Let \(M\) be a real \(n\times n\) matrix.
Prove that \[ M=A+S+cI, \] where \(A\) is antisymmetric, \(S\) is symmetric with \(\operatorname{tr}S=0\), and \(c\) is a scalar.
Prove that, with this notation, \[ \operatorname{tr}(M^2) =\operatorname{tr}(A^2)+\operatorname{tr}(S^2) +\frac1n(\operatorname{tr}M)^2. \]
Solution
1. Define\[ A\coloneqq\frac{M-M^T}{2}, \qquad T\coloneqq\frac{M+M^T}{2}. \]Then \(A\) is antisymmetric, \(T\) is symmetric, and\[ M=A+T. \]
One has \[ A^T=-A, \qquad T^T=T, \] by direct transposition, and adding the two definitions gives \(A+T=M\).
2. Set\[ c\coloneqq\frac{\operatorname{tr}M}{n}, \qquad S\coloneqq T-cI. \]Then \(S\) is symmetric, \(\operatorname{tr}S=0\), and\[ M=A+S+cI. \]
Since \(T\) and \(I\) are symmetric, so is \(S\). Also \[ \operatorname{tr}T = \frac12\left(\operatorname{tr}M+\operatorname{tr}M^T\right) = \operatorname{tr}M. \] Therefore \[ \operatorname{tr}S = \operatorname{tr}M-cn =0. \] Finally, \(T=S+cI\), so step 1 gives the claimed decomposition.
3. One has\[ \operatorname{tr}(AS)=0. \]
Using invariance of trace under transpose and cyclicity of trace, \[ \operatorname{tr}(AS) = \operatorname{tr}((AS)^T) = \operatorname{tr}(S^TA^T) = -\operatorname{tr}(SA) = -\operatorname{tr}(AS). \] Hence \(2\operatorname{tr}(AS)=0\).
4. One also has\[ \operatorname{tr}A=0, \qquad \operatorname{tr}S=0. \]
2. For the first,\[ \operatorname{tr}A = \operatorname{tr}A^T = -\operatorname{tr}A, \]so \(\operatorname{tr}A=0\).
5. One has\[ \operatorname{tr}(M^2) = \operatorname{tr}(A^2)+\operatorname{tr}(S^2)+nc^2. \]
From step 2, \[ M=A+S+cI. \] Expanding \(M^2\) and taking traces gives \[ \begin{aligned} \operatorname{tr}(M^2) &=\operatorname{tr}(A^2)+\operatorname{tr}(S^2)+nc^2\\ &\quad+\operatorname{tr}(AS+SA) +2c\operatorname{tr}A +2c\operatorname{tr}S. \end{aligned} \] By step 3, \[ \operatorname{tr}(AS+SA)=2\operatorname{tr}(AS)=0, \] and step 4 kills the remaining cross terms.
6. Therefore\[ \boxed{ \operatorname{tr}(M^2) = \operatorname{tr}(A^2)+\operatorname{tr}(S^2) +\frac1n(\operatorname{tr}M)^2}. \]
By step 2, \[ c=\frac{\operatorname{tr}M}{n}, \] so \[ nc^2=\frac1n(\operatorname{tr}M)^2. \] Substitute this into step 5.
7. Q.E.D.
Step 2 proves part 1, and step 6 proves part 2.