# Tag Info

11

tl;dr If these functions cannot decide, they will simply return False. A False result means that the selected equality testing method wasn't able to prove equality, but it does not mean that it was able to prove inequality. Interpret the result relative to the used SameTest option value. I will try to explain what I think is happening, though some of ...

9

Here's a nice one-liner: TransformationFunction[{{1, 0, 0, 0}, {0, 0, -1, 0}, {0, 1, 0, 3}, {0, 0, 0, 1}}] // InverseFunction TransformationFunction[{{1, 0, 0, 0}, {0, 0, 1, -3}, {0, -1, 0, 0}, {0, 0, 0, 1}}] Note that TransformationFunction[] is the head of the results returned by geometric *Transform functions, which take a homogeneous transformation ...

8

After some investigation I can mostly explain the (correct) behavior. First off, we set this up using machine doubles. After computing the eigensystems we reorder by eigenvalues. While there is the possibility of messing up conjugate pairs (wherein right and left eigenvalues get ordered differently due to small numeric discrepancies), I checked and this ...

7

This is related to the Orderless attribute of Times and Plus. These attributes could be removed permanently with some hacks, but that would break Mathematica. If you only want to display the result in a certain way, but not do calculations with it, it may be safe to remove those attributes temporarily using Block. Block[{Plus, Times}, With[{result = ...

6

If you have a homogenous transformation matrix of the form $$\begin{bmatrix} \mathrm{R_{3 \times 3}} & \mathrm{d}_{3 \times 1} \\ 0_{1\times 3} & 1_{1\times 1} \end{bmatrix}$$ Then the inverse is given by $$\begin{bmatrix} \mathrm{R}^{-1} & -\mathrm{R}^{-1}\mathrm{d} \\ 0 & 1 \end{bmatrix}$$ Therefore, if your ...

4

The formula for spectral norm you are using is meant to be the formal mathematical definition of the quantity. However this is restrictive for practical use as symbolic norm calculation on high dimensions are very cumbersome. The formulation you might be looking for is the following. Here $\mu_{2}$ is the logarithmic two norm. \mu_{2}(A) := ...

2

One possibility is to use a rectangular matrix with one block being the identity matrix. Let us consider the example in block matrix form. \$\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{pmatrix} \cdot \begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} ...

1

n=2; A = ConstantArray[1, {n, n}]; B = ConstantArray[2, {n, n}]; Y = ConstantArray[3, {n, n}]; Z = ConstantArray[0, {n, n}]; ClearAll[a, b, y, z] m = 10; mat=Normal@SparseArray[{{i_,i_}->a,{i_,j_}/;i-j==1->b,{i_,j_}/;j-i 1->y,{i_,j_}/;Abs[i-j]>1->z},{m,m}] ArrayFlatten[mat /. {a -> A, b -> B, y -> Y, z -> Z}]

1

Try constraining the domain to Reals Solve[{a == 0, b == 0}, {M, P},Reals] You may want to look also into Reduce Reduce[{a == 0, b == 0, M > 0, P > 0}, {M, P}, Reals]

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