Show that A+A^T is symmetric for any 2 x 2 matrix A. Is this true for any square matrix?

The correct answer and explanation is:

Correct Answer:

Yes, for any square matrix AA, the matrix A+ATA + A^T is symmetric.


Explanation (300 words):

To determine whether A+ATA + A^T is symmetric, we must recall the definition of a symmetric matrix:

A matrix MM is symmetric if MT=MM^T = M, where MTM^T is the transpose of MM.

Let’s test this with a general 2 × 2 matrix: A=[abcd]A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}

Then, the transpose of AA is: AT=[acbd]A^T = \begin{bmatrix} a & c \\ b & d \\ \end{bmatrix}

Now, compute A+ATA + A^T: A+AT=[abcd]+[acbd]=[a+ab+cc+bd+d]=[2ab+cb+c2d]A + A^T = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} + \begin{bmatrix} a & c \\ b & d \\ \end{bmatrix} = \begin{bmatrix} a + a & b + c \\ c + b & d + d \\ \end{bmatrix} = \begin{bmatrix} 2a & b + c \\ b + c & 2d \\ \end{bmatrix}

This matrix is symmetric because the transpose of A+ATA + A^T is: (A+AT)T=[2ab+cb+c2d]T=[2ab+cb+c2d]=A+AT(A + A^T)^T = \begin{bmatrix} 2a & b + c \\ b + c & 2d \\ \end{bmatrix}^T = \begin{bmatrix} 2a & b + c \\ b + c & 2d \\ \end{bmatrix} = A + A^T

Therefore, A+ATA + A^T is symmetric.

General Case: Any Square Matrix

The above result isn’t limited to 2 × 2 matrices. For any square matrix AA (i.e., of size n×nn \times n), we use the fact that: (A+AT)T=AT+(AT)T=AT+A=A+AT(A + A^T)^T = A^T + (A^T)^T = A^T + A = A + A^T

Since the transpose of A+ATA + A^T equals itself, it is symmetric.

Conclusion:

Yes, A+ATA + A^T is symmetric for any square matrix AA, including 2 × 2 matrices.

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