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0 pointsgugagore4y ago0 comments

> I actually had the latter one in mind: The one given by {Ax | x^T x = 1}.

I am near certain that you did not mean this.

> When the semi-axes are parallel to the coordinate axes, the matrix is diagonal.

Let D =

the matrix

  | 2 0 | 
  | 0 1 |

And let R be any 2-by-2 rotation matrix.

A = D * R would give you a locus of points (under {Ax | x^T x = 1}) that has the semi-axes parallel to the coordinate axes. In general, A is not diagonal.

For the condition you mention (axis-aligned semi-axes of ellipsoid iff matrix is diagonal) is true for {x | x^T * A * x = 1} .

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ogogmad4y ago

I only consider positive semi-definite symmetric matrices. A = D * R is almost certainly not such a matrix. The positive semi-definite symmetric case is the only one needed to compute SVDs.

A proof of convergence for the algorithm is only known for certain families of matrices. These include the positive-definite symmetric matrices, all symmetric matrices (once suitable improvements to the algorithm are made), the Hermitian matrices, and perhaps some others. But a proof that's valid for all matrices isn't known, even though the algorithm (when improved using Wilkinson shifts) appears to converge everywhere.

gugagoreOP4y ago

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?

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.

j / k navigate · click thread line to collapse

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ogogmad4y ago

I only consider positive semi-definite symmetric matrices. A = D * R is almost certainly not such a matrix. The positive semi-definite symmetric case is the only one needed to compute SVDs.

gugagoreOP4y ago

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?

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?

j / k navigate · click thread line to collapse