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0 pointsgugagore4y ago0 comments

Ah, okay. Sure. And PSD matrix A is diagonalizeable with orthogonal matrices i.e.

  A = Q^T * D * Q

The ellipsoid visualization you meant is insensitive to the Q term (any rotation of a sphere is the same sphere), but not to the Q^T term.

What I am still failing to understand is this sentence from the article:

> The basic QR algorithm can be visualized in the case where A is a positive-definite symmetric matrix.

It sounds like you can visualize the iterates A_{k} no matter what. Is the problem that there isn't a fixed point when the ellipsoid is axis-aligned?

The article has:

> Under certain conditions,[4] the matrices Ak converge to a triangular matrix, the Schur form of A.

So am I to understand that for A positive semi-definite, the A_{k} converges to a diagonal matrix?

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ogogmad4y ago

>> Under certain conditions,[4] the matrices Ak converge to a triangular matrix, the Schur form of A.

> So am I to understand that for A positive semi-definite, the A_{k} converges to a diagonal matrix?

Yes. Each iteration of QR (and LR) turns a PSD matrix into another PSD matrix. A PSD matrix in Schur form is necessarily a diagonal matrix.

>> The basic QR algorithm can be visualized in the case where A is a positive-definite symmetric matrix.

> It sounds like you can visualize the iterates A_{k} no matter what. Is the problem that there isn't a fixed point when the ellipsoid is axis-aligned?

My visualization technique assumes that the matrix is PSD. In the PSD case, there is a one-to-one correspondence between ellipsoids and PSD matrices. I don't know what happens if you apply it to non-PSD matrices, but the one-to-one correspondence gets lost.

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