Real Symmetric Matrix Part 2

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December 30th 2009, 09:07 PM
sym0110

Real Symmetric Matrix Part 2

This is a more general case on Real Symmetric Matrices(RSM)
What is sufficient knowledge of a RSM's eigenvalue and eigenvector that is required to define a RSM uniquely?
For example, suppose we know that A is a 3 by 3 RSM, and it has eigenvalues t=2,t=-1,t=-1, and when t=2, its corresponding eigenvector is (1,1,1)^T
can we define A uniquely? And if so, how?
Thx for any reply or discussion
December 31st 2009, 05:03 AM
HallsofIvy

Quote:

Originally Posted by sym0110

This is a more general case on Real Symmetric Matrices(RSM)
What is sufficient knowledge of a RSM's eigenvalue and eigenvector that is required to define a RSM uniquely?
For example, suppose we know that A is a 3 by 3 RSM, and it has eigenvalues t=2,t=-1,t=-1, and when t=2, its corresponding eigenvector is (1,1,1)^T
can we define A uniquely? And if so, how?
Thx for any reply or discussion

Certainly just knowing one eigenvector is not enough. If you knew three independent eigenvectors, then you would know that, using those eigenvectors as basis, the matrix is $\begin{bmatrix}2 & 0 & 0\\ 0 & -1 & 0\\0 & 0 & -1\end{bmatrix}$ and you could then use the known eigenvectors to write A in the standard basis.

But since -1 is a double eigenvalue, it is quite possible that there are not two independent eigenvectors for the eigenvalue -1. In that case, the matrix could be written in Jordan Normal Form, as $\begin{bmatrix}2 & 0 & 0 \\ 0 & -1 & 1\\0 & 0 & -1\end{bmatrix}$ in some basis but knowing only two eigenvectors would not give you the matrix in the standard basis.
December 31st 2009, 06:48 AM
sym0110

Quote:

Originally Posted by HallsofIvy

Certainly just knowing one eigenvector is not enough. If you knew three independent eigenvectors, then you would know that, using those eigenvectors as basis, the matrix is $\begin{bmatrix}2 & 0 & 0\\ 0 & -1 & 0\\0 & 0 & -1\end{bmatrix}$ and you could then use the known eigenvectors to write A in the standard basis.

But since -1 is a double eigenvalue, it is quite possible that there are not two independent eigenvectors for the eigenvalue -1. In that case, the matrix could be written in Jordan Normal Form, as $\begin{bmatrix}2 & 0 & 0 \\ 0 & -1 & 1\\0 & 0 & -1\end{bmatrix}$ in some basis but knowing only two eigenvectors would not give you the matrix in the standard basis.

HallsofIvy, surely for real symmetric matrices, the plane of eigenvalues -1,-1 would be perpendicular to that of 2? Does that then uniquely defines what the matrix has to be?