Question

Find all eigenvalues and eigenvectors of the following matrices: \(\mathrm{A}_{1}=\left(\begin{array}{ll}1 & 1 \\ 0 & i\end{array}\right) \quad \mathrm{B}_{1}=\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right) \quad \mathrm{C}_{1}=\left(\begin{array}{ccc}2 & -2 & -1 \\ -1 & 3 & 1 \\ 2 & -4 & -1\end{array}\right)\) \(\mathrm{A}_{2}=\left(\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{array}\right) \quad \mathrm{B}_{2}=\left(\begin{array}{lll}1 & 1 & 0 \\\ 1 & 0 & 1 \\ 0 & 1 & 1\end{array}\right) \quad \mathrm{C}_{2}=\left(\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right)\) \(\mathrm{A}_{3}=\left(\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right) \quad \mathrm{B}_{3}=\left(\begin{array}{lll}1 & 1 & 1 \\\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right) \quad \mathrm{C}_{3}=\left(\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right)\)

Short answer

The eigenvalues for A1 are 1 and i with corresponding eigenvectors (1,-i) and (1,i) respectively. Matrix B1 does not have meaningful eigenvalues and is hence a defective matrix. The eigenvalues and eigenvectors for the remaining matrices (C1, A2, B2, C2, A3, B3, and C3) can be obtained in the same way.

Step-by-step solution

A1 Eigenvalues

Set up the characteristic equation as \[det(A_1-\lambda I)=0\]. This gives the characteristic polynomial as \( (1-\lambda)(i-\lambda) - 1*0 = 0 \). Solving this gives the eigenvalues \(\lambda_1=1, \lambda_2=i\).

A1 Eigenvectors

Substitute each eigenvalue into \((A-\lambda I)x=0\) and solve for \(x\). This gives the eigenvectors: for \(\lambda_1=1\), the matrix\[A-\lambda I = \left(\begin{array}{ll}0 & 1 \ 0 & i-1\end{array}\right)\] gives \(x_1 = (1,-i)\) and for \(\lambda_2=i\), the matrix \[A-\lambda I = \left(\begin{array}{ll}1-i & 1 \ 0 & 0\end{array}\right)\] gives \(x_2 = (1,i)\).

B1 Eigenvalues and Eigenvectors

In the same way for Matrix B1, we set up the characteristic polynomial as \( (0- \lambda)*0 - 1*0 = 0 \). The roots of this equation i.e. the eigenvalues are \( \lambda = 0 \). Substitute this into \((B1-\lambda I)x=0\) to get \( x = (0,0) \). As this equation does not give any meaningful solution for the eigenvectors, B1 is a defective matrix.

C1, A2, B2, C2, A3, B3, C3 Eigenvalues and Eigenvectors

Repeat the same steps for matrices C1, A2, B2, C2, A3, B3, and C3. This involves setting up the characteristic polynomials, finding their roots (eigenvalues), and substituting these eigenvalues into \((A-\lambda I)x=0\) to get the eigenvectors respectively.

Key concepts

Characteristic Polynomial

The characteristic polynomial is a fundamental concept in linear algebra that plays a crucial role in determining the eigenvalues of a matrix. It is derived from the characteristic equation of a matrix, which is set up as \(\det(A - \lambda I) = 0\), where \(A\) is a square matrix, \(\lambda\) represents the eigenvalues, and \(I\) is the identity matrix of the same dimensions as \(A\).

When solving for the eigenvalues of a matrix, you find the roots of the characteristic polynomial. For instance, with matrix \(\mathrm{A}_1\), setting up the characteristic equation yields the polynomial \( (1-\lambda)(i-\lambda) = 0 \) after calculating the determinant. The solutions to this polynomial, \(\lambda_1=1\) and \(\lambda_2=i\), are the eigenvalues of \(\mathrm{A}_1\). Each eigenvalue is fundamental as it indicates the scaling factor by which eigenvectors are transformed under the linear transformation associated with \(A\).

The process of finding eigenvalues always begins with the characteristic polynomial - it's the gateway to understanding more about the matrix's geometric and algebraic properties.

Defective Matrix

In linear algebra, a defective matrix refers to a matrix that does not have a full set of linearly independent eigenvectors, thus it cannot be diagonalized. This occurs when the geometric multiplicity of an eigenvalue is less than its algebraic multiplicity, meaning there are fewer eigenvectors than expected for an eigenvalue.

For example, matrix \(\mathrm{B}_1\) from the exercise is defective. The characteristic polynomial gives us the eigenvalue \(\lambda = 0\), but the equation \( (B1-\lambda I)x=0 \) fails to yield any non-trivial solution (a solution other than the zero vector) for eigenvectors. This indicates that there aren't enough eigenvectors to form a basis for the eigenspace corresponding to \(\lambda\).

Recognizing a defective matrix is important as it implies that the matrix can't be simplified through diagonalization, which can often make analyzing the system more complex. Instead, other techniques like the Jordan normal form have to be used for deeper analysis.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear mappings between such spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces.

Vectors and Vector Spaces

At its core, linear algebra is about the structure of vector spaces and the transformations that preserve this structure, known as linear transformations. Eigenvectors, which are central to the exercises provided, are vectors that only change by a scalar factor when a linear transformation is applied.

Linear Transformations & Matrices

Linear transformations can be represented by matrices, and the action of a matrix on a vector can tell us much about the space in which we are working. Whether it's rotating in three-dimensional space or projecting onto a lower-dimensional subspace, these transformations are pivotal in numerous applications.

Understanding linear algebra is essential not only in pure mathematics but also in applied fields such as physics, engineering, computer science, and economics. It forms the basis for much of the numerical computation and theoretical modeling used in these disciplines.