Review for Quiz 3

Lecture 34

Minjae Park

Auburn University
MATH 2660 - Spring 2026

April 8, 2026

Attendance

Scan the QR code or go to join.iclicker.com/MBNJ.

Log in with our institution (Auburn - Mathematics & Statistics).

Quiz information

  • Policy update: You may submit the quiz only once.
  • Please bring an electronic device that can access WebAssign.
  • Be logged in and ready before 11:00 AM to avoid any technical issues.
  • The quiz is closed book. No materials are allowed, including the course website.
    • You may use blank scratch paper and a pen, or a tablet/iPad for writing.
    • If using a tablet, only a blank writing app (white page) is allowed—no other apps or materials may be open.
  • A traditional calculator is allowed, though it will likely not be necessary.
  • Headphones are allowed if music helps you focus, but the volume must be low enough to hear announcements and not distract others. If it becomes disruptive, you will be asked to turn it off.

Topics

  1. Applications of determinants
    • Cramer’s rule (1)
    • Cross product (1)
  2. Eigenvalues (1)
  3. Eigenvectors (1)
  4. Diagonalization (1)
  5. Differential equations (1)
  6. True/False questions

There will be a total of 6 questions plus several True/False questions.
You should aim to spend no more than 7 minutes per question.

Cramer’s Rule

  • Let \(A=[\vec{u}_1\ \dots\ \vec{u}_n]\) be an invertible \(n\times n\) matrix (so \(\det(A)\neq0\)).
  • Consider the linear system \[A\vec{x}=\vec{b},\] where \(\vec{x}=\langle x_1,\dots,x_n \rangle\).
  • Cramer’s Rule provides an explicit formula for each coordinate: \[ x_i=\frac{\det(A_i)}{\det(A)}, \] where \(A_i\) is obtained from \(A\) by replacing its \(i\)-th column with \(\vec{b}\).

Cross Product

  • Let \(\vec{u}=\langle u_1,u_2,u_3 \rangle\) and \(\vec{v}=\langle v_1,v_2,v_3 \rangle\) be vectors in \(\mathbb R^3\).
  • Often we want a vector perpendicular to both \(\vec{u}\) and \(\vec{v}\), i.e., orthogonal to the plane spanned by \(\vec{u}\) and \(\vec{v}\).
  • This appears in applications such as torque in physics and magnetic or electric forces.
  • We define the cross product using the formal determinant \[ \vec{u}\times\vec{v} = \begin{vmatrix} \vec{e}_1 & u_1 & v_1 \\ \vec{e}_2 & u_2 & v_2 \\ \vec{e}_3 & u_3 & v_3 \end{vmatrix}, \] where \(\vec{e}_1,\vec{e}_2,\vec{e}_3\) are the standard basis vectors of \(\mathbb R^3\) and \(\vec{u},\vec{v}\) appear as columns.

Eigenvalues and eigenvectors

Let \(A\) be an \(n\times n\) matrix. If a nonzero vector \(\vec{u}\in\mathbb{R}^n\) satisfies \[ A\vec{u} = \lambda \vec{u} \] for some \(\lambda\in\mathbb{R}\), then:

  • \(\vec{u}\) is called an eigenvector of \(A\)
  • \(\lambda\) is called an eigenvalue of \(A\)
  • Geometric interpretation: the action of \(A\) along the direction of \(\vec{u}\) is scaling by the factor \(\lambda\)

Characteristic equation

Let \(A\) be an \(n\times n\) matrix. The equation \[ \det(A-\lambda I_n)=0 \] is called the characteristic equation of \(A\). It is a polynomial equation of degree \(n\) in \(\lambda\).

  • If \(\lambda\) is a root of the characteristic equation, then \((A-\lambda I_n)\vec{u}=\vec{0}\) has a nonzero solution, so \(\lambda\) is an eigenvalue of \(A\).

Diagonalization

  • Let \(A\) be an \(n\times n\) matrix.
  • Suppose that \(\{\vec{u}_1,\dots,\vec{u}_n\}\) is an eigenbasis of \(A\) with eigenvalues \(\lambda_1,\dots,\lambda_n\).
  • Let \[ P=[\vec{u}_1 \ \cdots \ \vec{u}_n], \quad D=\operatorname{diag}(\lambda_1,\dots,\lambda_n). \]
  • Then \(A\) is diagonalizable and \[ A=PDP^{-1} \] is called a diagonalization of \(A\).
  • Geometric intuition: in the eigenbasis, \(A\) acts simply by scaling each coordinate by \(\lambda_i\).

Applications

  • Suppose \(A=PDP^{-1}\).
  • Matrix powers: \[ A^k = PD^kP^{-1}, \quad D^k=\operatorname{diag}(\lambda_1^k,\dots,\lambda_n^k). \]
  • Matrix exponential: \[ e^A = Pe^DP^{-1}, \quad e^D=\operatorname{diag}(e^{\lambda_1},\dots,e^{\lambda_n}). \]
  • Recall: \[ e^A = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots \]

Solving linear ODEs

  • Higher-order linear ODEs can be rewritten as a system of linear ODEs: \[ Y'(t)=AY(t). \]

  • The solution is given by \[ Y(t)=e^{At}Y(0). \]

  • If \(A\) is diagonalizable (\(A=PDP^{-1}\)), then \[ e^{At}=Pe^{Dt}P^{-1}, \] which makes computation much easier because \(e^{Dt}\) is just exponentials of diagonal entries.

  • In particular, if \(\vec{u}_1,\dots,\vec{u}_n\) are eigenvectors with eigenvalues \(\lambda_1,\dots,\lambda_n\), then the general solution is \[ Y(t)=c_1 e^{\lambda_1 t}\vec{u}_1 + \cdots + c_n e^{\lambda_n t}\vec{u}_n. \]

  • If an initial condition \(Y(0)\) is given, the constants \(c_1,\dots,c_n\) can be determined by solving a linear system.

Some statements

  • If all eigenvalues are distinct, then an eigenbasis exists, and the matrix is diagonalizable.
  • If a matrix is symmetric, then an orthonormal eigenbasis exists, and the matrix is orthogonally diagonalizable (i.e., \(A = P D P^T\) with \(P\) orthogonal).
  • If there are repeated eigenvalues, the matrix may or may not be diagonalizable.
  • Every matrix \(A\) has a singular value decomposition \(A = U \Sigma V^T\), where \(U\) and \(V\) are orthogonal matrices.
  • Geometrically, SVD can be interpreted as: rotation/flip \(\rightarrow\) stretch \(\rightarrow\) rotation/flip.
  • SVD is similar in spirit to diagonalization, but it always exists—even when a matrix is not diagonalizable.
  • Even if a matrix is diagonalizable, its SVD and diagonalization are generally different.

Practice questions

Please review the questions in Quiz 3 Review!