RCT-100 0 converse. If it is true, prove it. If it is false, give a counterexample. (5pt)
This Theorem 1.3.7: For any given matrix A, if A is a normal matrix, then A is an orthonormal matrix.
I needed to verify if the statement "For any given matrix A, if A is a normal matrix, then A is an orthonormal matrix" is true or false. If it is true, I wanted to prove it. If it is, provide a counterexample. I began by analyzing what each of these terms means and the relationship between them.
First, I needed to understand what normal and orthonormal matrices are.
I knew that in linear algebra, a normal matrix is one that commutes with its conjugate transpose, meaning AA* = A*A, where A* is the conjugate transpose of A. Orthonormal matrices, on the other, are matrices whose rows and columns are orthonormal vectors, meaning they are unit vectors and that are orthogonal to each other.
I also noted that every orthonormal matrix is normal, but not every normal matrix is orthonormal.
Given this, the statement seems to be false. A normal matrix doesn't have to be orthonormal. I need to provide a counterexample to show this.
I thought of matrices that are normal but not orthonormal. One choice is a diagonal matrix with distinct diagonal entries. Since it's diagonal, it commutes with its conjugate transpose, so it's normal. But its rows and columns aren't unit vectors, so it's not orthonormal.
Example: Consider the diagonal matrix A = | 3 0 |
| 0 4 |
Since A is diagonal, AA* = A*A = | 3 0 || 3 0 | = | 9 0 | = | 0 4 || 0 4 | | 0 16 | and A*A = | 3 0 || 3 0 | = | 9 0 | = | 0 4 || 0 4 | | 0 16 |, so AA* = A*A, proving that A is normal.
But the rows and columns of A are not unit vectors, e.g., the first row = [3, 0] has norm √(3² + 0²) = 3, not 1. So A is not orthonormal. This proves that the statement is false.
So, the statement is false, and the counterexample is A = | 3 0 |
| 0 4 |.
But to make sure, I checked if A is normal and if A is orthonormal.
AA* = | 3 0 || 3 0 | = | 9 0 | = | 0 4 || 0 4 | | 0 16 | and A*A = | 3 0 || 3 0 | = | 9 0 | = | 0 4 || 0 4 | | 0 16 |, so AA* = A*A, proving that A is normal.
Now, is A orthonormal? The first row is [3, 0], which has norm √(3² + 0²) = 3, not 1. So the first row is not a unit vector. Therefore, A is not orthonormal.
This confirms that the statement is false, and the counterexample is valid.
Answer: So, the statement is false, and the counterexample is A = | 3 0 |
| 0 4 |.
4月 19日 2009年
English Subtitles
中文字幕
日本語字幕
Subtitle Indonesia
Deutsche Untertitel
Sous-titres Français
