MatrixSymbol simplifications

Questions
,
1

I’m exploring the new 14.1 MatrixSymbol functionality, hoping to eventually simplify complex and lengthy matrix expressions. But I’m running into limitations even with simple expressions. For example:

A = MatrixSymbol["A", {n, n}];
FullSimplify[A + Transpose[A], Assumptions -> A == Transpose[A]]
Out[18]= 2 Transpose[(MatrixSymbol["A", {n, n}])]
(**  OK !!  **)

FullSimplify[A - Transpose[A], Assumptions -> A == Transpose[A]]
Out[19]= 0   
(** should we not see O_n,n ? **)

FullSimplify[Inverse[A] . Transpose[A], Assumptions -> A == Transpose[A]]
Out[20]= Inverse[(MatrixSymbol[
    "A", {n, n}])] . Transpose[(MatrixSymbol["A", {n, n}])]  
 (**I was hoping to see an identity matrix **)

What am I missing?

2

Somehow from

FullSimplify[A + Transpose[A], Assumptions -> A == Transpose[A]]

I would expect

2 MatrixSymbol["A", {n, n}]

which is simpler than

2 Transpose[(MatrixSymbol["A", {n, n}])]
3

I think you have two main alternatives:

  1. Use a MatrixSymbol object with explicit symmetry in its two levels:

    In[1]:= A = MatrixSymbol[“A”, {n, n}, Symmetric[{1, 2}]]

    In[2]:= A + Transpose[A] // TensorReduce
    Out[2]= 2 MatrixSymbol[“A”, {n, n}, Complexes, Symmetric[{1, 2}]]

    In[3]:= A - Transpose[A] // TensorReduce
    Out[3]= SymbolicZerosArray[{n, n}]

    In[4]:= Inverse[A] . Transpose[A] // TensorReduce
    Out[4]= MatrixPower[MatrixSymbol[“A”, {n, n}, Complexes, Symmetric[{1, 2}]], 0]

  2. or declare assumptions for a symbol B (I use a separate symbol here to avoid Element resolving to True immediately):

    In[5]:= $Assumptions = Element[B, Arrays[{n, n}, Symmetric[{1, 2}]]];

    In[6]:= B + Transpose[B] // TensorReduce
    Out[6]= 2 B

    In[7]:= B - Transpose[B] // TensorReduce
    Out[7]= SymbolicZerosArray[{n, n}]

    In[8]:= Inverse[B] . Transpose[B] // TensorReduce
    Out[8]= MatrixPower[B, 0]

Though correct, MatrixPower[squarematrix, 0] should be converted into IdentityMatrix[n].

4

Thanks Jose – this is very helpful. A follow-up question, extending the above… Would it be possible using Mathematica 14.1 capabilities to automatically carry out the calculation on the 4th line below, eliminating b, y, A, and Inv[A] in favor of x and c? Here A is an n x n matrix, b, c, x and y are n-vectors. It seems that to do this one needs (Full)Simplify/Eliminate/Reduce type commands that work directly on matrices. Thanks!

Screenshot2024-10-01191914.jpg
Screenshot2024-10-01191914.jpg227×300 9.43 KB

5

I think this type of computation requires guidance, mainly to specify what needs to be replaced by what. For example, I’d do something like this:

$Assumptions = {
   Element[A, Matrices[{n, n}, Symmetric[{1, 2}]]],
   Element[x | y | b | c, Vectors[{n}]]
};

rules = {
   x -> Inverse[A] . b,
   y -> Inverse[A] . c
};

In[3]:= b . y == c . x /. rules // TensorReduce
Out[3]= True

Note that the traditional notation that transposes the vector u on the LHS of a two-vector inner product uT . v can be represented more easily in WL as Dot[u, v].

6

Thanks Jose, I appreciate it. While this is very useful, the fact that I need to know the result before verifying it is a limiting factor.

I think it would be great if the WL (Full)Simplify and Eliminate commands could be extended to transform matrix statements. Perhaps LLMs can help? Just for the heck, I entered the following prompt in Claude.ai:

Prompt: Consider an nxn matrix A, and n-vectors b, c, x and y. Assume that A.x = b, A.y = c, and Transpose(A) = A. Develop an alternative expression for Transpose[b].y that does not involve b or y, and is as simple as possible, avoiding A if at all possible.

Response: Let’s approach this step-by-step:

We’re given that A.x = b
This means b = A.x
We’re also given that A.y = c
This means y = A^(-1).c (assuming A is invertible)
We’re told that Transpose(A) = A
This means A is symmetric
We need to find an expression for Transpose[b].y
Let’s start with Transpose[b].y:
Transpose[b].y = (A.x)^T.y (substituting from step 1)
= x^T.A^T.y (property of matrix transpose)
= x^T.A.y (because A is symmetric)
= x^T.c (substituting from step 2)

Therefore, Transpose[b].y = x^T.c

Perhaps WL can call an LLM under the hood to produce candidate results that next are (automatically) verified using the 14.1 approach you described?

7

Thank you for your suggestions. Simplification of symbolic array expressions is currently being greatly improved for version WL 14.2. We will explore the type of computation you propose.