32.2. Basis of an Eigenspace

To understand the basis of eigenvectors, we focus on finding a basis for each eigenspace $E_\lambda$.

Let’s suppose:

• A is an n×n matrix
• $\lambda$ is an eigenvalue of A
• We solve $(A−\lambda I)v=0$

This chunk introduces the concept of finding a basis of eigenvectors for a given eigenvalue $\lambda$. First, we determine what it means to have a basis: it is a set of vectors that spans the vector space formed by the eigenspace corresponding to $\lambda$. To find eigenvectors, we start with the matrix $A$ and the eigenvalue $\lambda$ and solve the equation $(A−\lambda I)v=0$, where $I$ is the identity matrix. This equation is a homogeneous system of linear equations, and its solutions will give us the eigenvectors that form our basis.

Imagine you are teaching a music class and want to teach a particular song. The notes you choose to play are analogous to the eigenvectors, while the full song represents the eigenspace. The eigenvectors must be unique but together create the complete music piece—this is the basis of your song.

The solutions to this homogeneous system form a vector space, and the vectors that span this space are the eigenvectors corresponding to $\lambda$.

When we solve the equation $(A−\lambda I)v=0$, the collection of solutions (eigenvectors) we find constitutes a vector space—specifically the eigenspace for that eigenvalue $\lambda$. This means that any linear combination of these eigenvectors will also be in this vector space, thereby fulfilling one of the necessary conditions for a valid vector space: closure under addition and scalar multiplication.

Think about filling a jar with water. Each drop of water represents an eigenvector. When combined, they fill the jar (the vector space). Just like you can mix and match drops to fill the jar to various levels, eigenvectors can be linearly combined to produce any vector in the eigenspace.

A basis of the eigenspace is a linearly independent set of eigenvectors that spans the entire eigenspace.

Once we find the eigenvectors corresponding to the eigenvalue $\lambda$, we need to check if they form a basis. A basis must not only span the entirety of the eigenspace but must also consist of linearly independent vectors. This means no vector in the basis can be expressed as a linear combination of the others. If the eigenvectors satisfy these criteria, they collectively provide a complete way to construct any vector within that eigenspace.

Consider a set of colored pencils. If every pencil (eigenvector) represents a unique color, then together they can create any color you want (span the eigenspace) as long as no two pencils are the same color (linearly independent). This way, you have a full palette to work with!

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