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Fill in open entries in decompositions table.

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Jitse Niesen 2010-06-30 10:41:23 +01:00
parent 1b8277fc2a
commit 096c13ea6d
1 changed files with 8 additions and 8 deletions
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doc/TopicLinearAlgebraDecompositions.dox
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@ -109,7 +109,7 @@ namespace Eigen {
<td><em>Soon: blocking</em></td>
</tr>
<tr><td colspan="8">\n Singular values and eigenvalues decompositions</td></tr>
<tr><td colspan="9">\n Singular values and eigenvalues decompositions</td></tr>
<tr>
<td>SVD</td>
@ -167,7 +167,7 @@ namespace Eigen {
<td>Yes</td>
<td>Eigenvalues/vectors</td>
<td>-</td>
<td>TODO Jitse answer this</td>
<td>Average</td>
<td>-</td>
</tr>
@ -183,7 +183,7 @@ namespace Eigen {
<td>-</td>
</tr>
<tr><td colspan="8">\n Helper decompositions</td></tr>
<tr><td colspan="9">\n Helper decompositions</td></tr>
<tr>
<td>RealSchur</td>
@ -193,13 +193,13 @@ namespace Eigen {
<td>Yes</td>
<td>-</td>
<td>-</td>
<td>TODO Jitse answer this</td>
<td>Average</td>
<td>-</td>
</tr>
<tr>
<td>ComplexSchur</td>
<td>Square and real</td>
<td>Square</td>
<td>Slow-very slow<sup><a href="#note2">2</a></sup></td>
<td>Depends on condition number</td>
<td>Yes</td>
@ -211,7 +211,7 @@ namespace Eigen {
<tr>
<td>UpperBidiagonalization</td>
<td>rows >= columns</td>
<td>Rows >= columns</td>
<td>Fast</td>
<td>Good</td>
<td>-</td>
@ -250,7 +250,7 @@ namespace Eigen {
\b Notes:
<ul>
<li><a name="note1">\b 1: </a>There exist a couple of variants of the LDLT algorithm. Eigen's one produces a pure diagonal matrix, and therefore it cannot handle indefinite matrix, unlike Lapack's one which produces a block diagonal matrix.</li>
<li><a name="note2">\b 2: </a>Eigenvalues and Schur decompositions rely on iterative algorithms. Their convergence speed depends on how the eigenvalues are well separated.</li>
<li><a name="note2">\b 2: </a>Eigenvalues and Schur decompositions rely on iterative algorithms. Their convergence speed depends on how well the eigenvalues are separated.</li>
</ul>
\section TopicLinAlgTerminology Terminology
@ -267,7 +267,7 @@ namespace Eigen {
In the same vein, it is negative semi-definite if \f$ v^* A v \le 0 \f$ for any non zero vector \f$ v \f$ </dd>
<dt><b>Blocking</b></dt>
<dd>Means the algorithm can work per block, whence guarantying a good scaling of the performance for large matrices.</dd>
<dd>Means the algorithm can work per block, whence guaranteeing a good scaling of the performance for large matrices.</dd>
<dt><b>Meta-unroller</b></dt>
<dd>Means the algorithm is automatically and explicitly unrolled for very small fixed size matrices.</dd>
<dt><b></b></dt>