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given two matrices: $A, B \in M_{n\times n}(\mathbb{R})$ , let $B$ is a symmetric matrix ($B=B^{T}$),

We'll define a linear map: $T:M_{n\times n}(\mathbb{R})\longrightarrow M_{n\times n}(\mathbb{R})$

defined by:

for every $x \in M_{n\times n}(\mathbb{R})$ $$T(x)=\frac{X+X^{T}}{2} $$

prove: $$||A-B||_{F}\geq||A-T(A)||_{F}$$

as $||A||_F$ is the matrix norm.

I just can't understand in what way to determine that it's true. I've tried everything I know - without making significant progress.

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This follows from a sort of pythagorean theorem: For any real $n\times n$ matrix $A$ and real symmetric $n\times n$ matrix $B$, we have $$ ||A-B||^{2} = ||A-T(A)||^{2} + ||T(A)-B||^{2}. $$ To prove this, we need a fact that the norm $$ ||A||_{F} = \left(\sum_{1\leq i, j\leq n}a_{ij}^{2}\right)^{1/2} $$ is an $L^{2}$-norm induced from the inner product $$ \langle X, Y\rangle = \mathrm{Tr}(X^{T}Y) = \sum_{1\leq i, j\leq n}x_{ij}y_{ij}. $$ Then to show the pythagorean theorem, it is enough to show that $A-T(A)$ and $T(A)-B$ are orthogonal with respect to this inner product, i.e. $\langle A-T(A), T(A)-B\rangle=0$. However, for any real anti symmetric matrix $X$ ($X^{T} = -X$) and real symmetric matrix $Y$ ($Y^{T}=Y$), we can prove that $\langle X, Y \rangle = 0$. Since $A-T(A)$ is anti-symmetric and $T(A)-B$ is symmetric, we are done.

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