A matrix A is said to be skew symmetric if A^T = -A. Show that is a matrix is skew symmetric then its diagonal entries must all be 0.

A^T meant to be A transpose.

Let a(i,j) stand for the element of A on the ith row and jth column.

Let
A be a skew symmetric matrix.

By the definition of skew-symmetry,
a(j,i)=-a(i,j)

On the diagonal,
i=j
=> a(i,i)=-a(i,i)
=> a(i,i)=0

since x=-x => x=0

To show that if a matrix A is skew symmetric, then its diagonal entries must all be 0, we can directly prove this property by examining the definition of a skew symmetric matrix and the properties of transposition.

Let's assume that A is a skew symmetric matrix, which means A^T = -A.

Now, let's denote the element on the i-th row and j-th column of A as A[i, j].

To show that the diagonal entries of A must all be 0, we need to consider the case when i = j. In other words, we are examining the elements on the diagonal of the matrix.

For the diagonal entries, we have A[i, i]. Now, applying the definition of transposition, we know that the element on the i-th row and i-th column of A^T is A^T[i, i], which is equal to A[i, i] (since A^T is just the transpose of A).

Therefore, from A^T = -A, we have A^T[i, i] = -A[i, i].

Since A^T[i, i] = A[i, i], we can rewrite the equation as A[i, i] = -A[i, i].

Now, by solving this equation, we find that the only possible value for A[i, i] is 0, as the negative of 0 is still 0.

Hence, we can conclude that if a matrix A is skew symmetric, then its diagonal entries must all be 0.

To prove that a matrix A is skew symmetric, we need to show that its diagonal entries are all 0.

Let's consider a square matrix A of size n×n.

The transpose of A, denoted by A^T, is obtained by interchanging its rows with columns.

Since A is skew symmetric, we have A^T = -A.

Now, let's look at the diagonal entries of A. The diagonal of a square matrix consists of the elements where the row index is equal to the column index.

For example, in a 3×3 matrix A, the diagonal elements are A[1,1], A[2,2], and A[3,3].

To prove that the diagonal entries of A are all 0, we need to show that for each i from 1 to n, A[i,i] = 0.

Let's consider the i-th diagonal entry A[i,i].

Using the definition of the transpose, we have:

A^T[i,i] = -A[i,i]

Since A^T[i,i] represents the element in the i-th row and i-th column of A^T, and -A[i,i] represents the negative of the element in the i-th row and i-th column of A, we can conclude that:

A[i,i] = -A[i,i]

This implies that A[i,i] = 0.

Since this holds for every i from 1 to n, we can conclude that all the diagonal entries of A must be 0.

Therefore, if a matrix A is skew symmetric, its diagonal entries must all be 0.