Message Boards Message Boards

1
|
534 Views
|
6 Replies
|
8 Total Likes
View groups...
Share
Share this post:

MatrixSymbol simplifications

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?
POSTED BY: Eric Michielssen
6 Replies
Sort By:

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].
POSTED BY: Jose Martin-Garcia

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!

enter image description here
POSTED BY: Eric Michielssen

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].
POSTED BY: Jose Martin-Garcia

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?
POSTED BY: Eric Michielssen

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.
POSTED BY: Jose Martin-Garcia

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}])]
POSTED BY: Gianluca Gorni
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract