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<p>Some <strong>answers</strong> I particularly enjoyed writing:</p>
<ul>
<li><p><a href="https://math.stackexchange.com/a/1790810/339790">How to prove that a <span class="math-container">$3 \times 3$</span> Magic Square must have <span class="math-container">$5$</span> in its middle cell?</a>, Mathematics, May 18, 2016.</p>
</li>
<li><p><a href="https://cs.stackexchange.com/q/74153">How to reduce the low-rank matrix completion problem to integer programming?</a>, Computer Science, April 18, 2017.</p>
</li>
<li><p><a href="https://math.stackexchange.com/a/4258463/339790">Finding the gradient of the restricted function in terms of the gradient of the original function</a>, Mathematics, September 23, 2021.</p>
</li>
<li><p><a href="https://math.stackexchange.com/a/3920044/339790">Matrix exponential via Euler's formula</a>, Mathematics, November 23, 2020.</p>
</li>
<li><p><a href="https://math.stackexchange.com/a/2835002/339790">Second derivative of <span class="math-container">$\det ( \mathrm I_n + t \mathrm A )$</span></a>, Mathematics, June 28, 2018.</p>
</li>
<li><p><a href="https://mathoverflow.net/a/295232/91764">Toeplitz matrix nearest to a given (symmetric) matrix</a>, MathOverflow, March 15, 2018.</p>
</li>
<li><p><a href="https://math.stackexchange.com/q/2414847">Writing <span class="math-container">$(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$</span> as a sum of two squares of quadratic forms</a>, Mathematics, September 3-8, 2017.</p>
</li>
</ul>
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<p>Profile picture courtesy of <a href="https://groups.csail.mit.edu/mac/users/rauch/dapm/talk" rel="nofollow noreferrer">Erik Rauch</a>. For <a href="/questions/tagged/rank-1-matrices" class="post-tag" title="show questions tagged &#39;rank-1-matrices&#39;" aria-label="show questions tagged &#39;rank-1-matrices&#39;" rel="tag" aria-labelledby="tag-rank-1-matrices-tooltip-container">rank-1-matrices</a>, go <a href="https://math.meta.stackexchange.com/a/33289/339790">here</a>. One can label using, say, <code>\label{eq1}</code> and cite using <code>\eqref{eq1}</code>.</p>

Rodrigo de Azevedo