January 6th, 2018, 01:03 PM  #11 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 
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 nonsingular, 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 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 112 Thanks: 35 Math Focus: Algebraic Number Theory, Arithmetic Geometry  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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 
"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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 
Hmmm. For some reason I had wrong link to great video on diagonalization. Try again: 

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