It need not be, so the question makes no sense!
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A matrix A is orthogonal if itstranspose is equal to it inverse. So AT is the transpose of A and A-1 is the inverse. We have AT=A-1 So we have : AAT= I, the identity matrix Since it is MUCH easier to find a transpose than an inverse, these matrices are easy to compute with. Furthermore, rotation matrices are orthogonal. The inverse of an orthogonal matrix is also orthogonal which can be easily proved directly from the definition.
a square matrix that is equal to its transpose
The transpose of a matrix A is the matrix B that is obtained by swapping the rows and columns of A into the columns and rows of B. In algebraic form, if A = {aij} then B = {aji} is its transpose, where 1 ≤ i ≤ n and 1 ≤ j ≤ m.
The classical adjoint of a square matrix A the transpose of the matrix who (i, j) entry is the a i j cofactor.
Since the columns of AT equal the rows of A by definition, they also span the same space, so yes, they are equivalent.