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Eigenvalues and eigenvectors

Vectors and matrices · Chapter 4.5 · Lecture notes (PDF) · Exercises: Eigenvalues and eigenvectors · Slides (PDF) · App: Eigenvalues and definiteness

1. Eigenvalues

Definition 1: eigenvalue and eigenvector

Let \( \boldsymbol Q \in \R^{n\times n} \) be a square matrix. A scalar \( \lambda \in \R \) is called an eigenvalue of \( \boldsymbol Q \) if there exists a nonzero vector \( \boldsymbol v \in \R^{n} \setminus \{\boldsymbol 0\} \) such that

\[ \boldsymbol Q\,\boldsymbol v = \lambda\,\boldsymbol v. \]

In this case, \( \boldsymbol v \) is called an eigenvector associated with \( \lambda \).

  • Eigenvalues are solutions of the characteristic equation

    \[ \det(\boldsymbol Q - \lambda \boldsymbol I)=0. \]

  • If \( \boldsymbol Q \) is symmetric, then all its eigenvalues are real.

  • The function \( p(\lambda)=\det(\boldsymbol Q-\lambda\boldsymbol I) \) is a polynomial of degree \( n \) in \( \lambda \), called the characteristic polynomial of \( \boldsymbol Q \).

  • If \( \lambda \) is an eigenvalue, the eigenvectors associated with \( \lambda \) are the nonzero solutions \( \boldsymbol v \) of the homogeneous system

    \[ (\boldsymbol Q-\lambda\boldsymbol I)\,\boldsymbol v=\boldsymbol 0 . \]

  • If \( \boldsymbol v \) is an eigenvector associated with \( \lambda \), then also \( \alpha\,\boldsymbol v \) is an eigenvector associated with \( \lambda \), for every \( \alpha\in\R\setminus\{0\} \). Eigenvectors are therefore never unique: we just give one of them (“for example”).

A square symmetric matrix \( \boldsymbol Q \in \R^{n \times n} \) is

  • positive semidefinite if and only if all its eigenvalues are non-negative.

  • positive definite if and only if all its eigenvalues are strictly positive.

  • negative semidefinite if and only if all its eigenvalues are non-positive.

  • negative definite if and only if all its eigenvalues are strictly negative.

  • Recall that (semi)definiteness has been defined in the chapter on matrices through the sign of \( \boldsymbol x' \boldsymbol Q \, \boldsymbol x \): for instance, \( \boldsymbol Q \) is positive semidefinite if \( \boldsymbol x' \boldsymbol Q \, \boldsymbol x \ge 0 \) for all \( \boldsymbol x \in \R^n \). The box above gives an equivalent characterization in terms of eigenvalues.

  • One direction is easy to see: if \( \boldsymbol Q\boldsymbol v=\lambda\boldsymbol v \) with \( \boldsymbol v\neq\boldsymbol 0 \), then

    \[ \boldsymbol v' \boldsymbol Q \, \boldsymbol v=\lambda\,\boldsymbol v'\boldsymbol v=\lambda\sum_{i=1}^n v_i^2, \qquad \text{with} \quad \sum_{i=1}^n v_i^2>0, \]

    so \( \boldsymbol v' \boldsymbol Q \, \boldsymbol v \) and \( \lambda \) have the same sign. For example, if \( \boldsymbol Q \) is positive semidefinite, then every eigenvalue satisfies \( \lambda\ge 0 \).

  • A symmetric matrix which is neither positive semidefinite nor negative semidefinite is called indefinite: equivalently, it has at least one strictly positive and at least one strictly negative eigenvalue.

Example 1: eigenvalues and eigenvectors

Let us consider the matrix

\[ \boldsymbol Q= \begin{pmatrix} 2 & -1\\[0.5ex] -1 & 2 \end{pmatrix}. \]

The characteristic equation is:

\[ \det(\boldsymbol Q-\lambda \boldsymbol I) = \det\begin{pmatrix} 2-\lambda & -1\\[0.5ex] -1 & 2-\lambda \end{pmatrix} = (2-\lambda)^2-1 = \lambda^2-4\lambda+3. \]

Hence

\[ \det(\boldsymbol Q-\lambda \boldsymbol I)=0 \Longleftrightarrow (\lambda-1)(\lambda-3)=0, \]

and therefore the eigenvalues are:

\[ \lambda_1=1, \qquad \lambda_2=3. \]

An eigenvector associated with \(\lambda_1=1\) can be found by solving

\[ (\boldsymbol Q-\boldsymbol I)\boldsymbol v=\boldsymbol 0 \Longleftrightarrow \begin{pmatrix} 1 & -1\\[0.5ex] -1 & 1 \end{pmatrix} \boldsymbol v=\boldsymbol 0, \]

which gives (for example)

\[ \boldsymbol v_1= \begin{pmatrix} 1\\[0.5ex] 1 \end{pmatrix}. \]

Similarly, an eigenvector associated with \(\lambda_2=3\) satisfies

\[ (\boldsymbol Q-3\boldsymbol I)\boldsymbol v=\boldsymbol 0 \Longleftrightarrow \begin{pmatrix} -1 & -1\\[0.5ex] -1 & -1 \end{pmatrix} \boldsymbol v=\boldsymbol 0, \]

which gives (for example)

\[ \boldsymbol v_2= \begin{pmatrix} 1\\[0.5ex] -1 \end{pmatrix}. \]

Therefore,

\[ \boldsymbol Q\,\boldsymbol v_1 = 1\,\boldsymbol v_1, \qquad \boldsymbol Q\,\boldsymbol v_2 = 3\,\boldsymbol v_2. \]

We can verify it directly:

\[ \boldsymbol Q\,\boldsymbol v_1 = \begin{pmatrix} 2 & -1\\[0.5ex] -1 & 2 \end{pmatrix} \begin{pmatrix} 1\\[0.5ex] 1 \end{pmatrix} = \begin{pmatrix} 1\\[0.5ex] 1 \end{pmatrix} = 1\,\boldsymbol v_1, \qquad \boldsymbol Q\,\boldsymbol v_2 = \begin{pmatrix} 2 & -1\\[0.5ex] -1 & 2 \end{pmatrix} \begin{pmatrix} 1\\[0.5ex] -1 \end{pmatrix} = \begin{pmatrix} 3\\[0.5ex] -3 \end{pmatrix} = 3\,\boldsymbol v_2. \]

Since \( \boldsymbol Q \) is symmetric and both eigenvalues are strictly positive, the matrix \( \boldsymbol Q \) is positive definite.

2. Properties of eigenvalues

2.1 Triangular and diagonal matrices

Observation 1: eigenvalues of triangular and diagonal matrices

Let \( \boldsymbol Q \in \R^{n\times n} \) be an upper triangular, a lower triangular or a diagonal matrix. Then the eigenvalues of \( \boldsymbol Q \) are its diagonal entries \( q_{11}, q_{22}, \dots, q_{nn} \).

Proof

If \( \boldsymbol Q \) is triangular (or diagonal), then also \( \boldsymbol Q-\lambda\boldsymbol I \) is triangular (or diagonal), with diagonal entries \( q_{11}-\lambda, \dots, q_{nn}-\lambda \). Recall that the determinant of a triangular matrix is the product of its diagonal entries. Hence

\[ \det(\boldsymbol Q-\lambda\boldsymbol I)=\prod_{i=1}^{n}(q_{ii}-\lambda), \]

which is equal to zero if and only if \( \lambda=q_{ii} \) for some \( i\in\{1,2,\dots,n\} \). □

Example 2: eigenvalues of a triangular matrix

Let us consider the upper triangular matrix

\[ \boldsymbol U= \begin{pmatrix} 2 & 1 & 3\\[0.5ex] 0 & -1 & 4\\[0.5ex] 0 & 0 & 5 \end{pmatrix}. \]

The characteristic equation is

\[ \det(\boldsymbol U-\lambda\boldsymbol I) = \det\begin{pmatrix} 2-\lambda & 1 & 3\\[0.5ex] 0 & -1-\lambda & 4\\[0.5ex] 0 & 0 & 5-\lambda \end{pmatrix} = (2-\lambda)(-1-\lambda)(5-\lambda)=0, \]

and therefore the eigenvalues are the diagonal entries:

\[ \lambda_1=2, \qquad \lambda_2=-1, \qquad \lambda_3=5. \]

2.2 Trace, determinant and eigenvalues

Definition 2: trace of a matrix

Let \( \boldsymbol Q \in \R^{n\times n} \) be a square matrix. The trace of \( \boldsymbol Q \) is the sum of its diagonal entries:

\[ \operatorname{tr}(\boldsymbol Q)=\sum_{i=1}^{n} q_{ii}=q_{11}+q_{22}+\dots+q_{nn}. \]

  • The characteristic polynomial \( p(\lambda)=\det(\boldsymbol Q-\lambda\boldsymbol I) \) has degree \( n \). Over the complex numbers, a polynomial of degree \( n \) has exactly \( n \) roots \( \lambda_1,\lambda_2,\dots,\lambda_n \in \C \), provided that each root is counted as many times as its multiplicity (the number of times the factor \( (\lambda_i-\lambda) \) appears in the factorization). Hence

    \[ p(\lambda)=(\lambda_1-\lambda)\,(\lambda_2-\lambda)\cdots(\lambda_n-\lambda). \]

  • The real roots are the eigenvalues of Definition Definition 1. The roots which are not real are called complex eigenvalues of \( \boldsymbol Q \) (their eigenvectors have complex entries). If \( \boldsymbol Q \) is symmetric, all the roots are real.

Observation 2: determinant, trace and eigenvalues

Let \( \boldsymbol Q \in \R^{n\times n} \) and let \( \lambda_1,\lambda_2,\dots,\lambda_n \in \C \) be the roots of its characteristic polynomial, each repeated according to its multiplicity. Then

\[ \det(\boldsymbol Q)=\prod_{i=1}^{n}\lambda_i, \qquad\qquad \operatorname{tr}(\boldsymbol Q)=\sum_{i=1}^{n}\lambda_i. \]
Proof (Idea of the proof)

Setting \( \lambda=0 \) in \( p(\lambda)=(\lambda_1-\lambda)\cdots(\lambda_n-\lambda) \) gives \( \det(\boldsymbol Q)=p(0)=\lambda_1\,\lambda_2\cdots\lambda_n \). For the trace, we compare the coefficients of \( \lambda^{n-1} \): in the product \( (\lambda_1-\lambda)\cdots(\lambda_n-\lambda) \) this coefficient is \( (-1)^{n-1}\sum_{i=1}^n\lambda_i \), while in \( \det(\boldsymbol Q-\lambda\boldsymbol I) \) the power \( \lambda^{n-1} \) only comes from the product of the diagonal entries \( (q_{11}-\lambda)\cdots(q_{nn}-\lambda) \), and its coefficient is \( (-1)^{n-1}\sum_{i=1}^n q_{ii} \). □

Example 3: trace and determinant

  • For the matrix of Example Example 1 we have \( \lambda_1=1 \) and \( \lambda_2=3 \), and indeed

    \[ \operatorname{tr}\begin{pmatrix} 2 & -1\\[0.5ex] -1 & 2 \end{pmatrix} =2+2=4=1+3, \qquad \det\begin{pmatrix} 2 & -1\\[0.5ex] -1 & 2 \end{pmatrix} =4-1=3=1\cdot 3. \]

  • Multiplicities must be taken into account. The diagonal matrix

    \[ \boldsymbol D= \begin{pmatrix} 4 & 0 & 0\\[0.5ex] 0 & -3 & 0\\[0.5ex] 0 & 0 & 4 \end{pmatrix} \]

    has characteristic polynomial \( (4-\lambda)^2(-3-\lambda) \): the eigenvalue \( 4 \) has multiplicity \( 2 \) and the eigenvalue \( -3 \) has multiplicity \( 1 \). Hence \( \lambda_1=\lambda_2=4 \), \( \lambda_3=-3 \), and

    \[ \operatorname{tr}(\boldsymbol D)=4+4-3=5, \qquad \det(\boldsymbol D)=4\cdot 4\cdot(-3)=-48. \]

  • Complex roots must be taken into account. For the (non symmetric) matrix

    \[ \boldsymbol R= \begin{pmatrix} 0 & -1\\[0.5ex] 1 & 0 \end{pmatrix} \quad\text{we have}\quad \det(\boldsymbol R-\lambda\boldsymbol I)=\lambda^2+1, \]

    which has no real roots: \( \boldsymbol R \) has no (real) eigenvalues. Its complex eigenvalues are \( \lambda_1=i \) and \( \lambda_2=-i \), and indeed

    \[ \operatorname{tr}(\boldsymbol R)=0=i+(-i), \qquad \det(\boldsymbol R)=1=i\cdot(-i). \]

2.3 Eigenvectors of distinct eigenvalues

  • Recall that the vectors \( \boldsymbol v_1,\dots,\boldsymbol v_k \) are linearly independent if \( \alpha_1\boldsymbol v_1+\dots+\alpha_k\boldsymbol v_k=\boldsymbol 0 \) implies \( \alpha_1=\dots=\alpha_k=0 \).

Observation 3: eigenvectors of distinct eigenvalues

Let \( \lambda_1,\lambda_2,\dots,\lambda_k \) be distinct eigenvalues of \( \boldsymbol Q \in \R^{n\times n} \) and let \( \boldsymbol v_1,\boldsymbol v_2,\dots,\boldsymbol v_k \) be associated eigenvectors (\( \boldsymbol Q\boldsymbol v_i=\lambda_i\boldsymbol v_i \)). Then \( \boldsymbol v_1,\boldsymbol v_2,\dots,\boldsymbol v_k \) are linearly independent.

  • In particular, if \( \boldsymbol Q \in \R^{n\times n} \) has \( n \) distinct real eigenvalues, then one eigenvector for each eigenvalue gives \( n \) linearly independent vectors of \( \R^n \).

  • In Example Example 1, the eigenvectors \( \boldsymbol v_1=(1,1)' \) and \( \boldsymbol v_2=(1,-1)' \) are linearly independent, since neither of them is a multiple of the other.

3. A complete example with a \(3\times 3\) matrix

Example 4: eigenvalues and eigenvectors of a \(3\times 3\) matrix

Let us consider the symmetric matrix

\[ \boldsymbol Q= \begin{pmatrix} 2 & 1 & 0\\[0.5ex] 1 & 3 & 1\\[0.5ex] 0 & 1 & 2 \end{pmatrix}. \]

Characteristic polynomial. Expanding along the first row (Laplace expansion):

\[\begin{align*} \det(\boldsymbol Q-\lambda \boldsymbol I) &= \det\begin{pmatrix} 2-\lambda & 1 & 0\\[0.5ex] 1 & 3-\lambda & 1\\[0.5ex] 0 & 1 & 2-\lambda \end{pmatrix}\\[1ex] &= (2-\lambda)\big[(3-\lambda)(2-\lambda)-1\big] -1\cdot\big[1\cdot(2-\lambda)-1\cdot 0\big]\\[1ex] &= (2-\lambda)\big[(3-\lambda)(2-\lambda)-2\big] = (2-\lambda)\,(\lambda^2-5\lambda+4)\\[1ex] &= (2-\lambda)(\lambda-1)(\lambda-4). \end{align*}\]

Therefore the eigenvalues are

\[ \lambda_1=1, \qquad \lambda_2=2, \qquad \lambda_3=4. \]

Eigenvector for \( \lambda_1=1 \). We solve

\[ (\boldsymbol Q-\boldsymbol I)\boldsymbol v=\boldsymbol 0 \Longleftrightarrow \begin{pmatrix} 1 & 1 & 0\\[0.5ex] 1 & 2 & 1\\[0.5ex] 0 & 1 & 1 \end{pmatrix} \boldsymbol v=\boldsymbol 0 \Longleftrightarrow \left\{ \begin{array}{l} v_1+v_2=0\\[0.5ex] v_1+2v_2+v_3=0\\[0.5ex] v_2+v_3=0 \end{array} \right. \]

From the first and the third equation, \( v_1=-v_2 \) and \( v_3=-v_2 \) (the second equation is then satisfied). Choosing \( v_2=-1 \):

\[ \boldsymbol v_1= \begin{pmatrix} 1\\[0.5ex] -1\\[0.5ex] 1 \end{pmatrix}. \]

Eigenvector for \( \lambda_2=2 \). We solve

\[ (\boldsymbol Q-2\boldsymbol I)\boldsymbol v=\boldsymbol 0 \Longleftrightarrow \begin{pmatrix} 0 & 1 & 0\\[0.5ex] 1 & 1 & 1\\[0.5ex] 0 & 1 & 0 \end{pmatrix} \boldsymbol v=\boldsymbol 0 \Longleftrightarrow \left\{ \begin{array}{l} v_2=0\\[0.5ex] v_1+v_2+v_3=0 \end{array} \right. \]

hence \( v_2=0 \), \( v_3=-v_1 \) and, for example,

\[ \boldsymbol v_2= \begin{pmatrix} 1\\[0.5ex] 0\\[0.5ex] -1 \end{pmatrix}. \]

Eigenvector for \( \lambda_3=4 \). We solve

\[ (\boldsymbol Q-4\boldsymbol I)\boldsymbol v=\boldsymbol 0 \Longleftrightarrow \begin{pmatrix} -2 & 1 & 0\\[0.5ex] 1 & -1 & 1\\[0.5ex] 0 & 1 & -2 \end{pmatrix} \boldsymbol v=\boldsymbol 0 \Longleftrightarrow \left\{ \begin{array}{l} -2v_1+v_2=0\\[0.5ex] v_1-v_2+v_3=0\\[0.5ex] v_2-2v_3=0 \end{array} \right. \]

From the first and the third equation, \( v_2=2v_1 \) and \( v_3=v_2/2=v_1 \) (the second equation is then satisfied: \( v_1-2v_1+v_1=0 \)). Choosing \( v_1=1 \):

\[ \boldsymbol v_3= \begin{pmatrix} 1\\[0.5ex] 2\\[0.5ex] 1 \end{pmatrix}. \]

Checks. By Observation Observation 2,

\[ \operatorname{tr}(\boldsymbol Q)=2+3+2=7=1+2+4, \qquad \det(\boldsymbol Q)=2\cdot(6-1)-1\cdot(2-0)=8=1\cdot 2\cdot 4. \]

Since the three eigenvalues are distinct, by Observation Observation 3 the eigenvectors \( \boldsymbol v_1,\boldsymbol v_2,\boldsymbol v_3 \) are linearly independent. Finally, \( \boldsymbol Q \) is symmetric with strictly positive eigenvalues, hence it is positive definite.

4. Sylvester's criterion

  • Computing all the eigenvalues of a matrix can be difficult. For symmetric matrices, definiteness and semidefiniteness can also be checked by computing determinants of suitable square submatrices.

Definition 3: principal minors

Let \( \boldsymbol Q \in \R^{n\times n} \).

  • A principal minor of order \( k \) of \( \boldsymbol Q \) is the determinant of the \( k\times k \) submatrix obtained from \( \boldsymbol Q \) by keeping the rows and the columns with the same indices \( i_1<i_2<\dots<i_k \) (equivalently, by deleting the same \( n-k \) rows and columns).

  • The leading principal minor of order \( k \) is the principal minor obtained by keeping the first \( k \) rows and columns:

    \[ \Delta_k=\det \begin{pmatrix} q_{11} & \cdots & q_{1k}\\ \vdots & \ddots & \vdots\\ q_{k1} & \cdots & q_{kk} \end{pmatrix}, \qquad k=1,2,\dots,n. \]

  • A matrix \( \boldsymbol Q\in\R^{n\times n} \) has exactly \( n \) leading principal minors \( \Delta_1=q_{11},\ \Delta_2,\ \dots,\ \Delta_n=\det(\boldsymbol Q) \).

  • For \( n=3 \), the principal minors are: of order 1, \( q_{11},q_{22},q_{33} \); of order 2, the determinants of the submatrices with indices \( \{1,2\} \), \( \{1,3\} \), \( \{2,3\} \); of order 3, \( \det(\boldsymbol Q) \).

Theorem 1: Sylvester's criterion

Let \( \boldsymbol Q \in \R^{n\times n} \) be a symmetric matrix with leading principal minors \( \Delta_1,\Delta_2,\dots,\Delta_n \). Then

  • \( \boldsymbol Q \) is positive definite if and only if

    \[ \Delta_k>0, \qquad \forall k\in\{1,2,\dots,n\}; \]

  • \( \boldsymbol Q \) is negative definite if and only if the leading principal minors alternate in sign starting with a negative one:

    \[ \Delta_1<0,\quad \Delta_2>0,\quad \Delta_3<0,\quad \dots \qquad\text{i.e.}\qquad (-1)^k\,\Delta_k>0, \qquad \forall k\in\{1,2,\dots,n\}. \]

Observation 4: semidefinite matrices and principal minors

Let \( \boldsymbol Q \in \R^{n\times n} \) be a symmetric matrix. Then

  • \( \boldsymbol Q \) is positive semidefinite if and only if all its principal minors (not only the leading ones) are \( \ge 0 \);

  • \( \boldsymbol Q \) is negative semidefinite if and only if every principal minor of order \( k \) has the sign of \( (-1)^k \) or is zero, i.e., \( (-1)^k\,(\text{principal minor of order } k) \ge 0 \), for all \( k \).

For semidefiniteness it is not enough to check \( \Delta_k\ge 0 \) for the leading principal minors only.

Example 5: a positive definite matrix

For the matrix of Example Example 4

\[ \boldsymbol Q= \begin{pmatrix} 2 & 1 & 0\\[0.5ex] 1 & 3 & 1\\[0.5ex] 0 & 1 & 2 \end{pmatrix} \]

the leading principal minors are

\[ \Delta_1=2>0, \qquad \Delta_2=\det\begin{pmatrix} 2 & 1\\[0.5ex] 1 & 3 \end{pmatrix}=6-1=5>0, \qquad \Delta_3=\det(\boldsymbol Q)=8>0. \]

By Sylvester's criterion, \( \boldsymbol Q \) is positive definite, in agreement with its eigenvalues \( 1,2,4 \).

Example 6: an indefinite matrix

Let us consider

\[ \boldsymbol Q= \begin{pmatrix} 1 & 2\\[0.5ex] 2 & 1 \end{pmatrix}. \]

We have \( \Delta_1=1>0 \) and \( \Delta_2=1-4=-3<0 \). Hence:

  • \( \boldsymbol Q \) is not positive definite (\( \Delta_2<0 \)) and not negative definite (\( \Delta_1>0 \));

  • \( \boldsymbol Q \) is not positive semidefinite (the principal minor \( \Delta_2 \) of order 2 is negative) and not negative semidefinite (\( (-1)^2\Delta_2<0 \)).

Therefore \( \boldsymbol Q \) is indefinite. Indeed, recalling the definition via \( \boldsymbol x' \boldsymbol Q \, \boldsymbol x \):

\[ \boldsymbol x=\begin{pmatrix}1\\0\end{pmatrix}:\ \ \boldsymbol x' \boldsymbol Q \, \boldsymbol x=1>0, \qquad \boldsymbol x=\begin{pmatrix}1\\-1\end{pmatrix}:\ \ \boldsymbol x' \boldsymbol Q \, \boldsymbol x=1-4+1=-2<0. \]

The eigenvalues are the roots of \( (1-\lambda)^2-4=(\lambda-3)(\lambda+1) \), i.e., \( 3 \) and \( -1 \): one positive and one negative.

Example 7: a singular positive semidefinite matrix

Let us consider

\[ \boldsymbol Q= \begin{pmatrix} 1 & 1 & 0\\[0.5ex] 1 & 2 & 1\\[0.5ex] 0 & 1 & 1 \end{pmatrix}. \]

The leading principal minors are \( \Delta_1=1 \), \( \Delta_2=2-1=1 \) and \( \Delta_3=\det(\boldsymbol Q)=1\cdot(2-1)-1\cdot(1-0)=0 \). Since \( \Delta_3=0 \), \( \boldsymbol Q \) is not positive definite.

To decide whether it is positive semidefinite we check all principal minors:

  • order 1: \( q_{11}=1,\ q_{22}=2,\ q_{33}=1 \);

  • order 2: \( \det\begin{pmatrix}1 & 1\\ 1 & 2\end{pmatrix}=1,\ \ \det\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=1,\ \ \det\begin{pmatrix}2 & 1\\ 1 & 1\end{pmatrix}=1 \);

  • order 3: \( \det(\boldsymbol Q)=0 \).

All principal minors are \( \ge 0 \), hence \( \boldsymbol Q \) is positive semidefinite. Indeed

\[ \boldsymbol x' \boldsymbol Q \, \boldsymbol x = x_1^2+2x_1x_2+2x_2^2+2x_2x_3+x_3^2=(x_1+x_2)^2+(x_2+x_3)^2\ge 0, \]

and the eigenvalues are \( 0,1,3 \) (\( \boldsymbol v=(1,-1,1)' \) satisfies \( \boldsymbol Q\boldsymbol v=\boldsymbol 0=0\,\boldsymbol v \)).

Example 8: leading principal minors are not enough for semidefiniteness

Let us consider

\[ \boldsymbol Q= \begin{pmatrix} 0 & 0\\[0.5ex] 0 & -1 \end{pmatrix}. \]

The leading principal minors are \( \Delta_1=0\ge 0 \) and \( \Delta_2=0\ge 0 \). However \( \boldsymbol Q \) is not positive semidefinite: for

\[ \boldsymbol x=\begin{pmatrix}0\\1\end{pmatrix} \qquad\text{we have}\qquad \boldsymbol x' \boldsymbol Q \, \boldsymbol x=-1<0. \]

The principal minor \( q_{22}=-1 \) (which is not a leading one) detects this. In fact, the eigenvalues are \( 0 \) and \( -1 \), and \( \boldsymbol Q \) is negative semidefinite.

Try it in the lab

the same computation step by step: change the matrix and watch the steps change.

Exercises and lab