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Recall that an \(n \times n\) matrix \(A\) is diagonalizable if and only if it has \(n\) linearly independent eigenvectors. Moreover, the matrix \(P\) with
these eigenvectors as columns is a diagonalizing matrix for \(A\), that is
\begin{equation*} P^{-1}AP =D \end{equation*}
where \(D\) is a diagonal matrix with diagonal entries
consisting of the eigenvalues of \(A\).
Orthogonal Matrices
A collection of non-zero, pairwise orthogonal vectors in \(\RR ^n\) is called an orthogonal set of vectors. An orthogonal set of vectors is
called orthonormal if \(\norm {\vec {q}} = 1\) for each vector \(\vec {q}\) in the set. A set of orthogonal vectors \(\{\vec {v}_{1}, \vec {v}_{2}, \dots , \vec {v}_{k}\}\) can be “normalized”, i.e. converted into an
orthonormal set \(\left \{ \frac {1}{\norm {\vec {v}_{1}}}\vec {v}_{1}, \frac {1}{\norm {\vec {v}_{2}}}\vec {v}_{2}, \dots , \frac {1}{\norm {\vec {v}_{k}}}\vec {v}_{k} \right \}\). In particular, if a matrix \(A\) has \(n\) orthogonal eigenvectors, they can (by normalizing) be taken to be orthonormal.
The corresponding diagonalizing matrix (we will use \(Q\) instead of \(P\)) has orthonormal columns, and such matrices are very easy
to invert.
The following conditions are equivalent for an \(n \times n\) matrix \(Q\).
1.
\(Q\) is invertible and \(Q^{-1} = Q^{T}\).
2.
The rows of \(Q\) are orthonormal.
3.
The columns of \(Q\) are orthonormal.
First note that condition 1 is equivalent to \(Q^{T}Q = I\). Let \(\vec {q}_{1}, \vec {q}_{2}, \dots , \vec {q}_{n}\) denote the columns of \(Q\). Then \(\vec {q}_{i}^{T}\) is the \(i\)th row of \(Q^{T}\), so the \((i, j)\)-entry of \(Q^{T}Q\) is \(\vec {q}_{i} \dotp \vec {q}_{j}\). Thus \(Q^{T}Q = I\)
means that \(\vec {q}_{i} \dotp \vec {q}_{j} = 0\) if \(i \neq j\) and \(\vec {q}_{i} \dotp \vec {q}_{j} = 1\) if \(i = j\). Hence condition 1 is equivalent to 3. The proof of the equivalence of 1 and 2 is similar.
Orthogonal Matrices An \(n \times n\) matrix \(Q\) is called an orthogonal matrix if its columns are orthonormal vectors. (i.e. The matrix satisfies
one (and hence all) of the conditions in Theorem 1.)
The above definition calls a square matrix with orthonormal columns an orthogonal matrix. It is true that orthonormal matrix
might be a better name. But orthogonal matrix is standard.
The rotation matrix \(\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\) is orthogonal for any angle \(\theta \).
See Practice Problem ??.
Let \(A=\begin{bmatrix} 2&1&1\\-1&1&1\\0&-1&1 \end{bmatrix}\)
1.
Check that matrix \(A\) has rows that are orthogonal.
2.
Check that matrix \(A\) has columns that are NOT orthogonal.
3.
Check that matrix \(A\) has rows that are NOT orthonormal.
4.
Create a matrix \(Q\) by normalizing each of the rows of \(A\).
5.
Check that columns and rows of \(Q\) are orthonormal.
An \(n \times n\) matrix \(A\) is said to be orthogonally diagonalizable if an orthogonal matrix \(Q\) can be found such that \(Q^{-1}AQ = Q^{T}AQ\) is diagonal.
We have learned earlier that when we diagonalize a matrix \(A\), we write \(P^{-1}AP=D\) for some matrix \(P\) where \(D\) is diagonal, and the diagonal
entries are the eigenvalues of \(A\). We have also learned that the columns of the matrix \(P\) are the corresponding eigenvectors of \(A\).
So when a matrix is orthogonally diagonalizable, we are able to accomplish the diagonalization using a matrix \(Q\) consisting of \(n\)
eigenvectors that form an orthonormal basis for \(\RR ^n\). In the following section we will learn that the matrices that have this property
are precisely the symmetric matrices.
Symmetric Matrices
A symmetric matrix is a matrix which is equal to its transpose.
When we began our study of eigenvalues and eigenvectors, we saw examples of matrices with entries that were real
numbers with eigenvalues that were complex numbers. It can be shown that symmetric matrices only have real
eigenvalues.
respectively. Moreover,these eigenvectors are orthogonal. We have \(\norm {\vec {x}_{1}}^{2} = 6\), \(\norm {\vec {x}_{2}}^{2} = 5\), and \(\norm {\vec {x}_{3}}^{2} = 30\), so
Hence the distinct eigenvalues are \(0\) and \(9\) are of algebraic multiplicity \(1\) and \(2\), respectively. The
geometric multiplicities must be the same, for \(A\) is diagonalizable, being symmetric. It follows that \(\mbox {dim}(\mathcal {S}_0) = 1\) and \(\mbox {dim}(\mathcal {S}_9) = 2\). Gaussian elimination
gives
The eigenvectors in \(\mathcal {S}_{9}\) are both orthogonal to \(\vec {x}_{1}\) as Theorem 12 guarantees, but not to each other. However, an orthogonal
basis can be found using projections. (See Gram-Schmidt Orthogonalization)
is
an orthogonal matrix such that \(Q^{-1}AQ\) is diagonal.
It is worth noting that other, more convenient, diagonalizing matrices \(Q\) exist. For example, \(\vec {y}_{2} = \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix}\) and \(\vec {y}_{3} = \begin{bmatrix} -2 \\ 2 \\ 1 \end{bmatrix}\) lie in \(\mathcal {S}_{9}(A)\) and they are orthogonal.
Moreover, they both have norm \(3\) (as does \(\vec {x}_{1}\)), so
