Best,j3n nyIs B=A^2 skew-symmetric, symmetric or neither?

[j3n]

Hey,

I need help with thris proof.. :)

Say A is a skew-symmetric matrix. Is B=A^2 skew-symmetric, symmetric or neither. Prove your answer.



I know it's symmetric, but I'm having problems with subscriptmanship. Can anyone help me out?

Thanks,
j3n

 

[robphy]

Here's a starting point:
$$A=-A^\top$$

 

[tacman]

ny

Hi j3nny,

No problem, I can help you out with this proof. First, let's define what it means for a matrix to be skew-symmetric. A square matrix A is skew-symmetric if it satisfies the condition A^T = -A, where A^T is the transpose of A. This means that the entries on the main diagonal of A are all 0, and the entries below the main diagonal are the negatives of the corresponding entries above the main diagonal.

Now, let's consider the matrix B = A^2. Using the definition of matrix multiplication, we can write B as B = A*A. Since A is skew-symmetric, we know that A^T = -A. Therefore, we can rewrite B as B = A*(-A). Using the properties of matrix multiplication, we can also write this as B = -(A*A).

Now, since A*A is just another square matrix, we can use the definition of skew-symmetric matrices to rewrite it as A*A = -A^T. Substituting this back into our previous expression for B, we get B = -(-A^T) = A^T. This means that B is equal to its own transpose, which is the definition of a symmetric matrix.

To summarize, we have shown that if A is a skew-symmetric matrix, then B = A^2 is a symmetric matrix. Therefore, B is neither skew-symmetric nor symmetric, but rather it is a special type of matrix called a skew-symmetric matrix.

I hope this explanation helps and clarifies any confusion you had with the subscriptmanship. Let me know if you have any further questions. Good luck with your proof!