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There is a linear isomorphism between upper-triangular matrices and symmetric matrices given by $$f : \{\text{upper-triangular}\} \mapsto \{\mathrm{symmetric}\}, \; \; A \mapsto \frac{A + A^T}{2}.$$

On symmetric matrices, we can act by conjugation: $$A \mapsto B^T A B, \; \; B \in \mathbb{R}^{n \times n}.$$

Question: Is there a closed-form way to express the corresponding action on upper-triangular matrices? i.e. given an upper-triangular matrix $A$ and a matrix $B$, what is the upper-triangular matrix corresponding to $\frac{B^T A B + B^T A^T B}{2}$?

When $B$ is diagonal then it is just $B^T A B$ since this is upper-triangular again. In general it isn't clear to me.

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    Here's one of the differences between \mathrm{} and \text{}: $$\mathrm{upper-triangular}$$$$\text{upper-triangular}$$ I changed the question to use \text{}, so the hyphen looks like a hyphen and not like a minus sign. – 2017-01-07
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    Might the question be phrased as a request for the closed form of the upper-triangular part $A$ of $\dfrac{A+A^T} 2$ as a function of the latter matrix, or is there something more involved? $\qquad$ – 2017-01-07
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    Are you asking for a linear map $\phi$ on triangular matrices with $\phi(A) = B^T f(A) B$? But that's the closed form. I don't understand your question. – 2017-01-08
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    @user251257 I am looking for an expression directly in terms of $A$ and $B$, involving sums, multiplications, transposes, maybe diagonalizing (if necessary) but not just the expression $\phi(A) = f^{-1}(B^T f(A) B)$ as that does not work for me. – 2017-01-08
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    @MichaelHardy I think an answer to that would probably solve my question – 2017-01-08
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    I doubt you'll find anything much more illuminating then just writing down the formula for the $(i,j)$-th entry of $B^T \frac{A + A^T}{2} B$ in terms of entries of $A$ and $B$. – 2017-01-08

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