A First Course in Linear Algebra, 2017a

7.4. Orthogonality 401
In the above solution, the repeated eigenvalue implies that there would have been many other orthonormal
bases which could have been obtained. While we chose to take z = 0,y = 1, we could just as easily
have taken y = 0oreveny = z = 1. Any such change would have resulted in a different orthonormal set.
Recall the following definition.
Definition 7.59: Diagonalizable
An n × n matrix A is said to be non defective or diagonalizable if there exists an invertible matrix
P such that P −1 AP = D where D is a diagonal matrix.
As indicated in Theorem 7.54 if A is a real symmetric matrix, there exists an orthogonal matrix U
such that U T AU = D where D is a diagonal matrix. Therefore, every symmetric matrix is diagonalizable
because if U is an orthogonal matrix, it is invertible and its inverse is U T . In this case, we say that A is
orthogonally diagonalizable. Therefore every symmetric matrix is in fact orthogonally diagonalizable.
The next theorem provides another way to determine if a matrix is orthogonally diagonalizable.
Theorem 7.60: Orthogonally Diagonalizable
Let A be an n×n matrix. Then A is orthogonally diagonalizable if and only if A has an orthonormal
set of eigenvectors.
Recall from Corollary 7.55 that every symmetric matrix has an orthonormal set of eigenvectors. In fact
these three conditions are equivalent.
In the following example, the orthogonal matrix U will be found to orthogonally diagonalize a matrix.
Example 7.61: Diagonalize a Symmetric Matrix
⎡ ⎤
1 0 0
Let A = ⎢ 0 3 1
2 2
⎣
0 1 2
3
2
⎥
⎦ . Find an orthogonal matrix U such that U T AU is a diagonal matrix.
Solution. In this case, the eigenvalues are 2 (with multiplicity one) and 1 (with multiplicity two). First
we will find an eigenvector for the eigenvalue 2. The appropriate augmented matrixandresultingreduced
row-echelon form are given by
⎡
⎢
⎣
2 − 1 0 0 0
0 2− 3 2
− 1 2
0
0 − 1 2
2 − 3 2
0
⎤
⎡
⎥
⎦ →···→ ⎣
1 0 0 0
0 1 −1 0
0 0 0 0
⎤
⎦
and so an eigenvector is
⎡
⎣
0
1
1
⎤
⎦