## Python weird 2d matrix behaviour

Why does this code:

``````matrix = []

for a in range(2):
matrix.append([])
for b in range(2):
matrix[a].append(a)

print(matrix)
``````

Outputs this:

``````[
[0, 0],
[1, 1]
]
``````

But this code:

``````matrix = [[None] * 2] * 2

for a in range(2):
for b in range(2):
matrix[a][b] = a

print(matrix)
``````

Outputs this:

``````[
[1, 1],
[1, 1]
]
``````

In the second example wherever we assing ‘a’ in the matrix it changes to the last value of ‘a’ at the end of the loop. I don’t quite understand this behaviour. If some one can explain it please.

If I want to fill a matrix with the number of the row and column of each cell, what would be the best aproach? I feel like calling .append() it’s not the best when we already know the size of the matrix. Since creating the matrix with:

``````matrix = [[None] * col] * row
``````

and then filling it doesn’t work as I expected. What would be the best approach?

Something like this?

``````matrix = [[str(a)+str(b) for b in range(2)] for a in range(2)]
``````

## Graph implementation in c++ using adjacency matrix

Here is my code for implementing Graph using adjacency matrix. The code should be Object-oriented and there is a method `isPath()` checking if there is a connection between two nodes. Some advice?

``````#include
#include
#include

using namespace std;

class Graph
{
protected:
int value;
int **graphm;

public:
Graph(int value)
{
this->value = value;
graphm = new int*[value];
int k, j;
for (k = 0; k < value; k++)
{
graphm[k] = new int[value];
for (j = 0; j < value; j++)
{
graphm[k][j] = 0;
}
}
}

{
if (head > value || end > value || head < 0 || end < 0)
{
cout << "Invalid edge!n";
}
else
{
graphm[head - 1][end - 1] = 1;
graphm[end - 1][head - 1] = 1;
}
}

void display()
{
int i, p;

for (i = 0; i < value; i++)
{
for (p = 0; p < value; p++)
{
cout << graphm[i][p] << "   ";
}
cout << endl;
}
}

{
int k, o;
ofstream fullstore("Test.txt");

cout << graphm[head - 1][end - 1];
cout << endl;
if (!fullstore.is_open())
{
cout << "File can't be open";
}

if (graphm[head - 1][end - 1] == 1)
{
cout << "There is an edge between " << head << " and " << end << "n";
fullstore << head << ", " << end;
fullstore.close();
}
else
{
}
}
};

int main()
{
int vertex, numberOfEdges, i, head, end;
cout << "Enter number of nodes: ";
cin >> vertex;
numberOfEdges = vertex * (vertex - 1);
Graph g1(vertex);
for (int i = 0; i < numberOfEdges; i++)
{
cout << "Enter edge ex.1 2 (-1 -1 to exit): n";
if ((head == -1) && (end == -1))
{
break;
}
}
g1.display();
cout << endl;
g1.isPath(1, 3);

return 0;
}
``````

## Is a company with Tech Leads a matrix organization?

Employees of my company are grouped into pools depending on technical skills (e.g. front-end devs, back-end devs, QA engineers, etc). For each pool there’s a Tech Lead who doesn’t manage projects (or people) but provides general technical directions to the company: he decides technical stacks, standard solutions, best practices and so on.

To initiate a project, the Head of Delivery selects a PM and staffs a team by taking technicians from the pools (depending on the requirements). The PM has full authority over his colleagues except for the technical decisions.

Is my company still a projectized organization or do the Tech Leads make it a matrix organization?

Thank you very much

## How to write a matrix of \$ntimes 2\$ where the first column is made of only 1’s?

How to write a matrix of $$ntimes 2$$ where the first column is made of only 1’s?

Suppose I have a vector x={.24,…,10}.

How can I get this

$$left( begin{matrix} 1 & .24 \ 1 & 21 \ 1 & 33 \ 1 & 11 \ 1 & 10 end{matrix}right)$$

How can I do that?

I know how to write a ‘normal’ matrix, it’s MatrixForm[{1,2},{1,1}] but how to give the form above?

## How to interpret upper-triangular matrix of cosine similarities

in Spark, there is a `RowMatrix.columnSimilarities()` method (see http://spark.apache.org/docs/latest/api/java/org/apache/spark/mllib/linalg/distributed/RowMatrix.html#columnSimilarities()) that returns “An n x n sparse upper-triangular matrix of cosine similarities between columns of this matrix“.

How should I read it? If I try to implement an example from https://stackoverflow.com/a/1750187 as following:

``````JavaRDD rows = sc.parallelize(Arrays.asList(
new DenseVector(new double[]{2, 1, 0, 2, 0, 1, 1, 1}),
new DenseVector(new double[]{2, 1, 1, 1, 1, 0, 1, 1})
));

RowMatrix mat = new RowMatrix(rows.rdd());
List sims = mat.columnSimilarities().toRowMatrix().rows().toJavaRDD().collect();
for(Vector v: sims) {
System.out.println(v);
}
``````

I get this

``````(8,[6,7],[0.7071067811865475,0.7071067811865475])
(8,[1,2,3,4,5,6,7],[0.9999999999999998,0.7071067811865475,0.9486832980505137,0.7071067811865475,0.7071067811865475,0.9999999999999998,0.9999999999999998])
(8,[2,3,4,5,6,7],[0.7071067811865475,0.9486832980505137,0.7071067811865475,0.7071067811865475,0.9999999999999998,0.9999999999999998])
(8,[7],[0.9999999999999998])
(8,[4,5,6,7],[0.4472135954999579,0.8944271909999159,0.9486832980505137,0.9486832980505137])
(8,[6,7],[0.7071067811865475,0.7071067811865475])
(8,[3,4,6,7],[0.4472135954999579,1.0,0.7071067811865475,0.7071067811865475])
``````

How should I interpret it? How do I get the cosine angle 0.822 from this, as mentioned in the referenced StackOverflow post?

Thanks!

## Homomorphisms of a matrix ring.

I have the following question, What are all possible right $$R$$-module $$(R = mathbb{M}_2(mathbb{Z}))$$ homomorphisms from $$mathbb{M}_2(nmathbb{Z})$$ to $$mathbb{M}_2(mathbb{Z})$$. I tried to extend maps from $$nmathbb{Z}$$ to $$mathbb{Z}$$ defined as $$f_{n,a}(nx) = ax$$ and found that these are right module homomorphisms. But are these the only ones like in case of $$mathbb{Z}$$ ??

## Finding all the solutions of some matrix equations

I am trying to find all possible solutions to the following:

``````A = ( {
{a11, a12},
{a21, a22}
} );
B = ( {
{b11, b12},
{b21, b22}
} );

A.A
``````
``````{{a11^2 + a12 a21, a11 a12 + a12 a22}, {a11 a21 + a21 a22, a12 a21 + a22^2}}
``````
``````A.B
``````
``````{{a11 b11 + a12 b21, a11 b12 + a12 b22}, {a21 b11 + a22 b21, a21 b12 + a22 b22}}
``````
``````B.A
``````
``````{{a11 b11 + a21 b12, a12 b11 + a22 b12}, {a11 b21 + a21 b22, a12 b21 + a22 b22}}
``````
``````B.B
``````
``````{{b11^2 + b12 b21, b11 b12 + b12 b22}, {b11 b21 + b21 b22, b12 b21 + b22^2}}
``````

but I am unable to find all solutions, I have found few solutions using:

``````Solve[
{a11^2 + a12 a21 == a11, a11 a12 + a12 a22 == a12,
a11 a21 + a21 a22 == a21, a12 a21 + a22^2 == a22,
a11 b11 + a12 b21 == b11, a11 b12 + a12 b22 == b12,
a21 b11 + a22 b21 == b21, a21 b12 + a22 b22 == b22,
a11 b11 + a21 b12 == b11, a12 b11 + a22 b12 == b12,
a11 b21 + a21 b22 == b21, a12 b21 + a22 b22 == b22,
b11^2 + b12 b21 == a11, b11 b12 + b12 b22 == a12,
b11 b21 + b21 b22 == a21, b12 b21 + b22^2 == a22},
{a11, a12, a21, a22, b11, b12, b21, b22}]
``````

but not all. Hence, I want to find all solutions.

Also I am trying to solve a similar problem in the 5×5 case. I type the following:

``````A = ( {
{a11, a12, a13, a14, a15},
{a21, a22, a23, a24, a25},
{a31, a32, a33, a34, a35},
{a41, a42, a43, a44, a45},
{a51, a52, a53, a54, a55}
} );
B = ( {
{b11, b12, b13, b14, b15},
{b21, b22, b23, b24, b25},
{b31, b32, b33, b34, b35},
{b41, b42, b43, b44, b45},
{b51, b52, b53, b54, b55}
} );
C = ( {
{c11, c12, c13, c14, c15},
{c21, c22, c23, c24, c25},
{c31, c32, c33, c34, c35},
{c41, c42, c43, c44, c45},
{c51, c52, c53, c54, c55}
} );
D = ( {
{d11, d12, d13, d14, d15},
{d21, d22, d23, d24, d25},
{d31, d32, d33, d34, d35},
{d41, d42, d43, d44, d45},
{d51, d52, d53, d54, d55}
} );
E = ( {
{e11, e12, e13, e14, e15},
{e21, e22, e23, e24, e25},
{e31, e32, e33, e34, e35},
{e41, e42, e43, e44, e45},
{e51, e52, e53, e54, e55}
} );

Solve[
{A.A == A, A.B == B, A.C == C, A.D == D, A.E == E, B.A == B, B.B == C,
B.C == D, B.D == E, B.E == A, C.A == C, C.B == D, C.C == E, C.D == A,
C.E == B, D.A == D, D.B == E, D.C == A, D.D == B, D.E == C, E.A == E,
E.B == A, E.C == B, E.D == C, E.E == D}
``````

But it shows a red thing on the right hand side. What is wrong?