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2019-04-12 11:54:01 +0100 commented answer Unflatten a vector

and take the transpose then. that works.

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2019-04-09 15:22:18 +0100 asked a question Unflatten a vector

I have a vector say a $9 \times 1$ vector which looks like $$\begin{bmatrix} 2x +1 \newline x \newline 1 \newline x \newline x^2 + 2x \newline 2x \newline x \newline 2x^2 \newline 0 \newline \end{bmatrix}$$ with entries in $\mathbb{F}_3[x]$.

There are 9 rows in this matrix and i want to write a function which takes an $n^2 \times 1$ matrix and turns it into a $n \times n$ matrix. So, in this case, we want the function would turn the above vector into

$$ \begin{bmatrix} 2x+1 & x & 1 \newline x & x^2+2x & 2x \newline x & 2x^2 & 0 \newline \end{bmatrix} $$

In this Sage, I tried this:

sage: v = Matrix(GF(3)[x], [[2*x+1],[x],[1],[x],[x^2+2*x],[2*x],[x],[2*x^2],[0]])

Then, write the following:

sage: Matrix(v.base_ring(), 3, 3, v)

This gave an error:

inconsistent number of rows: should be 3 but got 1
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2019-03-06 03:44:27 +0100 asked a question What an efficient way to construct a n by n matrix with all entries -1?

you could write create an empty list and then, create a column of -1 and augment it to the previous columns, do this n times but is there a faster way?

2018-12-31 05:58:38 +0100 commented question Finding the kernel of a matrix in a non-integral domain

a and b are in F3. yes, i should have mentioned.

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2018-12-30 22:19:33 +0100 asked a question Finding the kernel of a matrix in a non-integral domain

I have been trying to find the kernel of the matrix in a quotient, for example. If we have the following quotient ring in sage:

R.<t> = PolynomialRing(GF(3),'t')
I = R.ideal([t^3])
S = R.quotient_ring(I);

and if I try to find the kernel of the matrix:

E = Matrix(S, ([[0+a*t+b*t^2, 1+a*t+b*t^2, 0+a*t+b*t^2, 0+a*t+b*t^2],
                        [0+a*t+b*t^2, 0+a*t+b*t^2, 0+a*t+b*t^2, 0+a*t+b*t^2],
                        [0+a*t+b*t^2, 0+a*t+b*t^2, 0+a*t+b*t^2, 1+a*t+b*t^2],
                        [0+a*t+b*t^2, 0+a*t+b*t^2,  0+a*t+b*t^2,0+a*t+b*t^2]]))
E.kernel()

It gives me the following error: NotImplementedError.

I guess this is because F3[x]/(x^3) is not an integral domain but I would like a way around it.

Thanks in advance.

2018-12-30 22:19:33 +0100 asked a question Finding the kernel of a non-integral domain

I have been trying to find the kernel of the matrix in a quotient, for example. If we have the following quotient ring in sage:

R.<t> = PolynomialRing(GF(3),'t') I = R.ideal([t^3]) S = R.quotient_ring(I);

and if I try to find the kernel of the matrix:

E = Matrix(S, ([[0+at+bt^2, 1+at+bt^2, 0+at+bt^2, 0+at+bt^2], [0+at+bt^2, 0+at+bt^2, 0+at+bt^2, 0+at+bt^2], [0+at+bt^2, 0+at+bt^2, 0+at+bt^2, 1+at+bt^2], [0+at+bt^2, 0+at+bt^2, 0+at+bt^2,0+at+bt^2]])) E.kernel()

It gives me the following error: NotImplementedError.

I guess this is because F3[x]/(x^3) is not an integral domain but I would like a way around it.

Thanks in advance.