Transpose of product

Dependencies:

1. Matrix

Let $R$ be a commutative semiring. Let $A$ and $B$ be matrices on $R$. Then $(AB)^T = B^TA^T$.

Proof

\begin{align} & (AB)^T[i, j] \\ &= (AB)[j, i] \\ &= \sum_k A[j, k] B[k, i] \\ &= \sum_k A^T[k, j] B^T[i, k] \\ &= \sum_k B^T[i, k] A^T[k, j] \tag{$R$ is commutative semiring} \\ &= (B^TA^T)[i, j] \end{align}

Therefore, $(AB)^T = B^TA^T$.

Dependency for:

1. Matrix of orthonormal basis change
2. Symmetric operator iff hermitian
3. Orthogonally diagonalizable iff hermitian
4. A is diagonalizable iff there are n linearly independent eigenvectors
5. All eigenvalues of a hermitian matrix are real
6. Matrices form an inner-product space
7. Bounding matrix quadratic form using eigenvalues

Info:

• Depth: 3
• Number of transitive dependencies: 4

Transitive dependencies:

1. Group
2. Ring
3. Semiring
4. Matrix