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Coordinated Linear Algebra

Anna Davis, Paul Zachlin, Allan Donsig, Levi Heath, Mikkel Munkholm

Section 10.4 Orthogonal Matrices and Symmetric Matrices

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 \(A\text{,}\) that is
\begin{equation*} P^{-1}AP \mbox{ is diagonal.} \end{equation*}
As we have seen, the nice bases of \(\R^n\) are the orthogonal ones, so a natural question is: which \(n \times n\) matrices have \(n\) orthogonal eigenvectors, so that columns of \(P\) form an orthogonal basis for \(\R^n\text{?}\) These turn out to be precisely the symmetric matrices (matrices for which \(A=A^T\)), and this is the main result of this section.

Subsection 10.4.1 Orthogonal Matrices

Recall that an orthogonal set of vectors is called orthonormal if \(\norm{\mathbf{q}} = 1\) for each vector \(\mathbf{q}\) in the set, and that any orthogonal set \(\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{k}\}\) can be ``normalized’’, i.e. converted into an orthonormal set
\begin{equation*} \left\{ \frac{1}{\norm{\mathbf{v}_{1}}}\mathbf{v}_{1}, \frac{1}{\norm{\mathbf{v}_{2}}}\mathbf{v}_{2}, \dots, \frac{1}{\norm{\mathbf{v}_{k}}}\mathbf{v}_{k} \right\}. \end{equation*}
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.

Proof.

Note that condition Item 1 is equivalent to \(Q^{T}Q = I\text{.}\) Let \(\mathbf{q}_{1}, \mathbf{q}_{2}, \dots, \mathbf{q}_{n}\) denote the columns of \(Q\text{.}\) Then \(\mathbf{q}_{i}^{T}\) is the \(i\)th row of \(Q^{T}\text{,}\) so the \((i, j)\)-entry of \(Q^{T}Q\) is \(\mathbf{q}_{i} \cdot \mathbf{q}_{j}\text{.}\) Thus \(Q^{T}Q = I\) means that \(\mathbf{q}_{i} \cdot \mathbf{q}_{j} = 0\) if \(i \neq j\) and \(\mathbf{q}_{i} \cdot \mathbf{q}_{j} = 1\) if \(i = j\text{.}\) Hence Item 1 is equivalent to Item 3.
The proof of the equivalence of Item 1 and Item 2 is similar.

Definition 10.4.2. Orthogonal Matrices.

An \(n \times n\) matrix \(Q\) is called an orthogonal matrix if it satisfies one (and hence all) of the conditions in Theorem 10.4.1.
Rotation matrices are perhaps the most canonical examples of orthogonal matrices, as discussed below.

Example 10.4.3.

The rotation matrix
\begin{equation*} \begin{bmatrix} \cos\theta \amp -\sin\theta \\ \sin\theta \amp \cos\theta \end{bmatrix} \end{equation*}
is orthogonal for any angle \(\theta\text{.}\)
Answer.
It is not enough that the rows of a matrix \(A\) are merely orthogonal for \(A\) to be an orthogonal matrix. Here is an example.

Exploration 10.4.1.

Let
\begin{equation*} A=\begin{bmatrix} 2\amp 1\amp 1\\-1\amp 1\amp 1\\0\amp -1\amp 1 \end{bmatrix}. \end{equation*}
Problem 10.4.5.
  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 \(A\text{.}\)
  5. Check that \(Q\) is an orthogonal matrix.
Answer.
You should get
\begin{equation*} Q = \begin{bmatrix} \frac{2}{\sqrt{6}} \amp \frac{1}{\sqrt{6}} \amp \frac{1}{\sqrt{6}} \\ \frac{-1}{\sqrt{3}} \amp \frac{1}{\sqrt{3}} \amp \frac{1}{\sqrt{3}} \\ 0 \amp \frac{-1}{\sqrt{2}} \amp \frac{1}{\sqrt{2}} \end{bmatrix}. \end{equation*}
This exploration can certainly be done by hand (although it takes some time), but feel free to use an online calculator, MathLab or any other programme.
We studied the idea of closure studying subspaces of \(\R^n\text{.}\) The next theorem tells us that orthogonal matrices are closed under matrix multiplication.

Subsection 10.4.2 Symmetric Matrices

We now shift our focus from orthogonal matrices to another important class of \(n \times n\) matrices called symmetric matrices. A symmetric matrix is a matrix which is equal to its transpose. We saw a few examples such when taking transpose matrices.
When we began our study of eigenvalues and eigenvectors, we saw numerous 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. We also learned that some matrices are diagonalizable while other matrices are not. It turns out that every symmetric matrix is diagonalizable. In fact, we can say more, but first we need the following definition.

Definition 10.4.7.

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 \(A\text{,}\) we write \(P^{-1}AP=D\) for some matrix \(P\) where \(D\) is diagonal, and the diagonal entries are the eigenvalues of \(A\text{.}\) We have also learned that the columns of the matrix \(P\) are the corresponding eigenvectors of \(A\text{.}\) 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 \(\R^n\text{.}\)
The following remarkable theorem shows that the matrices that have this property are precisely the symmetric matrices.

Proof.

If \(A\) is orthogonally diagonalizable, then it is an easy exercise to prove that it is symmetric. You are asked to do this in Exercise 10.4.3.1.
To prove the ``only if’’ part of this theorem, we assume \(A\) is symmetric, and we need to show it is orthogonally diagonalizable.
We proceed by induction on \(n\text{,}\) the size of the symmetric matrix. If \(n = 1\text{,}\) \(A\) is already diagonal. If \(n \gt 1\text{,}\) assume that we know the ``only if’’ statement holds for \((n - 1) \times (n - 1)\) symmetric matrices. Let \(\lambda_{1}\) be an eigenvalue of \(A\text{,}\) and let \(A\mathbf{x}_{1} = \lambda_{1}\mathbf{x}_{1}\text{,}\) where \(\norm{\mathbf{x}_{1}} = 1\text{.}\) Next, set \(\mathbf{q}_{1}=\mathbf{x}_{1}\text{,}\) and use the Gram-Schmidt algorithm to find an orthonormal basis \(\{\mathbf{q}_{1}, \mathbf{q}_{2}, \dots, \mathbf{q}_{n}\}\) for \(\R^n\text{.}\) Let
\begin{equation*} Q_{1} = \begin{bmatrix} | \amp | \amp \amp | \\ \mathbf{q}_1 \amp \mathbf{q}_2 \amp \cdots \amp \mathbf{q}_n \\ | \amp | \amp \amp | \end{bmatrix}, \end{equation*}
so that \(Q_{1}\) is an orthogonal matrix. We have
\begin{align*} Q_{1}^TAQ_{1} \amp = \begin{bmatrix} -- \amp \mathbf{q}_{1}^T \amp -- \\ -- \amp \mathbf{q}_{2}^T \amp -- \\ \amp \vdots \amp \\ -- \amp \mathbf{q}_{n}^T \amp -- \end{bmatrix} A \begin{bmatrix} | \amp | \amp \amp | \\ \mathbf{q}_1 \amp \mathbf{q}_2 \amp \cdots \amp \mathbf{q}_n \\ | \amp | \amp \amp | \end{bmatrix} \\ \amp = \begin{bmatrix} -- \amp \mathbf{q}_{1}^T \amp -- \\ -- \amp \mathbf{q}_{2}^T \amp -- \\ \amp \vdots \amp \\ -- \amp \mathbf{q}_{n}^T \amp -- \end{bmatrix} \begin{bmatrix} | \amp | \amp \amp | \\ A\mathbf{q}_1 \amp A\mathbf{q}_2 \amp \cdots \amp A\mathbf{q}_n \\ | \amp | \amp \amp | \end{bmatrix} \\ \amp = \begin{bmatrix} \lambda_{1} \amp B \\ \mathbf{0} \amp A_{1} \end{bmatrix}, \end{align*}
where the block \(B\) has dimensions \(1 \times (n-1)\text{,}\) and the block under \(\lambda_1\) is a \((n-1) \times 1\) zero matrix, because of the orthogonality of the basis.
Next, using the fact that \(A\) is symmetric, we notice that
\begin{equation*} (Q_{1}^TAQ_{1})^T = Q_{1}^T A^T (Q_{1}^T)^T = Q_{1}^TAQ_{1}, \end{equation*}
so \(Q_{1}^TAQ_{1}\) is symmetric. It follows that \(B\) is also a zero matrix and that \(A_{1}\) is symmetric. Since \(A_{1}\) is an \((n - 1) \times (n - 1)\) symmetric matrix, we may apply the inductive hypothesis, so there exists an \((n - 1) \times (n - 1)\) orthogonal matrix \(Q\) such that \(Q^{T}A_{1}Q = D_{1}\) is diagonal. We observe that
\begin{equation*} Q_{2} = \begin{bmatrix} 1 \amp 0\\ 0 \amp Q \end{bmatrix} \end{equation*}
is orthogonal, and we compute:
\begin{align*} (Q_{1}Q_{2})^TA(Q_{1}Q_{2}) \amp = Q_{2}^T(Q_{1}^TAQ_{1})Q_{2} \\ \amp = \begin{bmatrix} 1 \amp 0 \\ 0 \amp Q^T \end{bmatrix} \begin{bmatrix} \lambda_{1} \amp 0 \\ 0 \amp A_{1} \end{bmatrix}\begin{bmatrix} 1 \amp 0 \\ 0 \amp Q \end{bmatrix} \\ \amp = \begin{bmatrix} \lambda_{1} \amp 0 \\ 0 \amp D_{1} \end{bmatrix} \end{align*}
is diagonal. Because \(Q_{1}Q_{2}\) is orthogonal by Theorem 10.4.6Item 1, this completes the proof.
Because the eigenvalues of a real symmetric matrix are real, Theorem 10.4.8 is also called the Real Spectral Theorem, and the set of distinct eigenvalues is called the spectrum of the matrix. A similar result holds for matrices with complex entries, which we avoid digging into.

Example 10.4.9.

Find an orthogonal matrix \(Q\) such that \(Q^{-1}AQ\) is diagonal, where
\begin{equation*} A = \begin{bmatrix} 1 \amp 0 \amp -1 \\ 0 \amp 1 \amp 2 \\ -1 \amp 2 \amp 5 \end{bmatrix}\text{.} \end{equation*}
Answer.
The characteristic polynomial of \(A\) is (adding twice row 1 to row 2):
\begin{equation*} c_{A}(z) = \det \begin{bmatrix} z - 1 \amp 0 \amp 1 \\ 0 \amp z - 1 \amp -2 \\ 1 \amp -2 \amp z - 5 \end{bmatrix} = z(z - 1)(z - 6). \end{equation*}
Thus the eigenvalues are \(\lambda = 0\text{,}\) \(1\text{,}\) and \(6\text{,}\) and corresponding eigenvectors are
\begin{equation*} \mathbf{x}_{1} = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \; \mathbf{x}_{2} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} \; \mathbf{x}_{3} = \begin{bmatrix} -1 \\ 2 \\ 5 \end{bmatrix}, \end{equation*}
respectively. Moreover, by what at first appears to be remarkably good luck, these eigenvectors are orthogonal. We have \(\norm{\mathbf{x}_{1}}^{2} = 6\text{,}\) \(\norm{\mathbf{x}_{2}}^{2} = 5\text{,}\) and \(\norm{\mathbf{x}_{3}}^{2} = 30\text{,}\) so
\begin{equation*} P = \begin{bmatrix} | \amp | \amp | \\ \frac{1}{\sqrt{6}}\mathbf{x}_{1} \amp \frac{1}{\sqrt{5}}\mathbf{x}_{2} \amp \frac{1}{\sqrt{30}}\mathbf{x}_{3} \\ | \amp | \amp | \end{bmatrix} = \frac{1}{\sqrt{30}} \begin{bmatrix} \sqrt{5} \amp 2\sqrt{6} \amp -1 \\ -2\sqrt{5} \amp \sqrt{6} \amp 2 \\ \sqrt{5} \amp 0 \amp 5 \end{bmatrix}. \end{equation*}
is an orthogonal matrix. Thus \(P^{-1} = P^{T}\) and
\begin{equation*} P^TAP = \begin{bmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 6 \end{bmatrix}. \end{equation*}
Actually, the fact that the eigenvectors in Example 10.4.9 are orthogonal is no coincidence. These vectors certainly must be linearly independent (they correspond to distinct eigenvalues). We will see that the fact that the matrix is symmetric implies that the eigenvectors are orthogonal. To prove this we need the following useful fact about symmetric matrices.

Proof.

Recall that \(\mathbf{x} \cdot \mathbf{y} = \mathbf{x}^{T} \mathbf{y}\) for all columns \(\mathbf{x}\) and \(\mathbf{y}\text{.}\) Because \(A^{T} = A\text{,}\) we get
\begin{equation*} (A\mathbf{x}) \cdot \mathbf{y} = (A\mathbf{x})^T\mathbf{y} = \mathbf{x}^TA^T\mathbf{y} = \mathbf{x}^TA\mathbf{y} = \mathbf{x} \cdot (A\mathbf{y}). \end{equation*}

Proof.

Let \(A\mathbf{x} = \lambda \mathbf{x}\) and \(A\mathbf{y} = \mu \mathbf{y}\text{,}\) where \(\lambda \neq \mu\text{.}\) We compute
\begin{equation*} \lambda(\mathbf{x} \cdot \mathbf{y}) = (\lambda\mathbf{x}) \cdot \mathbf{y} = (A\mathbf{x}) \cdot \mathbf{y} = \mathbf{x} \cdot (A\mathbf{y}) = \mathbf{x} \cdot (\mu\mathbf{y}) = \mu(\mathbf{x} \cdot \mathbf{y}). \end{equation*}
Hence \((\lambda - \mu)(\mathbf{x} \cdot \mathbf{y}) = 0\text{,}\) and so \(\mathbf{x} \cdot \mathbf{y} = 0\) because \(\lambda \neq \mu\text{.}\)
Now the procedure for diagonalizing a symmetric \(n \times n\) matrix is clear. Find the distinct eigenvalues and find orthonormal bases for each eigenspace (the Gram-Schmidt algorithm may be needed when there is a repeated eigenvalue). Then the set of all these basis vectors is orthonormal by Theorem 10.4.12 and contains \(n\) vectors. Here is an example.

Example 10.4.13.

Orthogonally diagonalize the symmetric matrix
\begin{equation*} A = \begin{bmatrix} 8 \amp -2 \amp 2 \\ -2 \amp 5 \amp 4 \\ 2 \amp 4 \amp 5 \end{bmatrix}\text{.} \end{equation*}
Answer.
The characteristic polynomial is
\begin{equation*} c_{A}(z) = \det \begin{bmatrix} z-8 \amp 2 \amp -2 \\ 2 \amp z-5 \amp -4 \\ -2 \amp -4 \amp z-5 \end{bmatrix} = z(z-9)^2. \end{equation*}
Hence the distinct eigenvalues are \(0\) and \(9\) are of algebraic multiplicity \(1\) and \(2\text{,}\) respectively. The geometric multiplicities must be the same, for \(A\) is diagonalizable, being symmetric. It follows that \(\mbox{dim}(\mathcal{S}_0) = 1\) and \(\mbox{dim}(\mathcal{S}_9) = 2\text{.}\) Gaussian elimination gives
\begin{equation*} \mathcal{S}_{0}(A) = \mbox{span}\{\mathbf{x}_{1}\}, \ \mathbf{x}_{1} = \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix}, \ \mathcal{S}_{9}(A) = \mbox{span} \left\lbrace \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} \right \rbrace. \end{equation*}
The eigenvectors in \(\mathcal{S}_{9}\) are both orthogonal to \(\mathbf{x}_{1}\) as Theorem~Theorem 10.4.12 guarantees, but not to each other. However, the Gram-Schmidt process yields an orthogonal basis
\begin{equation*} \{\mathbf{f}_{2}, \mathbf{f}_{3}\} \mbox{ of } \mathcal{S}_{9}(A) \ \mbox{ where } \mathbf{f}_{2} = \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} \mbox{ and } \mathbf{f}_{3} = \begin{bmatrix} 2 \\ 4 \\ 5 \end{bmatrix}. \end{equation*}
Normalizing gives orthonormal vectors \(\{\frac{1}{3}\mathbf{x}_{1}, \frac{1}{\sqrt{5}}\mathbf{f}_{2}, \frac{1}{3\sqrt{5}}\mathbf{f}_{3}\}\text{,}\) so
\begin{equation*} Q = \begin{bmatrix} | \amp | \amp | \\ \frac{1}{3}\mathbf{x}_{1} \amp \frac{1}{\sqrt{5}}\mathbf{f}_{2} \amp \frac{1}{3\sqrt{5}}\mathbf{f}_{3} \\ | \amp | \amp | \end{bmatrix} = \frac{1}{3\sqrt{5}}\begin{bmatrix} \sqrt{5} \amp -6 \amp 2 \\ 2\sqrt{5} \amp 3 \amp 4 \\ -2\sqrt{5} \amp 0 \amp 5 \end{bmatrix} \end{equation*}
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, \(\mathbf{y}_{2} =[2,1,2]\) and \(\mathbf{y}_{3} = [-2,2-1]\) lie in \(\mathcal{S}_{9}(A)\) and they are orthogonal. Moreover, they both have norm \(3\) (as does \(\mathbf{x}_{1}\)), so
\begin{equation*} \hat{Q} = \begin{bmatrix} | \amp | \amp | \\ \frac{1}{3}\mathbf{x}_{1} \amp \frac{1}{3}\mathbf{y}_{2} \amp \frac{1}{3}\mathbf{y}_{3} \\ | \amp | \amp | \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 1 \amp 2 \amp -2 \\ 2 \amp 1 \amp 2 \\ -2 \amp 2 \amp 1 \end{bmatrix} \end{equation*}
is a nicer orthogonal matrix with the property that \(\hat{Q}^{-1}A\hat{Q}\) is diagonal.

Proof.

Let \(\mathbf{q}_{1}, \mathbf{q}_{2}, \dots, \mathbf{q}_{n}\) be orthonormal eigenvectors of \(A\) with corresponding eigenvalues \(\lambda_1, \lambda_2, \ldots, \lambda_n\) . We must show \(A\) is orthogonally diagonalizable. Let
\begin{equation*} Q = \begin{bmatrix} | \amp | \amp \amp | \\ \mathbf{q}_1 \amp \mathbf{q}_2 \amp \cdots \amp \mathbf{q}_n \\ | \amp | \amp \amp | \end{bmatrix} \end{equation*}
so that \(Q\) is orthogonal. We have
\begin{align*} AQ \amp = \begin{bmatrix} | \amp | \amp \amp | \\ A\mathbf{q}_1 \amp A\mathbf{q}_2 \amp \cdots \amp A\mathbf{q}_n \\ | \amp | \amp \amp | \end{bmatrix} \\ \amp = \begin{bmatrix} | \amp | \amp \amp | \\ \lambda_1 \mathbf{q}_{1} \amp \lambda_2 \mathbf{q}_{2} \amp \dots \amp \lambda_n \mathbf{q}_{n} \\ | \amp | \amp \amp | \end{bmatrix}=QD \end{align*}
where \(D\) is the diagonal matrix with diagonal entries \(\lambda_1, \lambda_2, \ldots, \lambda_n\text{.}\) But then \(Q^TAQ=D\text{,}\) proving this half of the theorem.
For the converse, if \(A\) is orthogonally diagonalizable, then by Theorem 10.4.8 it is symmetric. But then Theorem 10.4.12 tells us that eigenvectors corresponding to distinct eigenvalues are orthogonal. Because \(A\) is (orthogonally) diagonalizable, we know the geometric multiplicity of each eigenvalue is equal to its algebraic multiplicity. This implies that we can use Gram-Schmidt on each eigenspace of dimension \(\gt 1\) to get a full set of \(n\) orthogonal eigenvectors.

Exercises 10.4.3 Exercises

1.

Suppose \(A\) is orthogonally diagonalizable. Prove that \(A\) is symmetric. This is the easy direction of the "if and only if" in Theorem 10.4.8.

Exercise Group.

Normalize the rows to make each of the following matrices orthogonal.
2.
\begin{equation*} A = \begin{bmatrix} 1 \amp 1 \\ -1 \amp 1 \end{bmatrix}. \end{equation*}
3.
\begin{equation*} A = \begin{bmatrix} 1 \amp 2 \\ -4 \amp 2 \end{bmatrix}. \end{equation*}
4.
\begin{equation*} A = \begin{bmatrix} \cos\theta \amp -\sin\theta \amp 0 \\ \sin\theta \amp \cos\theta \amp 0 \\ 0 \amp 0 \amp 2 \end{bmatrix}. \end{equation*}
5.
\begin{equation*} A = \begin{bmatrix} -1 \amp 2 \amp 2 \\ 2 \amp -1 \amp 2 \\ 2 \amp 2 \amp -1 \end{bmatrix}. \end{equation*}

6.

If \(Q\) is a triangular orthogonal matrix, show that \(Q\) is diagonal and that all diagonal entries are \(1\) or \(-1\text{.}\)
Answer.
We have \(Q^{T} = Q^{-1}\text{;}\) the first step is to show that \(Q\) is lower triangular and also upper triangular, and so is diagonal. But then \(Q = Q^{T} = Q^{-1}\text{,}\) so \(Q^{2} = I\text{.}\) This implies that the diagonal entries of \(Q\) are all \(\pm 1\text{.}\)

7.

If \(Q\) is orthogonal, show that \(kQ\) is orthogonal if and only if \(k = 1\) or \(k = -1\text{.}\)

8.

If the first two rows of an orthogonal matrix are
\begin{equation*} \left [ \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right ] \quad \text{and} \quad \left [ \frac{2}{3}, \frac{1}{3}, \frac{-2}{3} \right ], \end{equation*}
find all possible third rows.

Exercise Group.

For each matrix \(A\text{,}\) find an orthogonal matrix \(Q\) such that \(Q^{-1}AQ\) is diagonal.
9.
\begin{equation*} A = \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix}. \end{equation*}
10.
\begin{equation*} A = \begin{bmatrix} 1 \amp -1 \\ -1 \amp 1 \end{bmatrix}. \end{equation*}
11.
\begin{equation*} A = \begin{bmatrix} 3 \amp 0 \amp 0 \\ 0 \amp 2 \amp 2 \\ 0 \amp 2 \amp 5 \end{bmatrix}. \end{equation*}
12.
\begin{equation*} A = \begin{bmatrix} 3 \amp 0 \amp 7 \\ 0 \amp 5 \amp 0 \\ 7 \amp 0 \amp 3 \end{bmatrix}. \end{equation*}
13.
\begin{equation*} A = \begin{bmatrix} 1 \amp 1 \amp 0 \\ 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 2 \end{bmatrix}. \end{equation*}
14.
As a challenge:
\begin{equation*} A = \begin{bmatrix} 5 \amp -2 \amp -4 \\ -2 \amp 8 \amp -2\\ -4 \amp -2 \amp 5 \end{bmatrix}. \end{equation*}
15.
As a challenge:
\begin{equation*} A = \begin{bmatrix} 3 \amp 5 \amp -1 \amp 1 \\ 5 \amp 3 \amp 1 \amp -1 \\ -1 \amp 1 \amp 3 \amp 5 \\ 1 \amp -1 \amp 5 \amp 3 \end{bmatrix}. \end{equation*}

16.

Show that the following are equivalent for a symmetric matrix \(A\text{.}\)
  1. \(A\) is orthogonal.
  2. \(A^{2} = I\text{.}\)
  3. All eigenvalues of \(A\) are \(\pm 1\text{.}\)
Hint.
For (b) if and only if (c), use Theorem 7.2.15.

17.

We call matrices \(A\) and \(B\) orthogonally similar (and write \(A \stackrel{\circ}{\sim} B\)) if \(B = P^{T}AP\) for an orthogonal matrix \(P\text{.}\)
  1. Show that \(A \stackrel{\circ}{\sim} A\) for all \(A\text{;}\) \(A \stackrel{\circ}{\sim} B \Rightarrow B \stackrel{\circ}{\sim} A\text{;}\) and \(A \stackrel{\circ}{\sim} B\) and \(B \stackrel{\circ}{\sim} C \Rightarrow A \stackrel{\circ}{\sim} C\text{.}\) (This means that ``orthogonally similar" is an equivalence relation.)
  2. Show that the following are equivalent for any pair of symmetric matrices \(A\) and \(B\text{:}\)
    1. \(A\) and \(B\) are similar.
    2. \(A\) and \(B\) are orthogonally similar.
    3. \(A\) and \(B\) have the same eigenvalues.

18.

Assume that \(A\) and \(B\) are orthogonally similar (Exercise 10.4.3.17).
  1. If \(A\) and \(B\) are invertible, show that \(A^{-1}\) and \(B^{-1}\) are orthogonally similar.
  2. Show that \(A^{2}\) and \(B^{2}\) are orthogonally similar.
  3. Show that, if \(A\) is symmetric, so is \(B\text{.}\)

19.

If \(A\) is symmetric, show that every eigenvalue of \(A\) is nonnegative if and only if \(A = B^{2}\) for some symmetric matrix \(B\text{.}\)

20.

Prove the converse of Theorem 10.4.10: If \((A\mathbf{x}) \cdot \mathbf{y} = \mathbf{x} \cdot (A\mathbf{y})\) for all \(n\)-columns \(\mathbf{x}\) and \(\mathbf{y}\text{,}\) then \(A\) is symmetric.

21.

Show that every eigenvalue of \(A\) is zero if and only if \(A\) is nilpotent (\(A^{k} = 0\) for some \(k \geq 1\)).

22.

If \(A\) has real eigenvalues, show that \(A = B + C\) where \(B\) is symmetric and \(C\) is nilpotent.

23.

Let \(Q\) be an orthogonal matrix.
  1. Show that \(\det Q = 1\) or \(\det Q = -1\text{.}\)
  2. Give \(2 \times 2\) examples of \(Q\) such that \(\det Q = 1\) and \(\det Q = -1\text{.}\)
  3. If \(\det Q = -1\text{,}\) show that \(I + Q\) has no inverse.
  4. If \(P\) is \(n \times n\) and \(\det P \neq (-1)^{n}\text{,}\) show that \(I - P\) has no inverse.
Hint.
For part (c): \(Q^{T}(I + Q) = (I + Q)^{T}\text{.}\)
For part (d): \(P^{T}(I - P) = -(I - P)^{T}\text{.}\)

24.

We call a square matrix \(E\) a projection matrix if \(E^{2} = E = E^{T}\text{.}\)
  1. If \(E\) is a projection matrix, show that \(Q = I - 2E\) is orthogonal and symmetric.
  2. If \(Q\) is orthogonal and symmetric, show that \\ \(E = \frac{1}{2}(I - Q)\) is a projection matrix.
  3. If \(Q\) is \(m \times n\) and \(Q^{T}Q = I\) (for example, a unit column in \(\R^n\)), show that \(E = QQ^{T}\) is a projection matrix.

25.

A matrix that we obtain from the identity matrix by writing its rows in a different order is called a permutation matrix (see Theorem 4.6.9). Show that every permutation matrix is orthogonal.

26.

If the rows \(\mathbf{r}_{1}, \dots, \mathbf{r}_{n}\) of the \(n \times n\) matrix \(A = \begin{bmatrix} a_{ij} \end{bmatrix}\) are orthogonal, show that the \((i, j)\)-entry of \(A^{-1}\) is \(\frac{a_{ji}}{\norm{\mathbf{r}_{j}}^2}\text{.}\)

27.

  1. Let \(A\) be an \(m \times n\) matrix. Show that the following are equivalent.
  2. Show that an \(n \times n\) matrix \(A\) has orthogonal rows if and only if \(A\) can be factored as \(A = DQ\text{,}\) where \(Q\) is orthogonal and \(D\) is diagonal and invertible.

28.

Let \(A\) be a skew-symmetric matrix; that is, \(A^{T} = -A\text{.}\) Assume that \(A\) is an \(n \times n\) matrix.
  1. Show that \(I + A\) is invertible.
  2. Show that \(Q = (I - A)(I + A)^{-1}\) is orthogonal.
  3. Show that every orthogonal matrix \(P\) such that \(I + P\) is invertible arises as in part (b) from some skew-symmetric matrix \(A\text{.}\)
Hint.
For (a): It suffices to show that \((I + A)\mathbf{x} = \mathbf{0}\text{,}\) \(\mathbf{x}\) in \(\R^n\text{,}\) implies \(\mathbf{x} = \mathbf{0}\text{.}\) Compute \(\mathbf{x} \cdot \mathbf{x} = \mathbf{x}^{T}\mathbf{x}\text{,}\) and use the fact that \(A\mathbf{x} = -\mathbf{x}\) and \(A^{2}\mathbf{x} = \mathbf{x}\text{.}\)
For (c): Solve \(P = (I - A)(I + A)^{-1}\) for \(A\text{.}\)

29.

Show that the following are equivalent for an \(n \times n\) matrix \(Q\text{.}\)
  1. \(Q\) is orthogonal.
  2. \(\norm{Q\mathbf{x}} = \norm{\mathbf{x}}\) for all \(\mathbf{x}\in\R^n\text{.}\)
  3. \(\norm{ Q\mathbf{x} - Q\mathbf{y}} = \norm{\mathbf{x} - \mathbf{y}}\) for all \(\mathbf{x}\text{,}\) \(\mathbf{y}\in \R^n\text{.}\)
  4. \((Q\mathbf{x}) \cdot (Q\mathbf{y}) = \mathbf{x} \cdot \mathbf{y}\) for all columns \(\mathbf{x}\text{,}\) \(\mathbf{y}\in\R^n\text{.}\)
Hint.
For (d) \(\Rightarrow\) (a), show that column \(i\) of \(Q\) equals \(Q\mathbf{e}_{i}\text{,}\) where \(\mathbf{e}_{i}\) is column \(i\) of the identity matrix.

Remark 10.4.15.

The above exercise shows that linear transformations with orthogonal standard matrices are distance-preserving (b,c) and angle-preserving (d).

30.

  1. Show that
    \begin{equation*} \begin{bmatrix} \cos\theta \amp -\sin\theta \\ \sin\theta \amp \cos\theta \end{bmatrix} \end{equation*}
    is an orthogonal matrix.
  2. Show that every \(2 \times 2\) orthogonal matrix has the form
    \begin{equation*} \begin{bmatrix} \cos\theta \amp -\sin\theta \\ \sin\theta \amp \cos\theta \end{bmatrix} \quad \text{or} \quad \begin{bmatrix} \cos\theta \amp \sin\theta \\ \sin\theta \amp -\cos\theta \end{bmatrix} \end{equation*}
    for some angle \(\theta\text{.}\)
Hint.
If \(a^{2} + b^{2} = 1\text{,}\) then \(a = \cos\theta\) and \(b = \sin\theta\) for some angle \(\theta\text{.}\)
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