2019-04-12 11:54:01 +0100 commented answer Unflatten a vector and take the transpose then. that works. 2019-04-10 03:43:45 +0100 received badge ● Scholar (source) 2019-04-09 15:22:58 +0100 received badge ● Editor (source) 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  2019-03-07 08:54:39 +0100 received badge ● Supporter (source) 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. 2018-12-31 00:53:25 +0100 received badge ● Student (source) 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. = 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. = 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.