# complete matrices question

• Jun 30th 2007, 05:14 PM
kkoutsothodoros
complete matrices question
I need to prove that if A is complete then S-1AS is complete where S is a nonsingular matrix. A complete matrix is one whose eigenvectors span Rn. Can anyone look over what i've done below. i believe it's correct but i have no confidence. i need a little reassurance. thanks!

S-1AS = B. A is complete. Every eigenvalue of A is also an eigenvalue of B. To show this we suppose Av = λv. Then (S-1A)v = λ(S-1)v. This is equivalent to (S-1ASS-1)v = (S-1AS)( S-1)v = λ(S-1)v. This shows that λ is also an eigenvalue of S-1AS and the eigenvector is S-1v. And so if v is an eigenvector of A the S-1v is an eigenvector of S-1AS. If we consider S-1V where V is the matrix whose columns are eigenvectors of A. Because by assumption A is complete V is invertible (since it's columns are linearly independent). But S-1 is invertible and so therefore is S-1V. The columns of S-1V are S-1vi for each eigenvector of A. But the S-1vi are the eigenvectors of S-1AS. Since S-1V is nonsingular it's columns are linearly independent and therefore span Rn. This is what it means for S-1AS to be complete and we are done.
• Jun 30th 2007, 05:32 PM
kkoutsothodoros
S-1 is the inverse of S. thus we are asked to show that any matrix similar to A is complete.
• Jun 30th 2007, 06:43 PM
JakeD
Quote:

Originally Posted by kkoutsothodoros
I need to prove that if A is complete then S-1AS is complete where S is a nonsingular matrix. A complete matrix is one whose eigenvectors span Rn. Can anyone look over what i've done below. i believe it's correct but i have no confidence. i need a little reassurance. thanks!

S-1AS = B. A is complete. Every eigenvalue of A is also an eigenvalue of B. To show this we suppose Av = λv. Then (S-1A)v = λ(S-1)v. This is equivalent to (S-1ASS-1)v = (S-1AS)( S-1)v = λ(S-1)v. This shows that λ is also an eigenvalue of S-1AS and the eigenvector is S-1v. And so if v is an eigenvector of A the S-1v is an eigenvector of S-1AS. If we consider S-1V where V is the matrix whose columns are eigenvectors of A. Because by assumption A is complete V is invertible (since it's columns are linearly independent). But S-1 is invertible and so therefore is S-1V. The columns of S-1V are S-1vi for each eigenvector of A. But the S-1vi are the eigenvectors of S-1AS. Since S-1V is nonsingular it's columns are linearly independent and therefore span Rn. This is what it means for S-1AS to be complete and we are done.

I think your argument is correct. Good job! I tried summarizing the argument, but this is probably too short as a proof.

Every eigenvalue $\displaystyle \lambda$ of $\displaystyle A$ with eigenvector $\displaystyle v$ is also an eigenvalue of $\displaystyle S^{-1}AS$ with eigenvector $\displaystyle S^{-1}v$. This follows from

\displaystyle \begin{aligned} \lambda (S^{-1} v) &= S^{-1}(\lambda v) \\ &= S^{-1}(Av) \\ &= S^{-1}A(SS^{-1})v \\ &= (S^{-1}AS)( S^{-1}v). \end{aligned}

Let $\displaystyle V$ be the matrix whose columns are the eigenvectors of $\displaystyle A.$ Then the above shows that $\displaystyle W = S^{-1}V$ has columns that are the eigenvectors of $\displaystyle S^{-1}AS.$

A matrix is complete if and only if its matrix of eigenvectors is invertible. So $\displaystyle V$ is invertible and since $\displaystyle S^{-1}$ is invertible, so is $\displaystyle W= S^{-1}V.$ Hence $\displaystyle S^{-1}AS$ is complete.
• Jun 30th 2007, 06:52 PM
kkoutsothodoros
thanks!

how do you get the nice math editing?
• Jun 30th 2007, 07:01 PM
JakeD
Quote:

Originally Posted by kkoutsothodoros
thanks!

how do you get the nice math editing?

You're welcome.

See the LaTeX tutorial.