Question

Exercise 3.3 .6 Find the characteristic polynomial of the \(n \times n\) identity matrix \(I\). Show that \(I\) has exactly one eigenvalue and find the eigenvectors.

Short answer

The characteristic polynomial is \((1 - \lambda)^n\). The identity matrix I has exactly one eigenvalue, \(\lambda = 1\), with every vector in \(\mathbb{R}^n\) as an eigenvector.

Step-by-step solution

Define the Identity Matrix

Consider the identity matrix \(I\) which is an \(n \times n\) matrix with 1s on the diagonal and 0s elsewhere. It looks like this: \(I = \begin{pmatrix} 1 & 0 & \cdots & 0 \ 0 & 1 & \cdots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \cdots & 1 \end{pmatrix}\).

Write the Characteristic Equation

The characteristic polynomial of a matrix \(A\) is given by \(\det(A - \lambda I)\), where \(\lambda\) is a scalar. For the identity matrix \(I\), the equation becomes \(\det(I - \lambda I) = \det((1 - \lambda)I)\).

Calculate the Determinant

Since \((1 - \lambda)I\) is a diagonal matrix where each diagonal entry is \(1 - \lambda\), its determinant is \((1 - \lambda)^n\). Thus, the characteristic polynomial is \((1 - \lambda)^n\).

Identify the Eigenvalue

The roots of the characteristic polynomial \((1 - \lambda)^n\) are where this polynomial equals zero. Solving \((1 - \lambda)^n = 0\), we get \(\lambda = 1\). Thus, \(\lambda = 1\) is an eigenvalue of \(I\).

Find the Eigenvectors

To find the eigenvectors, solve \((I - 1I)x = 0\), which simplifies to \(0 = 0\). This implies that any non-zero vector will satisfy this equation, meaning every vector in \(\mathbb{R}^n\) is an eigenvector of the identity matrix corresponding to the eigenvalue \(1\).

Key concepts

Identity Matrix

An identity matrix is a special kind of square matrix that plays a crucial role in linear algebra. It is named so because it acts like the "identity" element for matrix multiplication. That is, when any matrix is multiplied by the identity matrix, it remains unchanged.

The identity matrix is denoted by the symbol \(I\), and it is defined as a square matrix with ones on the diagonal and zeros elsewhere. For example, a 3x3 identity matrix looks like this:

• \(I = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}\)

Every identity matrix is unique for its size, and it serves as the basis for discussing concepts like eigenvalues and eigenvectors.

Eigenvalues

Eigenvalues are fundamental in understanding the behavior of matrices, especially in transformations and systems of linear equations. An eigenvalue, usually denoted by \(\lambda\), is a scalar that indicates how much the eigenvector associated with it is stretched during a transformation represented by a matrix.

To find the eigenvalues of a matrix, we look for values of \(\lambda\) that satisfy the characteristic equation \(\det(A - \lambda I) = 0\). For an identity matrix \(I\) of size \(n \times n\), this equation simplifies to \((1 - \lambda)^n = 0\). This reveals the eigenvalue of the identity matrix as \(\lambda = 1\).

Since the identity matrix doesn't change any vector, its only eigenvalue is \(1\), repeated \(n\) times, corresponding to the matrix's size.

Eigenvectors

Along with eigenvalues, eigenvectors provide deep insights into the structure of a matrix. An eigenvector corresponding to an eigenvalue \(\lambda\) is a non-zero vector that satisfies \((A - \lambda I)x = 0\).

For an identity matrix \(I\) where the eigenvalue is \(\lambda = 1\), the equation simplifies to \(I - I = 0\), leading to the solution \( 0 = 0\). This means any non-zero vector is an eigenvector for the identity matrix.

In essence, the entire vector space \(\mathbb{R}^n\) becomes the set of eigenvectors corresponding to the eigenvalue \(1\) for an identity matrix. This means that every possible direction in the vector space remains unchanged during the transformation represented by the identity matrix, emphasizing its neutral effect.