Thread: Real Symmetric Matrix Part 2

1. 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

2. 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.

3. 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?