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	<title>Comments on: linear algebra — are eigenvectors orthogonal?</title>
	<link>http://linux.dsplabs.com.au/linear-algebra-are-eigenvectors-orthogonal-p84/</link>
	<description>Diary of my Linux journeys. Everything Linux by Kamil Wójcicki</description>
	<pubDate>Thu, 09 Feb 2012 21:50:04 +0000</pubDate>
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		<title>By: Kamil</title>
		<link>http://linux.dsplabs.com.au/linear-algebra-are-eigenvectors-orthogonal-p84/#comment-1563</link>
		<author>Kamil</author>
		<pubDate>Mon, 31 Oct 2011 08:41:08 +0000</pubDate>
		<guid>http://linux.dsplabs.com.au/linear-algebra-are-eigenvectors-orthogonal-p84/#comment-1563</guid>
		<description>Thank you for the correction! I have updated the post.</description>
		<content:encoded><![CDATA[<p>Thank you for the correction! I have updated the post.</p>
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		<title>By: Anonymous</title>
		<link>http://linux.dsplabs.com.au/linear-algebra-are-eigenvectors-orthogonal-p84/#comment-1559</link>
		<author>Anonymous</author>
		<pubDate>Mon, 24 Oct 2011 04:34:41 +0000</pubDate>
		<guid>http://linux.dsplabs.com.au/linear-algebra-are-eigenvectors-orthogonal-p84/#comment-1559</guid>
		<description>"It follows that eigenvectors of a symmetric real matrix A (i.e., A=A^T) are perpendicular"

This is incorrect, the guarantee is that an orthogonal set of eigenvectors exist for symmetric matrices, but it is also possible to choose a non-orthogonal basis for subspaces corresponding to eigenvalues with multiplicity &#62; 1.  For example, the identity matrix in R^2 is symmetric, and all vectors in R^2 are its eigenvectors, we could choose any non-orthogonal pair to form the basis for the eigenvectors - if a given algorithm returned such an eigenvectors basis, this is perfectly correct, but we users may incorrectly assume basis is orthogonal.</description>
		<content:encoded><![CDATA[<p>&#034;It follows that eigenvectors of a symmetric real matrix A (i.e., A=A^T) are perpendicular&#034;</p>
<p>This is incorrect, the guarantee is that an orthogonal set of eigenvectors exist for symmetric matrices, but it is also possible to choose a non-orthogonal basis for subspaces corresponding to eigenvalues with multiplicity &gt; 1.  For example, the identity matrix in R^2 is symmetric, and all vectors in R^2 are its eigenvectors, we could choose any non-orthogonal pair to form the basis for the eigenvectors - if a given algorithm returned such an eigenvectors basis, this is perfectly correct, but we users may incorrectly assume basis is orthogonal.</p>
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