Story of the chapter. In many applied problems, a matrix is not merely a table of numbers. It is an energy machine. A covariance matrix measures variation, a graph Laplacian measures roughness on a network, a Hessian measures local curvature, and a Gram matrix measures similarity. These matrices have a special symmetry. Over the real numbers this symmetry is \(A^T=A\); over the complex numbers it is \(A^*=A\).
The reward for this symmetry is enormous: real eigenvalues, orthogonal eigenvectors, stable decompositions, and a direct connection between algebra and geometry.
The central message of this chapter is:
\[ \boxed{\text{Symmetric and Hermitian matrices are the matrices that behave like real geometric measurements.}} \]
They are the matrices for which eigenvalues measure true stretching, eigenvectors can be chosen orthonormally, and quadratic forms can be understood as sums of squared coordinates.
By the end of this chapter, you should be able to:
In earlier chapters, eigenvectors helped us understand how a linear map stretches special directions. But for a general matrix, eigenvectors may fail to form a basis, eigenvalues may be complex, and numerical behavior can be unstable. Symmetric and Hermitian matrices are different. They are the matrices where the eigenvalue story is as clean as possible.
Let \(A\in \mathbb{R}^{n\times n}\). The matrix \(A\) is symmetric if
\[ A^T=A. \]
Let \(A\in \mathbb{C}^{n\times n}\). The matrix \(A\) is Hermitian if
\[ A^*=A, \]
where
\[ A^*:=\overline{A}^{\,T} \]
is the conjugate transpose.
The matrix
\[ A=\begin{bmatrix}10&1\\1&6\end{bmatrix} \]
is symmetric. The matrix
\[ H=\begin{bmatrix}1&i\\-i&2\end{bmatrix} \]
is Hermitian because
\[ H^*=\begin{bmatrix}1&i\\-i&2\end{bmatrix}=H. \]
However,
\[ B=\begin{bmatrix}1&i\\ i&2\end{bmatrix} \]
is not Hermitian, because the off-diagonal entries are not complex conjugates in the correct positions.
For complex vectors, the standard inner product is
\[ \langle \vec{z},\vec{w}\rangle=\vec{z}^{\,*}\vec{w}. \]
The conjugate transpose is the operation that makes the identity
\[ \langle A\vec{z},\vec{w}\rangle=\langle \vec{z},A^*\vec{w}\rangle \]
true for all vectors \(\vec{z},\vec{w}\). Thus Hermitian matrices are the complex analogue of real symmetric matrices.
Every square matrix has a symmetric part and an antisymmetric part. This is a first example of how vector spaces of matrices split into meaningful pieces.
Define
\[ \operatorname{Sym}_n=\{A\in \mathbb{R}^{n\times n}:A^T=A\} \]
and
\[ \operatorname{Skew}_n=\{A\in \mathbb{R}^{n\times n}:A^T=-A\}. \]
Every real square matrix has a unique decomposition as
\[ M=S+K, \]
where \(S\) is symmetric and \(K\) is skew-symmetric. More precisely,
\[ \mathbb{R}^{n\times n}=\operatorname{Sym}_n\oplus \operatorname{Skew}_n, \]
and
\[ S=\frac{M+M^T}{2}, \qquad K=\frac{M-M^T}{2}. \]
Set
\[ S=\frac{M+M^T}{2}, \qquad K=\frac{M-M^T}{2}. \]
Then
\[ S^T=\frac{M^T+M}{2}=S, \]
so \(S\) is symmetric. Also,
\[ K^T=\frac{M^T-M}{2}=-K, \]
so \(K\) is skew-symmetric. Clearly \(M=S+K\).
For uniqueness, suppose
\[ M=S_1+K_1=S_2+K_2, \]
where \(S_1,S_2\) are symmetric and \(K_1,K_2\) are skew-symmetric. Then
\[ S_1-S_2=K_2-K_1. \]
The left side is symmetric, and the right side is skew-symmetric. A matrix that is both symmetric and skew-symmetric must satisfy \(X^T=X\) and \(X^T=-X\), so \(X=-X\) and \(X=0\). Hence \(S_1=S_2\) and \(K_1=K_2\).
Let
\[ M=\begin{bmatrix} 1&4&0\\ 2&3&5\\ -1&7&2 \end{bmatrix}. \]
Then
\[ \frac{M+M^T}{2} = \begin{bmatrix} 1&3&-1/2\\ 3&3&6\\ -1/2&6&2 \end{bmatrix} \]
and
\[ \frac{M-M^T}{2} = \begin{bmatrix} 0&1&1/2\\ -1&0&-1\\ -1/2&1&0 \end{bmatrix}. \]
The first part is symmetric, the second part is skew-symmetric, and their sum is \(M\).
Define
\[ \mathcal{H}_n=\{A\in \mathbb{C}^{n\times n}:A^*=A\} \]
and
\[ \mathcal{K}_n=\{A\in \mathbb{C}^{n\times n}:A^*=-A\}. \]
Matrices in \(\mathcal{K}_n\) are called skew-Hermitian.
Every complex square matrix has the decomposition
\[ M=\frac{M+M^*}{2}+\frac{M-M^*}{2}, \]
where \(\frac{M+M^*}{2}\) is Hermitian and \(\frac{M-M^*}{2}\) is skew-Hermitian.
Let
\[ H=\frac{M+M^*}{2}, \qquad K=\frac{M-M^*}{2}. \]
Then
\[ H^*=\frac{M^*+M}{2}=H \]
and
\[ K^*=\frac{M^*-M}{2}=-K. \]
Thus \(H\) is Hermitian, \(K\) is skew-Hermitian, and \(M=H+K\).
The first miracle is that Hermitian matrices may have complex entries, but their eigenvalues are real.
Let \(A\in \mathbb{C}^{n\times n}\) be Hermitian. If
\[ A\vec{z}=\lambda \vec{z} \]
for some nonzero \(\vec{z}\in \mathbb{C}^n\), then \(\lambda\in \mathbb{R}\).
Since \(A\vec{z}=\lambda \vec{z}\),
\[ \vec{z}^{\,*}A\vec{z}=\vec{z}^{\,*}(\lambda \vec{z}) =\lambda \vec{z}^{\,*}\vec{z}. \]
Therefore
\[ \lambda=\frac{\vec{z}^{\,*}A\vec{z}}{\vec{z}^{\,*}\vec{z}}. \]
The denominator is \(\|\vec{z}\|^2>0\). The numerator is real because
\[ \overline{\vec{z}^{\,*}A\vec{z}} =(\vec{z}^{\,*}A\vec{z})^* =\vec{z}^{\,*}A^*\vec{z} =\vec{z}^{\,*}A\vec{z}. \]
Thus \(\lambda\) is real.
If \(A\in \mathbb{R}^{n\times n}\) is symmetric, then all eigenvalues of \(A\) are real.
A real symmetric matrix is Hermitian when viewed as a complex matrix. Therefore the previous proposition applies.
Let \(A\in \mathbb{C}^{n\times n}\) be Hermitian. Suppose
\[ A\vec{z}=\lambda \vec{z}, \qquad A\vec{w}=\mu \vec{w}, \]
where \(\lambda\ne \mu\). Then
\[ \langle \vec{z},\vec{w}\rangle=0. \]
Using \(A=A^*\), we have
\[ \langle A\vec{z},\vec{w}\rangle = \langle \vec{z},A\vec{w}\rangle. \]
Since \(\lambda\) and \(\mu\) are real,
\[ \lambda\langle \vec{z},\vec{w}\rangle = \mu\langle \vec{z},\vec{w}\rangle. \]
Thus
\[ (\lambda-\mu)\langle \vec{z},\vec{w}\rangle=0. \]
Because \(\lambda\ne \mu\), we get \(\langle \vec{z},\vec{w}\rangle=0\).
The spectral theorem is the main structural result of this chapter. It says that symmetric and Hermitian matrices are exactly the matrices that can be diagonalized by an orthonormal change of basis.
A real matrix \(A\in \mathbb{R}^{n\times n}\) is orthogonally diagonalizable if
\[ A=QDQ^T, \]
where \(Q\) is orthogonal and \(D\) is diagonal.
A complex matrix \(A\in \mathbb{C}^{n\times n}\) is unitarily diagonalizable if
\[ A=UDU^*, \]
where \(U\) is unitary and \(D\) is diagonal.
A real matrix \(A\in \mathbb{R}^{n\times n}\) is symmetric if and only if it is orthogonally diagonalizable. That is,
\[ A^T=A \quad\Longleftrightarrow\quad A=QDQ^T \]
for some orthogonal matrix \(Q\) and real diagonal matrix \(D\).
If \(A=QDQ^T\), then
\[ A^T=(QDQ^T)^T=QD^TQ^T=QDQ^T=A, \]
so \(A\) is symmetric.
Conversely, if \(A\) is symmetric, choose a unit eigenvector \(\vec{u}_1\). The orthogonal complement \(\vec{u}_1^\perp\) is invariant under \(A\). Indeed, if \(\vec{x}\perp \vec{u}_1\), then
\[ (A\vec{x})\cdot \vec{u}_1 =\vec{x}\cdot A\vec{u}_1 =\lambda_1\vec{x}\cdot \vec{u}_1 =0. \]
Thus \(A\) restricts to a symmetric operator on a space of dimension \(n-1\). Induction gives an orthonormal eigenbasis. Placing these eigenvectors in the columns of \(Q\) gives \(A=QDQ^T\).
A complex matrix \(A\in \mathbb{C}^{n\times n}\) is Hermitian if and only if it is unitarily diagonalizable with real eigenvalues:
\[ A^*=A \quad\Longleftrightarrow\quad A=UDU^*, \]
where \(U\) is unitary and \(D\) is diagonal with real diagonal entries.
The proof is the complex analogue of the real proof. Hermitian matrices have real eigenvalues. Choose a unit eigenvector \(\vec{u}_1\). The orthogonal complement of \(\vec{u}_1\) is invariant under \(A\) because \(A=A^*\). Restrict \(A\) to that orthogonal complement and use induction. This gives an orthonormal eigenbasis of \(\mathbb{C}^n\), hence \(A=UDU^*\).
For a real symmetric matrix \(A\):
For a Hermitian matrix, the same steps work with complex inner products, producing \(A=UDU^*\).
Suppose
\[ A=QDQ^T, \qquad Q=[\vec{u}_1\ \vec{u}_2\ \cdots\ \vec{u}_n], \qquad D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n). \]
Then
\[ A=\lambda_1\vec{u}_1\vec{u}_1^T+\cdots+\lambda_n\vec{u}_n\vec{u}_n^T. \]
Each matrix \(\vec{u}_i\vec{u}_i^T\) is an orthogonal projection onto the line \(\operatorname{span}\{\vec{u}_i\}\).
For Hermitian matrices,
\[ A=UDU^* \]
becomes
\[ A=\lambda_1\vec{u}_1\vec{u}_1^*+\cdots+\lambda_n\vec{u}_n\vec{u}_n^*. \]
Let \(\vec{u}\in \mathbb{C}^n\) be a unit vector. Then
\[ P=\vec{u}\vec{u}^{\,*} \]
is the orthogonal projection matrix onto \(\operatorname{span}\{\vec{u}\}\).
For any \(\vec{z}\in \mathbb{C}^n\),
\[ P\vec{z}=\vec{u}\vec{u}^{\,*}\vec{z} =\vec{u}\langle \vec{u},\vec{z}\rangle. \]
This lies in \(\operatorname{span}\{\vec{u}\}\). Also,
\[ \vec{z}-P\vec{z} \]
is orthogonal to \(\vec{u}\) because
\[ \langle \vec{u},\vec{z}-P\vec{z}\rangle = \langle \vec{u},\vec{z}\rangle- \langle \vec{u},\vec{u}\langle \vec{u},\vec{z}\rangle\rangle =0. \]
Thus \(P\vec{z}\) is exactly the orthogonal projection of \(\vec{z}\) onto the line spanned by \(\vec{u}\).
Consider
\[ A=\begin{bmatrix} 1&1&7\\ 1&7&1\\ 7&1&1 \end{bmatrix}. \]
One spectral decomposition is
\[ A=9\vec{u}_1\vec{u}_1^T+6\vec{u}_2\vec{u}_2^T-6\vec{u}_3\vec{u}_3^T, \]
where
\[ \vec{u}_1=\frac{1}{\sqrt{3}} \begin{bmatrix}1\\1\\1\end{bmatrix}, \qquad \vec{u}_2=\frac{1}{\sqrt{6}} \begin{bmatrix}-1\\-2\\1\end{bmatrix}, \qquad \vec{u}_3=\frac{1}{\sqrt{2}} \begin{bmatrix}-1\\0\\1\end{bmatrix}. \]
This example shows a symmetric matrix as a weighted sum of orthogonal projection matrices.
Symmetric and Hermitian matrices appear naturally when a matrix is used to produce a scalar energy.
Let \(A\in \mathbb{R}^{n\times n}\) be symmetric. The quadratic form defined by \(A\) is
\[ q(\vec{x})=\vec{x}^{\,T}A\vec{x}, \qquad \vec{x}\in \mathbb{R}^n. \]
It is a second-degree polynomial with no linear or constant terms.
If
\[ A=\begin{bmatrix}1&3\\3&2\end{bmatrix}, \]
then
\[ q(\vec{x})=\vec{x}^{\,T}A\vec{x}=x_1^2+6x_1x_2+2x_2^2. \]
The coefficient \(6\) appears because the off-diagonal contribution is
\[ 3x_1x_2+3x_2x_1=6x_1x_2. \]
Let \(A\in \mathbb{C}^{n\times n}\) be Hermitian. The Hermitian form defined by \(A\) is
\[ q(\vec{z})=\vec{z}^{\,*}A\vec{z}, \qquad \vec{z}\in \mathbb{C}^n. \]
This number is always real.
Let \(\alpha=\vec{z}^{\,*}A\vec{z}\). Then
\[ \overline{\alpha} =(\vec{z}^{\,*}A\vec{z})^* =\vec{z}^{\,*}A^*\vec{z} =\vec{z}^{\,*}A\vec{z} =\alpha. \]
So \(\alpha\) is real.
Let
\[ A=\begin{bmatrix}2&i\\-i&3\end{bmatrix}, \qquad \vec{z}=\begin{bmatrix}z_1\\z_2\end{bmatrix}. \]
Then
\[ \vec{z}^{\,*}A\vec{z} =2|z_1|^2+3|z_2|^2+i\overline{z_1}z_2-i\overline{z_2}z_1. \]
The cross terms combine to a real number.
Positive definite matrices are the matrices whose energy is strictly positive except at the origin.
Let \(A\in \mathbb{R}^{n\times n}\) be symmetric.
\[ \vec{x}^{\,T}A\vec{x}>0 \]
for every nonzero \(\vec{x}\in \mathbb{R}^n\).
\[ \vec{x}^{\,T}A\vec{x}\ge 0 \]
for every \(\vec{x}\in \mathbb{R}^n\).
Let \(A\in \mathbb{C}^{n\times n}\) be Hermitian.
\[ \vec{z}^{\,*}A\vec{z}>0 \]
for every nonzero \(\vec{z}\in \mathbb{C}^n\).
\[ \vec{z}^{\,*}A\vec{z}\ge 0 \]
for every \(\vec{z}\in \mathbb{C}^n\).
Let \(A\) be real symmetric or complex Hermitian. Then:
By the spectral theorem,
\[ A=UDU^*, \]
where \(U\) is unitary and
\[ D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n) \]
with real eigenvalues. Let \(\vec{y}=U^*\vec{z}\). Since \(U\) is unitary, \(\vec{z}\ne 0\) if and only if \(\vec{y}\ne 0\). Then
\[ \vec{z}^{\,*}A\vec{z} =\vec{z}^{\,*}UDU^*\vec{z} =(U^*\vec{z})^*D(U^*\vec{z}) =\vec{y}^{\,*}D\vec{y} =\sum_{i=1}^n \lambda_i |y_i|^2. \]
This is positive for all nonzero \(\vec{y}\) exactly when every \(\lambda_i>0\). It is nonnegative for every \(\vec{y}\) exactly when every \(\lambda_i\ge 0\).
On \(\mathbb{R}^3\):
Positive semidefinite matrices have a natural square root.
Suppose
\[ A=UDU^*, \qquad D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n), \qquad \lambda_i\ge 0. \]
Define
\[ D^{1/2}=\operatorname{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n}) \]
and
\[ A^{1/2}=UD^{1/2}U^*. \]
If \(A\) is Hermitian positive semidefinite, then
\[ A^{1/2}A^{1/2}=A. \]
Moreover, \(A^{1/2}\) is Hermitian positive semidefinite.
Compute
\[ A^{1/2}A^{1/2} =(UD^{1/2}U^*)(UD^{1/2}U^*) =UD^{1/2}D^{1/2}U^* =UDU^*=A. \]
Also,
\[ (A^{1/2})^*=A^{1/2}, \]
and the eigenvalues of \(A^{1/2}\) are \(\sqrt{\lambda_i}\ge 0\).
Gram matrices record inner products among a collection of vectors. They are one of the most common sources of positive semidefinite matrices.
Let \(\vec{v}_1,\ldots,\vec{v}_m\) be vectors in a real or complex inner product space. The Gram matrix is
\[ G=[g_{ij}], \qquad g_{ij}=\langle \vec{v}_i,\vec{v}_j\rangle. \]
Every Gram matrix is Hermitian positive semidefinite. If \(\vec{v}_1,\ldots,\vec{v}_m\) are linearly independent, then the Gram matrix is positive definite.
First,
\[ g_{ji}=\langle \vec{v}_j,\vec{v}_i\rangle =\overline{\langle \vec{v}_i,\vec{v}_j\rangle}, \]
so \(G^*=G\).
For any coefficient vector \(\vec{c}=(c_1,\ldots,c_m)^T\),
\[ \vec{c}^{\,*}G\vec{c} = \left\|\sum_{i=1}^m c_i\vec{v}_i\right\|^2 \ge 0. \]
Thus \(G\) is positive semidefinite. If the vectors are linearly independent and \(\vec{c}\ne 0\), then \(\sum_i c_i\vec{v}_i\ne 0\), so \(\vec{c}^{\,*}G\vec{c}>0\). Hence \(G\) is positive definite.
Let \(A\) be an \(m\times n\) real or complex matrix. Then \(A^*A\) is Hermitian positive semidefinite. If \(A\) has full column rank, then \(A^*A\) is positive definite.
First,
\[ (A^*A)^*=A^*A. \]
For any \(\vec{z}\in \mathbb{C}^n\),
\[ \vec{z}^{\,*}A^*A\vec{z} =(A\vec{z})^*(A\vec{z}) =\|A\vec{z}\|^2 \ge 0. \]
If \(A\) has full column rank, then \(A\vec{z}=0\) implies \(\vec{z}=0\). Therefore, for every nonzero \(\vec{z}\),
\[ \|A\vec{z}\|^2>0, \]
so \(A^*A\) is positive definite.
In real least squares, the normal equations are
\[ A^TA\vec{x}=A^T\vec{b}. \]
In complex least squares, they become
\[ A^*A\vec{x}=A^*\vec{b}. \]
The theorem explains why the normal-equation matrix is positive semidefinite, and why it is positive definite when the columns of \(A\) are independent.
Positive definite matrices are special enough to admit a stable triangular factorization.
If \(A\in \mathbb{R}^{n\times n}\) is symmetric positive definite, then
\[ A=LL^T, \]
where \(L\) is real lower triangular with positive diagonal entries.
If \(A\in \mathbb{C}^{n\times n}\) is Hermitian positive definite, then
\[ A=LL^*, \]
where \(L\) is lower triangular with positive real diagonal entries.
A principal minor of order \(k\) is the determinant of a \(k\times k\) submatrix obtained by keeping the same set of \(k\) rows and \(k\) columns.
The leading principal minor of order \(k\) is
\[ \det A_{1:k,1:k}. \]
A real symmetric matrix or complex Hermitian matrix is positive definite if and only if all leading principal minors are positive.
A real symmetric matrix or complex Hermitian matrix is positive semidefinite if and only if all principal minors are nonnegative.
Let
\[ A=\begin{bmatrix} 2&1&0\\ 1&3&1\\ 0&1&2 \end{bmatrix}. \]
The leading principal minors are
\[ \Delta_1=2>0, \]
\[ \Delta_2=\det\begin{bmatrix}2&1\\1&3\end{bmatrix}=5>0, \]
and
\[ \Delta_3=\det(A)=8>0. \]
So \(A\) is positive definite by Sylvester’s criterion.
Python is especially useful for this chapter because symmetric and Hermitian matrices have specialized algorithms. In NumPy, use np.linalg.eigh rather than np.linalg.eig when the matrix is symmetric or Hermitian.
import numpy as np
A = np.array([[1, 1, 7],
[1, 7, 1],
[7, 1, 1]], dtype=float)
# eigh is designed for symmetric/Hermitian matrices
lam, Q = np.linalg.eigh(A)
D = np.diag(lam)
print("Eigenvalues:", lam)
print("Q^T Q =")
print(np.round(Q.T @ Q, 6))
print("Reconstruction error:", np.linalg.norm(A - Q @ D @ Q.T))Eigenvalues: [-6. 6. 9.]
Q^T Q =
[[ 1. 0. -0.]
[ 0. 1. -0.]
[-0. -0. 1.]]
Reconstruction error: 1.1609663269362998e-14B = np.array([[2, 1, 0],
[1, 3, 1],
[0, 1, 2]], dtype=float)
lam_B = np.linalg.eigvalsh(B)
print("Eigenvalues of B:", lam_B)
print("Positive definite?", np.all(lam_B > 0))Eigenvalues of B: [1. 2. 4.]
Positive definite? TrueH = np.array([[2, 1+2j],
[1-2j, 5]], dtype=complex)
print("Hermitian check:", np.allclose(H.conj().T, H))
lam_H, U = np.linalg.eigh(H)
print("Eigenvalues:", lam_H)
print("Unitary check:", np.linalg.norm(U.conj().T @ U - np.eye(2)))
print("Reconstruction error:", np.linalg.norm(H - U @ np.diag(lam_H) @ U.conj().T))Hermitian check: True
Eigenvalues: [0.8074176 6.1925824]
Unitary check: 2.662405728838582e-16
Reconstruction error: 1.2673263468710813e-15C = np.array([[4, 2, 0],
[2, 5, 1],
[0, 1, 3]], dtype=float)
L = np.linalg.cholesky(C)
print("L =")
print(np.round(L, 6))
print("Reconstruction error:", np.linalg.norm(C - L @ L.T))L =
[[2. 0. 0. ]
[1. 2. 0. ]
[0. 0.5 1.658312]]
Reconstruction error: 0.0Decide whether each statement is true or false.
\[ (AB)^T=B^TA^T=BA. \]
Thus \(AB\) is symmetric if and only if \(AB=BA\). If \(A\) and \(B\) do not commute, \(AB\) need not be orthogonally diagonalizable.
Let
\[ A=\begin{bmatrix}a&c+di\\ c-di&b\end{bmatrix}, \qquad a,b,c,d\in \mathbb{R}. \]
Explain why \(A\) is Hermitian and write \(\vec{z}^{\,*}A\vec{z}\) for \(\vec{z}=(z_1,z_2)^T\).
The diagonal entries are real and the off-diagonal entries are conjugates of each other, so \(A^*=A\). We compute
\[ \vec{z}^{\,*}A\vec{z} =a|z_1|^2+b|z_2|^2+(c+di)\overline{z_1}z_2+(c-di)\overline{z_2}z_1. \]
The last two terms are conjugates, so their sum is twice the real part:
\[ \vec{z}^{\,*}A\vec{z} =a|z_1|^2+b|z_2|^2+2\operatorname{Re}\left((c+di)\overline{z_1}z_2\right). \]
Let \(A\) be an \(m\times n\) complex matrix. Prove that \(A^*A\) is invertible if and only if the columns of \(A\) are linearly independent.
The columns of \(A\) are linearly independent if and only if \(\ker(A)=\{0\}\). Also,
\[ A^*A\vec{z}=0 \quad\Longrightarrow\quad \vec{z}^{\,*}A^*A\vec{z}=0 \quad\Longrightarrow\quad \|A\vec{z}\|^2=0 \quad\Longrightarrow\quad A\vec{z}=0. \]
Thus \(\ker(A^*A)=\ker(A)\). Therefore \(A^*A\) is invertible if and only if \(\ker(A^*A)=\{0\}\), which is equivalent to \(\ker(A)=\{0\}\), which is equivalent to the columns of \(A\) being linearly independent.
For
\[ M=\begin{bmatrix}1&4\\2&3\end{bmatrix}, \]
find the symmetric and skew-symmetric parts of \(M\).
\[ S=\frac{M+M^T}{2} =\begin{bmatrix}1&3\\3&3\end{bmatrix}, \qquad K=\frac{M-M^T}{2} =\begin{bmatrix}0&1\\-1&0\end{bmatrix}. \]
Determine whether
\[ H=\begin{bmatrix}2&1+i\\1-i&4\end{bmatrix} \]
is Hermitian.
Yes. The diagonal entries are real and the off-diagonal entries are complex conjugates in symmetric positions. Therefore \(H^*=H\).
Classify the quadratic form
\[ q(x,y)=3x^2+2xy+3y^2. \]
The matrix is
\[ A=\begin{bmatrix}3&1\\1&3\end{bmatrix}. \]
Its eigenvalues are \(4\) and \(2\), both positive. Therefore the quadratic form is positive definite.
Use Sylvester’s criterion to test whether
\[ A=\begin{bmatrix}1&2\\2&1\end{bmatrix} \]
is positive definite.
The leading principal minors are
\[ \Delta_1=1>0, \qquad \Delta_2=\det(A)=1-4=-3<0. \]
Since the second leading principal minor is negative, \(A\) is not positive definite.
Let \(A\) be a matrix with full column rank. Explain why the normal-equation matrix \(A^TA\) is positive definite.
For any nonzero \(\vec{x}\),
\[ \vec{x}^{\,T}A^TA\vec{x}=\|A\vec{x}\|^2. \]
If \(A\) has full column rank, then \(A\vec{x}\ne 0\) whenever \(\vec{x}\ne 0\). Therefore \(\|A\vec{x}\|^2>0\), so \(A^TA\) is positive definite.
Use an AI tool as a study partner, not as a replacement for your reasoning. Ask for explanations, then verify the results by hand or in Python.
Ask:
Explain the spectral theorem for symmetric matrices at three levels: geometric intuition, computation, and proof.
Then compare the AI response to the theorem in this chapter. Identify one sentence you would rewrite more precisely.
Ask:
Give me one example of a symmetric positive definite matrix, one symmetric positive semidefinite matrix that is not positive definite, and one symmetric indefinite matrix. Verify each using eigenvalues.
Then check the examples in Python.
Ask:
A student claims that if \(A\) and \(B\) are symmetric, then \(AB\) is symmetric. Find the mistake and give a counterexample.
Then compute the counterexample by hand.
Ask:
Why are covariance matrices and Gram matrices positive semidefinite? Explain using both data and linear algebra.
Then write a short paragraph explaining the connection in your own words.
In this chapter, we learned that: