Creating anti-diagonal identity matrix

Question

I have an image that is stored in a DenseMatrix, using MathNet Numerics.

For rotating the image by 90 degrees counter-clockwise I want to get the transpose and then flip the result vertically by multiplying by the anti-diagonal identity matrix.

Is there a quick way to initialize that identity matrix?

For a 2x2 matrix that would look like:

0 1 
1 0

Update:

I ended up doing pretty much what @Joseph suggested. Turns out to be sufficiently fast.

public static Matrix<double> CreateAntiIdentityMatrix(int n)
{
    var output = Matrix<double>.Build.Dense(n, n, 0);
    for (int i = 0; i <= n - 1; i++)
    {
        output[i, n - i - 1] = 1;
    }
    return output;
}

score 2 · Accepted Answer · edited Aug 01 '19 at 15:12

2

Something like this should work:

var M = MathNet.Numerics.LinearAlgebra.Double.Matrix.Build.Dense(N, N, 0);
for (i = 0; i <= N - 1; i++)
{
    M(i, N - i - 1) = 1;
}

edited Aug 01 '19 at 15:12

xdtTransform

1,986
14
34

answered Jun 18 '18 at 14:25

Joseph

380
3
16

score 0 · Answer 2 · answered Aug 01 '19 at 14:46

@Joseph's way is fast. But I'd like to introduce a way which is expressively shows MathNet functionality:

var size = 3;
var diagonal = DenseMatrix.CreateDiagonal(size, size, 1);
Console.WriteLine(diagonal);

var reversedColumns = diagonal.EnumerateColumns().Select(c => c.Reverse());
var anti = DenseMatrix.OfColumns(reversedColumns);
Console.WriteLine(anti);

To get an anti-diagonal matrix one can take a diagonal one and reflect it by width (reverse columns).

Result is:

DenseMatrix 3x3-Double
1  0  0
0  1  0
0  0  1

DenseMatrix 3x3-Double
0  0  1
0  1  0
1  0  0

Creating anti-diagonal identity matrix

2 Answers2