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I'd like to define a symbolic real positive definite matrix. For the 2 x 2 example, I thought I could define four real variables using $Assumptions = {a,b,c,d} \esc elem \esc Reals. And then define the positive definite matrix using MatrixPD = Transpose[{{a,b},{c,d}}].{{a,b},{c,d}}].

However PositiveDefiniteMatrixQ[MatrixPD] returns False. Is this because, I have not defined the assumptions sufficiently to guarantee that MatrixPD will be positive definite, or is it because of a limitation of PositiveDefiniteMatrixQ.

I also tried the following. The manaul includes the following symbolic example PositiveDefiniteMatrixQ[{{1, a}, {-Conjugate[a], 1}}] Since I had defined "a" to be real, I assumed I could modify this to be PositiveDefiniteMatrixQ[{{1, a}, {-a, 1}}] However, this returned False. Am I not asserting the assumption that "a" is real properly or is something else going wrong?

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    $\begingroup$ Assumptions are only taken into consideration by a few functions and not all; I doubt that PositiveDefiniteMatrixQ does. The docs for PositiveDefiniteMatrixQ say that it returns true only for explicitly positive definite matrices, which I take to mean numerical ones. $\endgroup$
    – MarcoB
    Feb 22, 2021 at 20:40
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    $\begingroup$ @MarcoB Thanks this is helpful. Another step, I notice is that the use of Transpose[A].A requires that A be invertible. So my question reduces to generating a symbolic invertible matrix. $\endgroup$
    – Kenric
    Feb 22, 2021 at 21:01
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    $\begingroup$ I think your assumptions allows for semi definite matrices: E.g. {a, b, c, d}={2, 3, 4, 6}; m={{20, 30}, {30, 45}} $\endgroup$ Feb 22, 2021 at 21:03
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    $\begingroup$ @DanielHuber this is a helpful example. I see that {{2,3},{4,6}} is not invertible, so indeed this is a missing requirement. Do you have a suggestion for generating a symbolic invertible matrix. $\endgroup$
    – Kenric
    Feb 22, 2021 at 21:07
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    $\begingroup$ I'm not sure how general this is but one approach to generating an invertible matrix is to generate lower and upper matrices. Then A = LU is invertible. $\endgroup$
    – Kenric
    Feb 22, 2021 at 21:13

1 Answer 1

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Any positive real matrix has positive eigenvalues, so you can construct the matrix by:

n = 2;
diag = Array[e, n];
b = RandomReal[{-1, 1}, {n, n}];
sym = FullSimplify[Transpose[b] . DiagonalMatrix[diag] . b]

So sym is the general form of the matrix. To be positive definite, you then need ensure only that the n elements of diag (in the n=2 case, this is e[1] and e[2]) are positive.

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