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### Geometry problem inspired by Babylonian tablets (positive matrix factorization)

quite a mess. Given positive definite quadratic form $a x^2 \pm 2bxy + c y^2$ so that $ac>b^2,$ take $r > 0,$ with $$0 < r < \min \left( \frac{b}{a}, \frac{c}{b} \right)$$ Then ...
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### Why am I getting t is a vector when solving for t in the vector equation of a 3d line.

The trouble comes when you go from $$(x-x_0, y-y_0, z-z_0) = t(a,b,c)$$ to $$t = \frac{(x-x_0,y-y_0,z-z_0)}{(a,b,c)} = \big(\frac{x-x_0}{a}, \frac{y-y_0}{b}, \frac{z-z_0}{c}\big).$$ We cannot divide ...

### Finding how to decompose a polynomial with respect to a polynomials' basis

$x^2+x^3,\; \mathbf{B} = \langle \,\underbrace{1}_{p_1}\,, \,\underbrace{1+x}_{p_2}\,, \,\underbrace{1+x+x^2}_{p_3}\,, \,\underbrace{1+x+x^2+x^3}_{p4}\, \rangle$ This one is "easily solved by ...
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### Determining a value to form a projector

In your second-to-last step, you got rid of the left side but did nothing to the right side, which is a mistake. After that, it needs to be clearer why you're concluding $2x-1=0$ from your projector ...
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### Blend multiple shapes.

Finally got it working. pseudocode: ...
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### Geometry problem inspired by Babylonian tablets (positive matrix factorization)

From a different angle, let $\,u=x + iy\,$, $\,v=w + iz\,$, then the equations are $\,|u|^2=a\,$, $\,|v|^2 = c\,$, $\,\text{Im}(uv) = b\,$. The latter can be written as $\,u v - \bar u \bar v = 2ib\,$,...
1 vote
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Another solution that also uses spaces of matrices, but which is more simple and valid for any field $k$. It is based on the equality: $$\begin{pmatrix} 1\\ 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \... 1 vote ### Invertibility of linear maps (Exercise 3D.4 of Axler's "Linear Algebra Done Right") Call K the kernel (aka null space) of T_1 and T_2. Both ranges T_1V and T_2V are isomorphic to V/K by the first-isomorphism theorem (the rank-nullity theorem is just the dimensional ... 1 vote Accepted ### Compact SVD for A_{i, j} = \sin(i) \cdot \cos(j) They are already given to you. Just take$$a=\sqrt{\sum_{i=0}^{n-1}\sin^2i},\qquad\qquad b=\sqrt{\sum_{j=0}^{n-1}\cos^2j}, U=\frac1a\begin{bmatrix}\sin 0\\ \vdots\\ \sin(n -1)\end{bmatrix},\qquad\...

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