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For Cartesian components, ${\partial }_j{x}_i={\delta }_{ij}$, the Kronecker delta. Differentiating $r^2=x_i{x}_i$ givesPut $s=k_{\ell }{x}_{\ell }$. Using the product rule and ${\partial }_j{r}^{-q}=-q{x}_j{r}^{-q-2}$ gives the velocity gradientThis is the rate of strain and vorticity of a Stokeslet, apart from the overall physical normalization. Its symmetric and antisymmetric parts of a matrix areHere $E_{ji}=E_{ij}$ and $W_{ji}=-W_{ij}$, and $d=E+W$. Taking the trace gives the divergenceFor the curl, use the Levi-Civita symbol. Because $d_{kj}={\partial }_j{u}_k$, the symmetric part makes no contribution, andThus $\nabla \times \mathbf{u}=2\mathbf{k}\times \mathbf{x}/r^3$. All derivative identities hold away from the singular origin.

DefineThey sum to $t_{ij}$, with $s_{ji}=s_{ij}$ and $a_{ji}=-a_{ij}$. Interchanging two slots of a tensor still produces a tensor, so these are a symmetric second-rank tensor and an antisymmetric second-rank tensor, not merely symmetric and antisymmetric arrays in a selected frame.

For uniqueness, if $t=s+a=s^{\prime }+a^{\prime }$, then $s-s^{\prime }=a^{\prime }-a$. The common difference is both symmetric and antisymmetric, so each of its components equals its own negative and vanishes over $\mathbb{R}$. Thus the decomposition is unique. These are the symmetric and antisymmetric parts of a matrix, with their tensor transformation law preserved.

The unique symmetric and antisymmetric parts of a matrix are $S=(M+M^T)/2$ and $A=(M-M^T)/2$, givingTheir matrix products with their transposes are $S^{T}S=(a+b{)}^{2}I/4$ and $A^{T}A=(a-b{)}^{2}I/4$. Each is an orthogonal matrix precisely when the corresponding factor is one. Thus $a+b=\pm 2$ and $a-b=\pm 2$, with independent sign choices, and the simultaneous possibilities are

True. The symmetric and antisymmetric parts of a matrix are, in Einstein summation convention,They satisfy $S_{ji}=S_{ij}$, $A_{ji}=-A_{ij}$ and $S_{ij}+A_{ij}=M_{ij}$. This works over the real or complex numbers; it uses division by two.

For any real square matrix $N$, the symmetric and antisymmetric parts of a matrix give $N=(N+N^T)/2+(N-N^T)/2$, whose summands are respectively a symmetric matrix and an antisymmetric matrix. Here it givesTransposing the first summand leaves it unchanged; transposing the second changes its sign. Entrywise addition recovers the original matrix. The decomposition is unique because a matrix that is both symmetric and antisymmetric must be zero.

The symmetric and antisymmetric parts of a matrix areIndeed $S+A=M$, while the transpose identities $(M^T{)}^T=M$ give $S^T=S$ and $A^T=-A$. Thus $S$ is a symmetric matrix and $A$ is an antisymmetric matrix. The decomposition is also unique: a matrix that is both symmetric and antisymmetric must equal its own negative and hence vanish.

Let $S$ be a real symmetric matrix and $A$ a real antisymmetric matrix of the same size. Invariance of the trace under transpose and the cyclic trace identity giveHence $\mathrm{tr}(SA)=0$. Equivalently these two matrices are orthogonal for the Frobenius inner product, since $⟨S,A{⟩}_F=\mathrm{tr}(S^{T}A)=\mathrm{tr}(SA)$. Together with the unique symmetric and antisymmetric parts of a matrix, this is an orthogonal decomposition of the real square matrices.