Diagonalizability dependent on nullity and inner product?

  • Thread starter meboxley
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  • #1
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Hello everyone!

I am doing some work involves proving that a matrix A (A=U(V^T)) is diagonalizable. In this situation, U and V are non zero vectors in R^n.

So far, I have proven that U is an eigenvector for A, and that the corresponding eigenvalue is the inner product of U and V ((U^T)V). [Here I multiplied both sides by U and then did some simple equation manipulations.] Also, I have confirmed that the nullity of A = n-1 (via proof that rank(A)=1). [Since the rows of A are multiples of the rows of V^T, then the row space of A has dimension 1 --> rank(A)=1.]

I know that I need to use the things that I have already proven, and I also know that U(dot)V cannot equal zero. However, I am a bit stuck on how the information that I have constructed already shows that A will have n linearly independent eigenvectors. (I have also already done enough work to see that this is the only condition possible to be proven that implies that A is diagonalizable - I think!)

Any help with connecting all of this would be great! Thanks!
 

Answers and Replies

  • #2
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You're pretty much there. The nullity is the dimension of the kernel, the set of vectors annihilated by the matrix: in other words, the set of eigenvectors with zero eigenvalue. Since the nullity is n-1, you have n-1 linearly independent zero eigenvectors. You just need one more LI eigenvector to span the space, and any eigenvector with nonzero eigenvalue must by LI of the kernel. And you've got such an eigenvector, so you're done.
 
  • #3
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What about

[tex]UV^T=\left(\begin{array}{c} 1\\ 0\end{array}\right)\left(\begin{array}{cc} 0 & 1\end{array}\right)=\left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right)[/tex]?

This is not a diagonalizable matrix. Or did I misunderstand your question?
 
  • #4
160
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Hidden in the OP is the assumption that U and V are not orthogonal, and this is sufficient. It guarantees that U has a nonzero e'value.
 

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