7 Eigenvalues and Eigenvectors

Invariant directions and scaling

An eigenvector is a nonzero vector whose direction is preserved by a linear transformation. If $A$ is a matrix, the defining equation is

where $\lambda$ is the associated eigenvalue. The transformation may stretch, shrink, or reverse the vector, but it does not move it to a different line through the origin.

The value of $\lambda$ describes the effect on the eigenvector:

• If $\lambda>1$, the vector is stretched.

• If $0<\lambda<1$, the vector is shrunk without reversing direction.

• If $\lambda<0$, the vector is reversed and scaled.

• If $\lambda=0$, the vector is sent to the zero vector.

The zero vector is not an eigenvector because it satisfies $A0=\lambda 0$ for every scalar $\lambda$. Rearranging the defining equation gives

Therefore, finding eigenvectors for a known eigenvalue means finding the nonzero vectors in the nullspace of $A-\lambda I$.

For example, let

Then

Thus the coordinate-axis directions are preserved, with eigenvalues $3$ and $-2$.

Takeaway: Begin with $Av=\lambda v$, and remember that eigenvectors are nonzero vectors while eigenvalues are the corresponding scalars.

Finding eigenvalues and eigenvectors

The equation $(A-\lambda I)v=0$ has a nonzero solution exactly when the matrix $A-\lambda I$ is singular. The determinant criterion for singularity gives

Equivalently, one may use the characteristic polynomial

The roots of this polynomial are the eigenvalues of $A$. For an $n\times n$ matrix, the characteristic polynomial has degree $n$, counting repeated roots.

A reliable computation procedure is:

1. Form $p_A(\lambda)=\det(\lambda I-A)$.

2. Solve $p_A(\lambda)=0$ to find the eigenvalues.

3. For each eigenvalue $\lambda$, solve $(A-\lambda I)v=0$.

4. Describe all nonzero vectors in the resulting solution space.

Consider

Its characteristic polynomial is

The eigenvalues are therefore $5$ and $2$.

For $\lambda=5$,

which gives $y=x$. The eigenvectors are the nonzero multiples of

For $\lambda=2$,

which gives $y=-2x$. The eigenvectors are the nonzero multiples of

Takeaway: The determinant finds possible eigenvalues; nullspace calculations then find the corresponding eigenvectors.

Eigenspaces and multiplicity

For an eigenvalue $\lambda$, the eigenspace is

It contains every solution of $(A-\lambda I)v=0$, including the zero vector. The nonzero vectors in this subspace are the eigenvectors associated with $\lambda$.

For the matrix

the eigenspaces found from the previous calculation are

The dimension of an eigenspace is its geometric multiplicity:

The algebraic multiplicity is the number of times the eigenvalue appears as a root of the characteristic polynomial. For every eigenvalue,

A repeated root can have either one independent eigenvector or several. This distinction is essential: algebraic multiplicity records how often an eigenvalue appears in the polynomial, while geometric multiplicity records how many independent eigenvector directions it actually provides.

Takeaway: Use the characteristic polynomial to determine algebraic multiplicity and the nullspace to determine geometric multiplicity.

Independence and diagonalizability

Eigenvectors associated with distinct eigenvalues are linearly independent. Consequently, if an $n\times n$ matrix has $n$ distinct eigenvalues in the underlying field, it has $n$ linearly independent eigenvectors.

A set of $n$ linearly independent vectors in $\mathbb F^n$ forms a basis. Therefore, distinct eigenvalues provide a basis in which the matrix acts by independent scalar multiplication along the basis directions.

More generally, a matrix is diagonalizable over $\mathbb F$ exactly when $\mathbb F^n$ has a basis of eigenvectors. Equivalent conditions include

and, when the characteristic polynomial splits into linear factors over $\mathbb F$,

for every eigenvalue $\lambda$.

The field matters. A real matrix may have complex eigenvalues and therefore fail to be diagonalizable over $\mathbb R$, even though it may be diagonalizable over $\mathbb C$. Also, a repeated eigenvalue does not automatically prevent diagonalization.

For example,

has algebraic multiplicity $2$ for the eigenvalue $2$, and its eigenspace is $E_2=\mathbb R^2$, which has dimension $2$. It is therefore diagonalizable.

In contrast,

has characteristic polynomial $(\lambda-2)^2$, but

forces $y=0$. Hence

which has dimension $1$, less than the algebraic multiplicity $2$. This matrix is not diagonalizable.

Takeaway: Diagonalizability depends on having enough independent eigenvectors, not merely on having eigenvalues.

Constructing a diagonalization

To diagonalize a matrix, first compute its eigenvalues and then determine a basis for each eigenspace. If the resulting eigenvectors provide $n$ linearly independent vectors for an $n\times n$ matrix, place them into the columns of $P$. Place the corresponding eigenvalues in the same order along the diagonal of $D$.

The procedure is:

1. Compute the characteristic polynomial and its roots.

2. Record the algebraic multiplicity of each eigenvalue.

3. For each eigenvalue $\lambda$, compute a basis for $E_\lambda=\ker(A-\lambda I)$.

4. Compare geometric and algebraic multiplicities.

5. If the total number of independent eigenvectors is $n$, form $P$ from those eigenvectors.

6. Form $D$ with the matching eigenvalues on its diagonal.

7. Verify

   $$A=PDP^{-1}.$$

For

choose the eigenvectors

corresponding respectively to eigenvalues $5$ and $2$. Then

and

Diagonalization is especially useful for powers. Since

one obtains

where

Thus a matrix-power problem can be reduced to raising individual eigenvalues to powers.

Final checklist: identify the roots, compute each eigenspace, compare multiplicities, assemble matching eigenvectors and eigenvalues, and verify the factorization.