- #1
meboxley
- 1
- 0
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!
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!