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Chapter 7: Problem 61

Let \(V\) be the vector space of \(m \times n\) matrices over \(\mathbf{R}\). Show that \(\langle A, B\rangle=\operatorname{tr}\left(B^{T} A\right)\) defines an inner product in \(V\).

Short Answer

Expert verified
We can show that \(\langle A, B\rangle=\operatorname{tr}\left(B^{T} A\right)\) defines an inner product in vector space \(V\) by verifying the properties: 1. Positivity: \(\langle A, A \rangle \ge 0\) and \(\langle A, A \rangle = 0\) if and only if \(A = 0\) 2. Conjugate symmetry: \(\langle A, B \rangle = \langle B, A \rangle\) 3. Linearity in the first argument: \(\langle cA_1 + A_2, B \rangle = c \langle A_1, B \rangle + \langle A_2, B \rangle\) We showed the positivity by demonstrating that the trace expression is non-negative and equals zero only when the matrix is a zero matrix. Conjugate symmetry was verified by showing that the trace of the matrix product is the same under cyclic permutation. Finally, linearity in the first argument was established by verifying that the trace expression is linear with respect to matrix addition and scalar multiplication. Therefore, \(\operatorname{tr}\left(B^{T} A\right)\) defines an inner product on the vector space \(V\) of \(m \times n\) matrices over \(\mathbf{R}\).

Step by step solution

01

Positivity

Verify the positivity. Given that \(\langle A, A \rangle = \textrm{tr}(A^T A)\), we need to show that this expression is non-negative: \[\textrm{tr}(A^T A) = \sum_{i=1}^m \sum_{j=1}^n (A^T A)_{ij} = \sum_{i=1}^m \sum_{j=1}^n (A^T)_{ij} A_{ij} = \sum_{i=1}^m \sum_{j=1}^n A_{ji} A_{ij}\] Since each term \(A_{ji} A_{ij}\) is a square of a real number, their sum is non-negative. The expression \(\langle A, A \rangle\) is equal to \(0\) if and only if all the matrix elements \(A_{ij}\) are equal to \(0\), which means that the matrix \(A\) is the zero matrix.
02

Conjugate Symmetry

Verify the conjugate symmetry. We need to show that \(\langle A, B \rangle = \langle B, A \rangle\), which means \(\textrm{tr}(B^T A) = \textrm{tr}(A^T B)\). Recall that the trace of a matrix is the sum of its diagonal elements, and the trace of a product of two matrices is unchanged under cyclic permutation: \[\textrm{tr}(B^T A) = \textrm{tr}((A B^T)^T) = \textrm{tr}(A^T B)\]
03

Linearity in the first argument

Verify the linearity in the first argument. We must show that \(\langle cA_1 + A_2, B \rangle = c \langle A_1, B \rangle + \langle A_2, B \rangle\). Given that \(\langle A, B \rangle = \textrm{tr}(B^T A)\), the left side of the equality becomes: \[\textrm{tr}(B^T (cA_1 + A_2)) = \textrm{tr}(c B^T A_1 + B^T A_2)\] Since the trace of a matrix is linear with respect to matrix addition and scalar multiplication: \[\textrm{tr}(c B^T A_1 + B^T A_2) = c \textrm{tr}(B^T A_1) + \textrm{tr}(B^T A_2) = c \langle A_1, B \rangle + \langle A_2, B \rangle\] Having verified all the necessary properties, we can conclude that \(\textrm{tr}(B^T A)\) defines an inner product on the vector space V of \(m \times n\) matrices over \(\mathbf{R}\).

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