Investigate the conditions for the equality of the image of (A-λ₁I) and the kernel of (A-λ₂I) for a matrix A with distinct eigenvalues. | Step-by-Step Solution

Hello! This is a linear algebra problem that connects eigenvalues, kernels, and images in a really elegant way.

1. What We're Solving:

We want to find when the image (column space) of the matrix (A-λ₁I) equals the kernel (null space) of the matrix (A-λ₂I), where A is a 2×2 matrix with two distinct eigenvalues λ₁ and λ₂.

2. The Approach:

This problem explores a relationship between different subspaces. We'll use dimensional analysis and properties of eigenspaces to understand when these two subspaces can be equal. The key insight is that for a 2×2 matrix, subspaces have limited possible dimensions!

3. Step-by-Step Solution:

Step 1: Understand what we know

• A is 2×2 with distinct eigenvalues λ₁ and λ₂
• Since eigenvalues are distinct, A is diagonalizable
• Each eigenspace has dimension 1 (since we're in 2D with 2 distinct eigenvalues)

Step 2: Analyze the dimensions For any 2×2 matrix M, we have: dim(ker(M)) + dim(Im(M)) = 2

For (A-λ₁I):

• Since λ₁ is an eigenvalue, (A-λ₁I) is not invertible
• So dim(ker(A-λ₁I)) ≥ 1
• Since eigenvalues are distinct, dim(ker(A-λ₁I)) = 1
• Therefore: dim(Im(A-λ₁I)) = 2 - 1 = 1

Similarly for (A-λ₂I): dim(ker(A-λ₂I)) = 1

Step 3: Check if equality is possible Now we have:

• dim(Im(A-λ₁I)) = 1
• dim(ker(A-λ₂I)) = 1

Since both subspaces are 1-dimensional in a 2-dimensional space, they could be equal!

Step 4: Find the specific condition Let's consider what these subspaces actually are:

• ker(A-λ₂I) is the eigenspace for λ₂
• Im(A-λ₁I) is the orthogonal complement to ker(A-λ₁I) in the context of the transformation

The key insight: Im(A-λ₁I) = ker(A-λ₂I) if and only if the eigenspace for λ₂ is orthogonal to the eigenspace for λ₁.

Step 5: When does this happen? This occurs when the eigenvectors corresponding to λ₁ and λ₂ are orthogonal to each other.

4. The Answer:

Im(A-λ₁I) = ker(A-λ₂I) if and only if the eigenvectors of A corresponding to λ₁ and λ₂ are orthogonal.

In other words, this equality holds precisely when A is not just diagonalizable, but orthogonally diagonalizable (i.e., when A is a normal matrix, which for real matrices means A is symmetric).

5. Memory Tip:

Think of it this way: "Images and kernels play nicely together when the matrix is 'well-behaved' (symmetric/normal). The eigenvectors being perpendicular creates a perfect harmony between these subspaces!"

Great question! This problem beautifully illustrates how geometric properties (orthogonal eigenvectors) connect to algebraic properties (equality of subspaces). Keep exploring these connections - they're at the heart of advanced linear algebra!