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Exercise 1929Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Symmetric and skew-symmetric matrices

#symmetric#skew-symmetric#direct sum

Statement

Let S={M∈Mn(R):MT=M}S=\{M\in\mathcal{M}_n(\mathbb{R}) : M^T=M\} and A={M∈Mn(R):MT=−M}A=\{M\in\mathcal{M}_n(\mathbb{R}) : M^T=-M\}. Show that SS and AA are subspaces and that Mn(R)=S⊕A\mathcal{M}_n(\mathbb{R})=S\oplus A.

Hint

Show S∩A={0}S\cap A=\{0\} and every matrix decomposes as M=M+MT2+M−MT2M=\frac{M+M^T}{2}+\frac{M-M^T}{2}.

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Solution

Subspaces. SS and AA are defined by linear conditions (MT−M=0M^T-M=0 and MT+M=0M^T+M=0), hence are kernels of linear maps, hence subspaces.

Intersection. If M∈S∩AM\in S\cap A, then MT=MM^T=M and MT=−MM^T=-M, so M=−MM=-M, i.e. M=0M=0.

Decomposition. For any MM:
M=M+MT2⏟∈S+M−MT2⏟∈A.M = \underbrace{\frac{M+M^T}{2}}_{\in S} + \underbrace{\frac{M-M^T}{2}}_{\in A}.
Indeed, the first term is symmetric and the second is skew-symmetric.

Conclusion. Mn(R)=S⊕A\mathcal{M}_n(\mathbb{R})=S\oplus A. Their dimensions are n(n+1)2\frac{n(n+1)}{2} and n(n−1)2\frac{n(n-1)}{2}, summing to n2n^2.