32.10. Extended Example: 3×3 Matrix

Let us consider the matrix:

$A = \begin{bmatrix} 2 & 0 & 0 \\\ 0 & 3 & 4 \\\ 0 & 4 & 3 \end{bmatrix}$

Step 1: Characteristic Polynomial

$det(A-\lambda I)=det \begin{bmatrix} 2-\lambda & 0 & 0 \\\ 0 & 3-\lambda & 4 \\\ 0 & 4 & 3-\lambda \end{bmatrix}$

Expanding along the first row:

$(2-\lambda) \cdot det \begin{bmatrix} 3-\lambda & 4 \\\ 4 & 3-\lambda \end{bmatrix}$

So, eigenvalues are: $\lambda = 2, \lambda = 7, \lambda = -1$

In this step, we start with a 3x3 matrix and seek to find its characteristic polynomial. The characteristic polynomial is obtained by subtracting an eigenvalue (represented as λ) multiplied by the identity matrix I from our original matrix A and calculating the determinant of the resulting matrix. This determinant is set to zero to find the eigenvalues. In this case, we find that our eigenvalues are λ = 2, λ = 7, and λ = -1.

Think of the characteristic polynomial like a recipe where the ingredients (our matrix A and λ) must blend perfectly to create the dish (which in this case is the behavior of the matrix determined by its eigenvalues). When we find the perfect mix and set our recipe (the determinant) to zero, we uncover the tasty secrets (eigenvalues) that our matrix holds.

Step 2: Find Eigenvectors For ( \lambda=2:\n $A - 2I = \begin{bmatrix} 0 & 0 & 0 \\\ 0 & 1 & 4 \\\ 0 & 4 & 1 \end{bmatrix}$ Solve: $(A-2I)v=0$ This implies: $v_1 = 0, v_2 = 0, v_3 = free \Rightarrow v = \begin{bmatrix} 1 \\\ 0 \\\ 0 \end{bmatrix}$ Thus, basis of eigenvectors for ( \lambda=2: {\begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}}

After identifying the eigenvalue λ = 2, we proceed to find the corresponding eigenvector. This involves substituting λ back into the modified matrix (A - 2I) and then solving the resulting system of equations for the vector v. The solution provides us with the basis for eigenvectors related to this eigenvalue. Here, we find that the eigenvector for λ = 2 is simply [1, 0, 0], which indicates a direction in our vector space.

Finding eigenvectors can be compared to determining the specific route to take when navigating a city. Just like there are multiple roads leading to your destination, each eigenvalue provides a unique direction (or eigenvector) that exhibits how the matrix will behave. When the eigenvalue is 2, this particular route corresponds simply to one straight road marked as [1, 0, 0].

Similarly solve for $\lambda=7$ and $\lambda=-1$:

In this part of the exercise, we apply the same method to determine the eigenvectors associated with the remaining eigenvalues, λ = 7 and λ = -1. By updating our calculations based on these eigenvalues and solving the equations formed from (A - λI)v = 0, we will find specific directions (eigenvectors) for each eigenvalue. Although the specific calculations are not shown here, they follow the pattern established in finding the eigenvector for λ = 2.

Imagine you are exploring different paths in a park. Each path represents an eigenvalue leading you through different sections of the park. Just like you would take time to explore each path based on the unique scenery it offers, we explore each eigenvalue to uncover its unique eigenvectors, understanding how they each contribute to the overall layout (or behavior) of the park (or matrix).

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