \theta & \cos^2 \theta & 0\end{array}\right) + \left(\begin{array}{ccc}0 & \cos^2 \theta & \sin^2 \theta \\ \sin^2 \theta & 0 & \cos^2 \theta \\ \cos^2 \theta & \sin^2 \theta & 0\end{array}\right)\)

\(A + A^T = \left(\begin{array}{ccc}0+0 & \sin^2 \theta + \cos^2 \theta & \cos^2 \theta + \sin^2 \theta \\ \cos^2 \theta + \sin^2 \theta & 0+0 & \sin^2 \theta + \cos^2 \theta \\ \sin^2 \theta + \cos^2 \theta & \cos^2 \theta + \sin^2 \theta & 0+0\end{array}\right)\)

Using the trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\), the sum becomes:

\(A + A^T = \left(\begin{array}{ccc}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right)\)

Finally, we calculate P using the formula \(P = \frac{1}{2}(A + A^T)\):

\(P = \frac{1}{2} \left(\begin{array}{ccc}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right) = \left(\begin{array}{ccc}\frac{1}{2}(0) & \frac{1}{2}(1) & \frac{1}{2}(1) \\ \frac{1}{2}(1) & \frac{1}{2}(0) & \frac{1}{2}(1) \\ \frac{1}{2}(1) & \frac{1}{2}(1) & \frac{1}{2}(0)\end{array}\right)\)

\(P = \left(\begin{array}{ccc}0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0\end{array}\right)\)

Verifying the Result

To verify that P is indeed a symmetric matrix, we check if \(P^T = P\). The transpose of P is:

\(P^T = \left(\begin{array}{ccc}0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0\end{array}\right)^T = \left(\begin{array}{ccc}0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0\end{array}\right)\)

Since \(P^T = P\), P is a symmetric matrix.

The calculated matrix for P is:

\(\left(\begin{array}{ccc}0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0\end{array}\right)\)

Comparing this with the given options, we find that it matches Option 1.

Revision Table: Key Matrix Definitions

Additional Information: Properties of Matrix Decomposition

Any square matrix A can be uniquely written as the sum of a symmetric matrix P and a skew-symmetric matrix Q. This decomposition is unique because if A = P + Q where P is symmetric and Q is skew-symmetric, then:

\(A^T = (P+Q)^T = P^T + Q^T = P - Q\)

We now have two equations:

1. A = P + Q
2. \(A^T\) = P - Q

Adding these two equations gives:

\(A + A^T = (P+Q) + (P-Q) = 2P\)

So, \(P = \frac{1}{2}(A + A^T)\).

Subtracting the second equation from the first gives:

\(A - A^T = (P+Q) - (P-Q) = 2Q\)

So, \(Q = \frac{1}{2}(A - A^T)\).

These unique formulas demonstrate how the symmetric and skew-symmetric parts are derived and why the decomposition is unique for any square matrix.