Debasish - 2 years ago 97

Scala Question

I am trying to inverse a spark rowmatrix. The function I am using is below.

`def computeInverse(matrix: RowMatrix): BlockMatrix = {`

val numCoefficients = matrix.numCols.toInt

val svd = matrix.computeSVD(numCoefficients, computeU = true)

val indexed_U = new IndexedRowMatrix(svd.U.rows.zipWithIndex.map(r => new IndexedRow(r._2, r._1)))

val invS = DenseMatrix.diag(new DenseVector(svd.s.toArray.map(x => if(x == 0) 0 else math.pow(x,-1))))

val V_inv = svd.V.multiply(invS)

val inverse = indexed_U.multiply(V_inv.transpose)

inverse.toBlockMatrix.transpose

}

The logic I am implementing is through SVD. An explanation of the process is

`U, Σ, V = svd(A)`

A = U * Σ * V.transpose

A.inverse = (U * Σ * V.transpose).inverse

= (V.transpose).inverse * Σ.inverse * U.inverse

Now U and V are orthogonal matrix

Therefore,

M * M.transpose = 1

Applying the above,

A.inverse = V * Σ.inverse * U.transpose

Let V * Σ.inverse be X

A.inverse = X * U.transpose

Now, A * B = ((A * B).transpose).transpose

= (B.transpose * A.transpose).transpose

Applying the same, to keep U as a row matrix, not a local matrix

A.inverse = X * U.transpose

= (U.transpose.transpose * X.transpose).transpose

= (U * X.transpose).transpose

The problem is with the input row matrix. For example

`1, 2, 3`

4, 5, 6

7, 8, 9

10,11,12

the inverse from the above code snippet and on using python numpy is different. I am unable to find out why is it so? Is it because of some underlying assumption made during svd calculation? Any help will be greatly appreciated. Thanks.

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Answer Source

The above code works properly. The reason I was getting this error was that I made the RowMatrix with a RDD[Vector]. Now, in spark things are sorted column wise to form a matrix, whereas in the case of numpy, array is converted row wise to a matrix

```
Array(1,2,3,4,5,6,7,8,9)
```

In Spark

```
1 4 7
2 5 8
3 6 9
```

In python, it is interpreted as

```
1 2 3
4 5 6
7 8 9
```

So, the test case was failing :|

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