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 January 6th, 2018, 01:03 PM #11 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Rank deficiency has nothing to do with diagonalizability.* Diagonalize $\displaystyle A=\begin{bmatrix} 1 & 2\\ 0& 0 \end{bmatrix}$ eigenvalues 0,1 eigenvectors $\displaystyle \begin{bmatrix} -2\\ 1 \end{bmatrix},\begin{bmatrix} 1\\0\end{bmatrix}$ $\displaystyle V=\begin{bmatrix} -2 &1 \\ 1 & 0 \end{bmatrix}, V^{-1}=\begin{bmatrix} 0& 1\\ 1& 2 \end{bmatrix}, V^{-1}AV=\begin{bmatrix} 0 &0 \\ 0 &1 \end{bmatrix}$ Background If $\displaystyle A$ has a complete set of eigenvalues it is diagonalizable: Let columns of $\displaystyle V$ be eigenvectors of $\displaystyle A$. Since $\displaystyle V$ is non-singular, find $\displaystyle V^{-1}$. Then: $\displaystyle V^{-1}AV = V^{-1}DV = DV^{-1}V = D$ If $\displaystyle A$ is diagonalizable it has a complete set of eigenvectors: $\displaystyle V^{-1}AV = D, AV = VD$ If $\displaystyle c_{i}$ is a column of $\displaystyle V$, $\displaystyle Ac_{i}=c_{i} \lambda_{i}$, $\displaystyle D$ consists of eigenvalues of $\displaystyle A$ and the $\displaystyle c_{i}$ are eigenvectors of $\displaystyle A$. *A defective matrix is one for which geometric multiplicity is less than algebraic multiplicity, in which case it doesn't have a complete set of eigenvalues. see: Eigenvectors of a distinct eigenvalue A great video on diagonalization is: https://www.google.com/search?ei=4hl...93.g16PkcYC8RU January 6th, 2018, 03:26 PM   #12
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Quote:
 Originally Posted by zylo Rank deficiency has nothing to do with diagonalizability.
He's not talking about whether A is rank deficient, but whether the matrices whose columns are eigenvectors of A are rank deficient. In your example, you relied on the fact that the matrix V formed by eigenvectors of A is invertible (i.e. is not rank deficient) to diagonalize A.

The condition that an n x n matrix A doesn't have n linearly independent eigenvectors can be rephrased by saying that every matrix whose columns are eigenvectors of A is rank deficient. Both of these are equivalent to A not being diagonalizable. January 8th, 2018, 05:22 AM #13 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 "In linear algebra, a defective matrix is a square matrix that does not have a complete basis of eigenvectors, and is therefore not diagonalizable. In particular, an n × n matrix is defective if and only if it does not have n linearly independent eigenvectors." "A defective matrix always has fewer than n distinct eigenvalues, since distinct eigenvalues always have linearly independent eigenvectors. In particular, a defective matrix has one or more eigenvalues λ with algebraic multiplicity m > 1 (that is, they are multiple roots of the characteristic polynomial), but fewer than m linearly independent eigenvectors associated with λ." https://en.wikipedia.org/wiki/Defective_matrix A matrix A whose eigenvectors form a rank deficient matrix is quite obtuse. If you didn't have a full complement of eigenvectors, you couldn't , or wouldn't, form such a matrix in the first place. January 8th, 2018, 12:21 PM #14 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Hmmm. For some reason I had wrong link to great video on diagonalization. Try again: Tags diagonal, exists, matrix, similar Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post silviatodorof Linear Algebra 2 March 22nd, 2015 06:28 AM annakar Linear Algebra 0 January 10th, 2013 10:58 AM lamhmh Applied Math 2 July 11th, 2011 01:12 PM

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