If A Commutes with Every Matrix B, Must A Be cI?

Suppose A is an n × n matrix and AB = BA for every n × n matrix B over the same field. Then A must be a scalar multiple of the identity matrix: A = cI. The key is to choose particular matrices B that reveal the entries of A.

This is a statement about commuting with every matrix of the same size. Commuting with just one matrix does not generally force a matrix to be scalar.

Proof using matrix units

For indices i and j, let Eᵢⱼ denote the matrix whose entry in row i, column j is 1, with every other entry 0. Since A commutes with every matrix, it commutes with each Eᵢⱼ.

First take B = Eᵢᵢ. Multiplication by Eᵢᵢ keeps column i of A in AEᵢᵢ, while EᵢᵢA keeps row i of A. The equality

$$AE_{ii}=E_{ii}A$$

therefore forces every off-diagonal entry in row i and column i to be zero. More explicitly, for p ≠ i, the entry in position (p,i) on the left is aₚᵢ, while it is zero on the right, so aₚᵢ = 0. For q ≠ i, the entry in position (i,q) on the right is aᵢq, while it is zero on the left, so aᵢq = 0. Varying i shows that A is diagonal.

Now take B = Eᵢⱼ with i ≠ j. Since A is diagonal, the (i,j) entry of AEᵢⱼ is aᵢᵢ, while the (i,j) entry of EᵢⱼA is aⱼⱼ. Commutativity gives

$$a_{ii}=a_{jj}.$$

This holds for every pair i ≠ j, so all diagonal entries are the same scalar c. Hence A = cI.

A worked 2 × 2 case

Write a general 2 × 2 matrix as

$$A=\begin{pmatrix}x&y\\z&w\end{pmatrix}.$$

If A commutes with every 2 × 2 matrix, it must commute with E₁₁ = [[1, 0], [0, 0]]. The two products are

$$AE_{11}=\begin{pmatrix}x&0\\z&0\end{pmatrix}, \qquad E_{11}A=\begin{pmatrix}x&y\\0&0\end{pmatrix}.$$

Equating them gives y = 0 and z = 0, so A is diagonal. It must also commute with E₁₂ = [[0, 1], [0, 0]]. For A = [[x, 0], [0, w]],

$$AE_{12}=\begin{pmatrix}0&x\\0&0\end{pmatrix}, \qquad E_{12}A=\begin{pmatrix}0&w\\0&0\end{pmatrix}.$$

These products are equal only if x = w. Thus A = xI, as claimed.

Why the converse holds

If A = cI, then for any n × n matrix B,

$$AB=(cI)B=cB=B(cI)=BA.$$

So scalar matrices do commute with every matrix; the proof above shows there are no others.

Common mistakes

• Checking only one matrix B. A matrix can commute with a particular nonscalar matrix without being scalar. The conclusion follows from commuting with all matrices, including every matrix unit.
• Forgetting to show the diagonal entries agree. Commuting with Eᵢᵢ eliminates off-diagonal entries, but it does not by itself make the diagonal entries equal. Use Eᵢⱼ for i ≠ j for that step.
• Mixing up the size or field. The equation assumes A and every B are n × n matrices over the same field. For n = 1, every 1 × 1 matrix is already a scalar multiple of I.