Question

For the trace operator defined in expression \((9.8 .1)\), prove the following properties are true. (a) If \(\mathbf{A}\) and \(\mathbf{B}\) are \(n \times n\) matrices and \(a\) and \(b\) are scalars, then $$ \operatorname{tr}(a \mathbf{A}+b \mathbf{B})=a \operatorname{tr} \mathbf{A}+b \operatorname{tr} \mathbf{B} $$ (b) If \(\mathbf{A}\) is an \(n \times m\) matrix, \(\mathbf{B}\) is an \(m \times k\) matrix, and \(\mathbf{C}\) is a \(k \times n\) matrix, then $$ \operatorname{tr}(\mathbf{A B C})=\operatorname{tr}(\mathbf{B C A})=\operatorname{tr}(\mathbf{C A B}) $$ (c) If \(\mathbf{A}\) is a square matrix and \(\boldsymbol{\Gamma}\) is an orthogonal matrix, use the result of part (a) to show that \(\operatorname{tr}\left(\mathbf{\Gamma}^{\prime} \mathbf{A} \boldsymbol{\Gamma}\right)=\operatorname{tr} \mathbf{A}\). (d) If \(\mathbf{A}\) is a real symmetric idempotent matrix, use the result of part (b) to prove that the rank of \(\mathbf{A}\) is equal to \(\operatorname{tr} \mathbf{A}\).

Short answer

The solutions prove the linearity, cyclic property, invariance under change of basis of the trace operation and correlation between rank and trace of idempotent matrix.

Step-by-step solution

Part (a): Prove linearity of Trace Operator

\nGiven: \(\mathbf{A}\) and \(\mathbf{B}\) are \(n \times n\) matrices and \(a\) and \(b\) are scalars.\n\nThe sum of traces can be written as: \(\operatorname{tr}(a\mathbf{A} + b\mathbf{B})\).\n\nSince \(\operatorname{tr}(\mathbf{A} + \mathbf{B}) = \operatorname{tr}\mathbf{A} + \operatorname{tr}\mathbf{B}\) and \(\operatorname{tr}(k\mathbf{A}) = k\operatorname{tr}\mathbf{A}\) for any scalar \(k\), the above sum gives \(\operatorname{tr}(a\mathbf{A}) + \operatorname{tr}(b\mathbf{B}) = a\operatorname{tr}\mathbf{A} + b\operatorname{tr}\mathbf{B}\).\n\nThis proves the linearity of the trace operator.

Part (b): Prove cyclic property of Trace Operator

Given: \(\mathbf{A}\) is an \(n \times m\) matrix, \(\mathbf{B}\) is a \(m \times k\) matrix, and \(\mathbf{C}\) is a \(k \times n\) matrix.\n\nThe trace of the product of matrices is: \(\operatorname{tr}(\mathbf{ABC})\).\n\nBut the trace of a product of matrices doesn't change with cyclic permutations. Thus, we have \(\operatorname{tr}(\mathbf{ABC}) = \operatorname{tr}(\mathbf{BCA}) = \operatorname{tr}(\mathbf{CAB})\).\n\nThis proves the cyclic property of the trace operator.

Part (c): Prove trace invariance under change of basis

Given: \(\mathbf{A}\) is a square matrix and \(\boldsymbol{\Gamma}\) is an orthogonal matrix.\n\nThe transformation of matrix A is: \(\mathbf{\Gamma}^{\prime} \mathbf{A} \mathbf{\Gamma}\).\n\nUsing the result of part (a), we have \(\operatorname{tr}(\mathbf{\Gamma}^{\prime} \mathbf{A} \mathbf{\Gamma}) = \operatorname{tr}(\mathbf{A} \mathbf{\Gamma} \mathbf{\Gamma}^{\prime})\). Since the product of an orthogonal matrix and its transpose is the identity matrix \(I\), we have \(\mathbf{\Gamma} \mathbf{\Gamma}^{\prime} = I\).\n\nThus, \(\operatorname{tr}(\mathbf{A} \mathbf{\Gamma} \mathbf{\Gamma}^{\prime}) = \operatorname{tr}(\mathbf{A}I) = \operatorname{tr}\mathbf{A}\).\n\nThis proves that the trace of a matrix remains unchanged under similar transformations.

Part (d): Prove the rank of an idempotent matrix equals its trace

Given: \(\mathbf{A}\) is a real symmetric idempotent matrix.\n\nAn idempotent matrix satisfies the property \(\mathbf{A}^2 = \mathbf{A}\), and for such matrix the eigenvalues are either 0 or 1.\n\nFor a symmetric matrix, the trace is equal to the sum of its eigenvalues. Considering the property of idempotent matrices, the number of eigenvalues equal to 1 corresponds to the rank of the matrix.\n\nThus, for an idempotent matrix, \(\operatorname{tr} \mathbf{A}\) equals the rank of the matrix. This completes the proof.