Linear Algebra 65 | Diagonalizable Matrices

Q: What goes right in the triangular $2\times2$ example where diagonalization succeeds?

In the successful $2\times2$ triangular case, there are two linearly independent eigenvectors—e.g., $(1,0)$ is an eigenvector and $(1,1)$ is another eigenvector. Together they span $\mathbb{C}^2$, so the eigenvector matrix $X$ is invertible. That guarantees the existence of a diagonal matrix $D$ such that $A=XDX^{-1}$, even though the original matrix is not diagonal.

Q: What is the quick criterion involving distinct eigenvalues?

If an $n\times n$ matrix has $n$ different eigenvalues, then each eigenvalue has algebraic multiplicity 1. Since geometric multiplicity is always at least 1 for an eigenvalue, the equality of geometric and algebraic multiplicities is forced. That guarantees diagonalizability without needing to compute eigenvectors.

TL;DR

A matrix $A$ is diagonalizable iff it has $n$ eigenvectors that form a basis of $C^{n}$ .

Briefing Cornell Notes

Briefing

Diagonalizable matrices are exactly the square matrices whose eigenvectors are rich enough to form a full coordinate system: if the eigenvectors span all of $C^{n}$ , then the matrix can be converted into a diagonal matrix by a change of basis. That matters because diagonal matrices make powers, functions, and long-term behavior of linear transformations far easier to compute—everything reduces to scaling along independent eigen-directions.

The key mechanism comes from rewriting vectors in an eigenvector basis. If $X$ is the matrix whose columns are eigenvectors of $A$ , then in that basis the action of $A$ becomes a diagonal scaling matrix $D$ . Concretely, the transformation takes the form $A = X D X^{- 1}$ , where the diagonal entries of $D$ are the eigenvalues (counted with multiplicity). The central question becomes whether one can choose $n$ eigenvectors that span $C^{n}$ . Equivalently, the eigenvector matrix $X$ must be invertible—meaning the eigenvectors are linearly independent and form a basis.

Examples clarify when this works. A diagonal matrix is automatically diagonalizable because its standard basis vectors are eigenvectors. A non-diagonal triangular matrix can also be diagonalizable: for a $2 \times 2$ upper-triangular matrix with eigenvectors such as $(1, 0)$ and $(1, 1)$ , there are enough independent eigenvectors to span $C^{2}$ , so a suitable $X$ and $D$ exist and the matrix can be diagonalized.

But diagonalization can fail even in simple cases. For a $2 \times 2$ triangular matrix with only one eigenvalue and an eigenspace that is only one-dimensional, there are not enough eigenvector directions to span $C^{2}$ . In that situation, the eigenvectors cannot form a basis, so no diagonalization of the form $A = X D X^{- 1}$ is possible.

To decide diagonalizability without guessing eigenvectors, the transcript emphasizes multiplicities. Algebraic multiplicity counts how many times an eigenvalue appears as a root of the characteristic polynomial, while geometric multiplicity counts the dimension of the eigenspace. A matrix is diagonalizable precisely when, for every eigenvalue, geometric multiplicity equals algebraic multiplicity (and the geometric multiplicities add up to $n$ ).

Two practical criteria follow. First, every normal matrix—such as self-adjoint (Hermitian) matrices—is diagonalizable, and its eigenvectors can be chosen as an orthonormal basis. Second, a quick sufficient test: if a $n \times n$ matrix has $n$ distinct eigenvalues (so each eigenvalue has algebraic multiplicity 1), then it is automatically diagonalizable. The overall takeaway is a clear checklist: diagonalization is about having enough independent eigenvectors, and several structural properties guarantee that requirement.

Cornell Notes

A square matrix $A$ is diagonalizable exactly when it has $n$ eigenvectors that form a basis of $C^{n}$ . If the eigenvectors are assembled as columns of $X$ , then $A$ can be written as $A = X D X^{- 1}$ , where $D$ is diagonal and its diagonal entries are the eigenvalues (with multiplicity). Diagonalization fails when the eigenspaces are too small—for instance, a $2 \times 2$ matrix with only one eigenvalue and a one-dimensional eigenspace cannot produce enough independent eigenvectors to span $C^{2}$ . The transcript links this to multiplicities: diagonalizability holds iff, for every eigenvalue, geometric multiplicity equals algebraic multiplicity. Normal matrices (including self-adjoint ones) always satisfy this, and having $n$ distinct eigenvalues is a simple sufficient condition.

Why does diagonalization reduce to finding enough eigenvectors?

Diagonalization requires a basis in which the linear map acts by simple scaling. If

X

collects

n

eigenvectors as columns, then

X

is invertible exactly when those eigenvectors are linearly independent and span

C^{n}

. In that eigenvector basis,

A

acts as

D

, a diagonal matrix whose diagonal entries are the eigenvalues, giving the identity

A = X D X^{- 1}

. If the eigenvectors don’t span

C^{n}

, no invertible

X

built from eigenvectors exists, so diagonalization fails.

What goes right in the triangular $2 \times 2$ example where diagonalization succeeds?

In the successful

2 \times 2

triangular case, there are two linearly independent eigenvectors—e.g.,

(1, 0)

is an eigenvector and

(1, 1)

is another eigenvector. Together they span

C^{2}

, so the eigenvector matrix

X

is invertible. That guarantees the existence of a diagonal matrix

D

such that

A = X D X^{- 1}

, even though the original matrix is not diagonal.

Why does the other $2 \times 2$ triangular example fail to be diagonalizable?

The failing case has only one eigenvalue and an eigenspace that is only one-dimensional. That means there is only one independent eigenvector direction (the eigenspace is the span of a single vector like

(1, 0)

). With only one eigenvector direction, the eigenvectors cannot span

C^{2}

, so they cannot form a basis. Since diagonalization needs an invertible eigenvector matrix

X

, no diagonal form

A = X D X^{- 1}

is possible.

How do algebraic and geometric multiplicities determine diagonalizability?

Algebraic multiplicity counts the eigenvalue’s multiplicity as a root of the characteristic polynomial. Geometric multiplicity is the dimension of the eigenspace, i.e., the number of independent eigenvectors available for that eigenvalue. Diagonalizability holds exactly when, for every eigenvalue

λ

, geometric multiplicity equals algebraic multiplicity. Because algebraic multiplicities sum to

n

, matching them ensures the total number of independent eigenvectors reaches

n

, letting them span

C^{n}

What special guarantee does normality provide?

Normal matrices—such as self-adjoint (Hermitian) matrices—are always diagonalizable. Moreover, their eigenvectors can be chosen to form an orthonormal basis (an orthonormal eigenbasis). This is stronger than mere diagonalizability because it ensures eigenvectors are not only complete but also orthogonal with respect to the standard inner product.

What is the quick criterion involving distinct eigenvalues?

If an

n \times n

matrix has

n

different eigenvalues, then each eigenvalue has algebraic multiplicity 1. Since geometric multiplicity is always at least 1 for an eigenvalue, the equality of geometric and algebraic multiplicities is forced. That guarantees diagonalizability without needing to compute eigenvectors.

Review Questions

State the definition of a diagonalizable matrix in terms of eigenvectors and spanning $C^{n}$ .
Explain the relationship between geometric multiplicity and algebraic multiplicity, and how their equality determines diagonalizability.
Give two sufficient conditions for diagonalizability mentioned in the transcript and describe why they work.

Key Points

1
A matrix $A$ is diagonalizable iff it has $n$ eigenvectors that form a basis of $C^{n}$ .
2
If $X$ is formed from eigenvectors as columns, diagonalization takes the form $A = X D X^{- 1}$ , with $D$ diagonal and its entries equal to eigenvalues (with multiplicity).
3
Diagonalization fails when eigenspaces are too small—e.g., a $2 \times 2$ matrix with only one eigenvalue and a one-dimensional eigenspace cannot span $C^{2}$ .
4
A matrix is diagonalizable exactly when, for every eigenvalue, geometric multiplicity equals algebraic multiplicity.
5
Normal matrices (including self-adjoint/Hermitian ones) are always diagonalizable, with an orthonormal eigenbasis.
6
Having $n$ distinct eigenvalues is a simple sufficient test for diagonalizability.

Highlights

Diagonalization is possible precisely when eigenvectors span the whole space, making the eigenvector matrix X invertible.

Even triangular matrices can be diagonalizable—success depends on whether there are enough independent eigenvectors.

A single eigenvalue with a one-dimensional eigenspace is enough to block diagonalization in C2.

Normal matrices guarantee diagonalizability and allow an orthonormal eigenbasis.

n distinct eigenvalues automatically imply diagonalizability, without computing eigenvectors.

Topics

Diagonalizable Matrices
Eigenvectors and Eigenvalues
Geometric vs Algebraic Multiplicity
Normal Matrices
Diagonalization Criteria