Scalar Matrices and Their Full Eigenspaces

The Special Case of a Scalar Matrix

For many matrices, finding eigenvectors means solving a system of equations, and the resulting eigenspace may occupy only part of the vector space. A scalar matrix is an important exception: its action simply multiplies every vector by the same number. That simple action has a strong consequence. Every nonzero vector is an eigenvector, so there is no preferred eigenbasis; any basis will do.

Throughout, let \(\mathbb{F}\) be either \(\mathbb{R}\) or \(\mathbb{C}\), let \(n\geq 1\), and consider matrices in \(\mathbb{F}^{n\times n}\). For a scalar \(c\in\mathbb{F}\), the scalar matrix with scalar \(c\) is \(c\mathbf{I}_n\). A scalar matrix has zero entries off the diagonal and has the same entry \(c\) in every diagonal position. The zero matrix is included: it is the scalar matrix \(0\mathbf{I}_n\).

The zero vector belongs to every eigenspace because it satisfies the homogeneous equation, but it is never an eigenvector: the definition of eigenvector requires a nonzero vector. A list is an eigenbasis when it is a basis of the whole space and each vector in the list is an eigenvector of the matrix. A basis is an ordered list that is linearly independent and spans the space.

One Eigenvalue and a Full Eigenspace

Proof. For every \(\mathbf{v}\in\mathbb{F}^n\), the scalar-matrix action gives \(\mathbf{A}\mathbf{v}=c\mathbf{v}\). In particular, there are nonzero vectors in \(\mathbb{F}^n\) because \(n\geq 1\), so \(c\) is an eigenvalue. Every nonzero vector satisfies the eigenvector equation for \(c\), and \(\mathbf{0}\) also belongs to the eigenspace. Thus \(E_c=\mathbb{F}^n\). Now suppose \(\lambda\) is any eigenvalue, with eigenvector \(\mathbf{v}\ne\mathbf{0}\). Then \(c\mathbf{v}=\lambda\mathbf{v}\), so \((c-\lambda)\mathbf{v}=\mathbf{0}\). At least one coordinate of \(\mathbf{v}\) is nonzero, so the corresponding coordinate equation forces \(c-\lambda=0\). Hence \(\lambda=c\); there is no other eigenvalue. \(\square\)

The proof uses only the eigenvector equation and the fact that a nonzero vector has at least one nonzero coordinate. In particular, it applies to the zero scalar matrix as well: its only eigenvalue is \(0\), and every nonzero vector is an eigenvector for \(0\).

Why Every Basis Is an Eigenbasis

Proof. Let \((\mathbf{b}_1,\ldots,\mathbf{b}_n)\) be any basis of \(\mathbb{F}^n\). Each \(\mathbf{b}_j\) is nonzero, since a list containing \(\mathbf{0}\) cannot be linearly independent. By the scalar-matrix equation, \(\mathbf{A}\mathbf{b}_j=c\mathbf{b}_j\) for every \(j\). Thus every basis vector is an eigenvector for \(c\). The list is already a basis of the whole space, so it meets both requirements in the definition of an eigenbasis. \(\square\)

This statement does not say that the zero vector is an eigenvector. It says that every vector in a basis is an eigenvector; basis vectors are nonzero by definition of linear independence. More broadly, any choice of basis vectors works because the eigenvector equation places no restriction on their directions.

The same fact appears in the matrix representation of a change of basis. Let \(\mathbf{P}\) be the matrix whose columns are any basis of \(\mathbb{F}^n\). Since those columns form a basis, \(\mathbf{P}\) is invertible. For a scalar matrix, multiplication gives

Thus changing basis leaves the matrix unchanged. In the notation of diagonalization, an eigenbasis produces a diagonal matrix of eigenvalues; here that diagonal matrix is \(c\mathbf{I}_n\), regardless of which basis is chosen. The conclusion follows from scalar multiplication, not from a special choice or ordering of eigenvectors.

Scalar Matrices Versus Repeated Eigenvalues

A repeated eigenvalue alone does not imply that the corresponding eigenspace is the whole space. The defining feature of a scalar matrix is that it acts by the same scalar on every vector, not merely that some eigenvalue occurs in more than one direction. The following example makes the distinction explicit.

A Characterization and a Common Mistake

Proof. If \(\mathbf{A}=\lambda\mathbf{I}_n\), the scalar-matrix theorem gives \(E_{\lambda}=\mathbb{F}^n\). Conversely, suppose \(E_{\lambda}=\mathbb{F}^n\). For every \(\mathbf{x}\in\mathbb{F}^n\), membership in the eigenspace means \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\). In particular, apply this equation to each standard basis vector \(\mathbf{e}_j\). The product \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{e}_j\) is column \(j\) of \(\mathbf{A}-\lambda\mathbf{I}_n\), so every column is zero. Hence \(\mathbf{A}-\lambda\mathbf{I}_n=\mathbf{O}\), or \(\mathbf{A}=\lambda\mathbf{I}_n\). \(\square\)

Common mistake: treating every scalar as an eigenvalue. For \(\mathbf{A}=4\mathbf{I}_2\), it is wrong to claim that \(7\) is an eigenvalue just because the matrix multiplies vectors by a number. Testing \(\mathbf{e}_1=(1,0)^{\top}\) gives \(\mathbf{A}\mathbf{e}_1=(4,0)^{\top}\), whereas \(7\mathbf{e}_1=(7,0)^{\top}\). These are unequal, and the theorem shows that \(4\) is the only eigenvalue. The eigenvalue must match the scalar by which the matrix acts.

A related notational mistake is to define the eigenspace as only the set of eigenvectors. That set omits \(\mathbf{0}\) and is not a subspace. The eigenspace includes \(\mathbf{0}\) by definition, while eigenvectors exclude it. Keeping these terms distinct makes the statement \(E_c=\mathbb{F}^n\) precise.

The central conclusion is unusually strong but easy to verify: a scalar matrix has exactly one eigenvalue, and its eigenspace fills the ambient space. Consequently, every basis is an eigenbasis and every change of basis leaves the matrix equal to the same scalar multiple of the identity.