32.5. Example

Let:

A= [4 1] [0 4]

Step 1: Characteristic Equation

det(A−λI)=det [4−λ 1 ] [0 4−λ]

So, λ=4 is a repeated eigenvalue (AM = 2).

In this step, we need to find the characteristic equation of the matrix A. This is done by calculating the determinant of the matrix (A - λI), where λ represents the eigenvalue and I is the identity matrix. The result shows that there is a repeated eigenvalue λ = 4, meaning the algebraic multiplicity (the number of times the eigenvalue appears) is 2.

Think about how many times your favorite item appears in a store. If there are two of the same product on the shelf, then in terms of eigenvalues, we can say that there is a repeated value. Here, our 'product' is the eigenvalue '4' that appears more than once.

Step 2: Eigenspace Solve (A−4I)v=0:

[0 1][x] = [0] [0 0] [y] = [0]

So, y=0

So, x∈R, and the eigenspace is:

E = Span{[1 0]}

Only one linearly independent eigenvector ⇒ GM = 1 < AM = 2 ⇒ Matrix is not diagonalizable.

Here, we are solving the equation derived from the eigenvalue we found earlier (λ = 4). By substituting λ into (A - λI) and solving for the vector v, we observe that y must equal 0. Consequently, we conclude that x can be any real number (x ∈ R). This gives us the eigenspace E, which spans in the direction of the vector [1, 0]. Since we found only one linearly independent eigenvector, the geometric multiplicity (GM) is 1, which is less than the algebraic multiplicity (AM) of 2. Therefore, this matrix cannot be diagonalized.

Imagine you have a group of people where only one person stands out because they are different. In this context, that 'one person' corresponds to the one linearly independent eigenvector we found, while the others who align with this person's direction represent higher multiplicities that don't offer any distinctiveness.

Basis of eigenvectors for λ=4:

{[1]}

0

This part states that the only basis of eigenvectors corresponding to the repeated eigenvalue λ=4 consists of the vector [1, 0]. This means any vector in the eigenspace can be expressed as a scalar multiple of this basis vector.

It's like having a single recipe that can be altered by changing just the quantity of one ingredient. Here, our basis vector [1, 0] is that primary recipe, and any vector in the eigenspace is simply a variation of it based on how much of it we 'use'.

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