Can you find the PseudoInverse of this matrix?

Questions
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1

Please execute:

alpha = 4; word = 3;

dict = Partition[Flatten[Tuples[IdentityMatrix[alpha], word]], (alpha*word)];

MatrixForm[dict]

Do you see an easy way to get the PseudoInverse without using Mathematica’s PseudoInverse function?

2

Other methods for finding the PseudoInverse require inverting the product of the matix with its transpose. That’s probably much more complex than the method used by the PseudoInverse function.

3

The product of the transpose of dict with dict is not of full rank and can’t be inverted.

In[26]:= test = {{1, 2}, {3, 5}, {7, 11}};

In[27]:= ps1 = PseudoInverse[test]

Out[27]= {{-(19/7), -(10/7), 8/7}, {12/7, 13/14, -(9/14)}}

In[29]:= ps2 = Inverse[Transpose[test].test].Transpose[test]

Out[29]= {{-(19/7), -(10/7), 8/7}, {12/7, 13/14, -(9/14)}}

In[30]:= alpha = 4; word = 3;

dict = Partition[
   Flatten[Tuples[IdentityMatrix[alpha], word]], (alpha*word)];

In[33]:= Dimensions[Transpose[dict].dict]

Out[33]= {12, 12}

In[35]:= MatrixRank[Transpose[dict].dict]

Out[35]= 10
4

Frank,

alpha=4; word=3 should yield a 64 x 12 matrix. I was trying to see someone would re-invent - or already knew - what you see on page 16 of:

Thanks,
Doug

5

The transpose of dict is 12 x 64, so Transpose[dict].dict is 12 x 12.