Matrices form an inner-product space
Dependencies:
- Matrices over a field form a vector space
- Inner product space
- Dot-product of vectors
- Trace of a matrix
- /complex-numbers/conjugation-is-homomorphic
- /complex-numbers/conjugate-product-abs
- Transpose of product
Let $F$ be a subfield of $\mathbb{C}$ (complex numbers). Then $\mathbb{M}_{m, n}(F)$ is an inner product space.
The inner product of matrices is given by $\operatorname{tr}(B^*A)$, where $A^*$ is the conjugate transpose of $A$.
\[ A^*[i, j] = \overline{A[j, i]} \implies A^* = \overline{A}^T = \overline{A^T} \]
If we only consider column vectors ($n=1$), \[ \langle \mathbf{u}, \mathbf{v} \rangle = \operatorname{tr}(\mathbf{v}^*\mathbf{u}) = \mathbf{v}^*\mathbf{u} = \mathbf{v} \cdot \mathbf{u} \] which is the dot product of $\mathbf{v}$ and $\mathbf{u}$.
For real-valued matrices, $\langle A, B \rangle = \operatorname{tr}(A^TB)$ is an equivalent definition of inner product.
Proof
Matrices over a field form a vector space. We only need to prove properties of the inner-product.
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Conjugate symmetry: \begin{align} \langle A, B \rangle &= \operatorname{tr}(B^*A) \\ &= \sum_{i=1}^n (B^*A)[i, i] \\ &= \sum_{i=1}^n \sum_{j=1}^m B^*[i, j]A[j, i] \\ &= \sum_{i=1}^n \sum_{j=1}^m \overline{B[j, i]}A[j, i] \\ &= \sum_{i=1}^n \sum_{j=1}^m \overline{B[j, i]}\;\overline{\overline{A[j, i]}} \tag{$\overline{\overline{z}} = z \forall z \in \mathbb{C}$} \\ &= \sum_{i=1}^n \sum_{j=1}^m \overline{B[j, i]\overline{A[j, i]}} \tag{conjugation is homomorphic} \\ &= \sum_{i=1}^n \sum_{j=1}^m \overline{\overline{A[j, i]}B[j, i]} \\ &= \sum_{i=1}^n \sum_{j=1}^m \overline{A^*[i, j]B[j, i]} \\ &= \sum_{i=1}^n \overline{(A^*B)[i, i]} \\ &= \overline{\sum_{i=1}^n (A^*B)[i, i]} \tag{conjugation is homomorphic} \\ &= \overline{\operatorname{tr}(A^*B)} = \overline{\langle B, A \rangle} \end{align}
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Linearity in first argument: \begin{align} \langle A+B, C \rangle &= \operatorname{tr}(C^*(A+B)) \\ &= \operatorname{tr}(C^*A + C^*B) \tag{distributivity} \\ &= \operatorname{tr}(C^*A) + \operatorname{tr}(C^*B) \\ &= \langle A, C \rangle + \langle B, C \rangle \end{align} \begin{align} \langle cA, B \rangle &= \operatorname{tr}(B^*(cA)) \\ &= \operatorname{tr}(c(B^*A)) \tag{associativity} \\ &= c\operatorname{tr}(B^*A) \\ &= c\langle A, B \rangle \end{align}
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Positive-definiteness: \begin{align} \langle A, A \rangle &= \operatorname{tr}(A^*A) \\ &= \sum_{i=1}^n (A^*A)[i, i] \\ &= \sum_{i=1}^n \sum_{j=1}^m A^*[i, j]A[j, i] \\ &= \sum_{i=1}^n \sum_{j=1}^m \overline{A[j, i]}A[j, i] \\ &= \sum_{i=1}^n \sum_{j=1}^m |A[j, i]|^2 \end{align} Therefore, $\langle A, A \rangle \ge 0$. \begin{align} \langle A, A \rangle = 0 &\iff \sum_{i=1}^n \sum_{j=1}^m |A[j, i]|^2 = 0 \iff \forall i, \forall j, |A[j, i]|^2 = 0 \\ &\iff \forall i, \forall j, A[j, i] = 0 \iff A = 0 \end{align}
Therefore, $\mathbb{M}_{m, n}(F)$ is an inner-product space.
If we only consider real-valued matrices, \[ \langle A, B \rangle = \operatorname{tr}(B^*A) = \operatorname{tr}(B^TA) = \operatorname{tr}((A^TB)^T) = \operatorname{tr}(A^TB) \]
Dependency for:
Info:
- Depth: 5
- Number of transitive dependencies: 14
Transitive dependencies:
- /complex-numbers/conjugate-product-abs
- /complex-numbers/conjugation-is-homomorphic
- Group
- Ring
- Vector
- Dot-product of vectors
- Field
- Vector Space
- Inner product space
- Semiring
- Matrix
- Transpose of product
- Trace of a matrix
- Matrices over a field form a vector space