How to Find the Rank of a Matrix A with A² = A

A matrix satisfying $A^2=A$ is idempotent: applying the corresponding linear transformation twice has the same effect as applying it once. This condition gives useful structure, but it does not by itself determine a single numerical rank. For an $n\times n$ matrix, the rank may be any integer from $0$ through $n$.

The equation does imply that, over the real or complex numbers, the rank equals the trace. If you have the entries of $A$, you can also find the rank directly by row reduction.

Why the equation does not determine one rank

For any $n\geq 1$, both the zero matrix and the identity matrix satisfy the condition:

$$0^2=0,\qquad I_n^2=I_n.$$

Their ranks are $0$ and $n$, respectively. More generally, for each integer $r$ between $0$ and $n$, the diagonal matrix

$$\begin{pmatrix} I_r & 0\\ 0 & 0 \end{pmatrix}$$

(where the zero block has size $n-r$) satisfies $A^2=A$ and has rank $r$. So without additional information, there is no unique rank to report.

The image and kernel split the space

Think of $A$ as a linear map from an $n$-dimensional vector space to itself. Every vector $x$ can be written as a sum of a vector in the image of $A$ and a vector in its kernel:

$$x=Ax+(x-Ax).$$

The first term is in the image. The second is in the kernel because

$$A(x-Ax)=Ax-A^2x=Ax-Ax=0.$$

Moreover, the image and kernel intersect only in the zero vector. If $y$ is in the image, then $y=Ax$ for some $x$, and idempotence gives $Ay=A^2x=Ax=y$. If $y$ is also in the kernel, then $Ay=0$, so $y=0$.

Thus the whole space is the direct sum of the image and kernel. The rank-nullity theorem then gives

$$\operatorname{rank}(A)+\operatorname{nullity}(A)=n.$$

In a basis formed first from a basis of the image and then from a basis of the kernel, the transformation has matrix $\operatorname{diag}(I_r,0)$, where $r=\operatorname{rank}(A)$. This explains why the rank can be any integer from $0$ to $n$.

Why rank equals trace over R or C

In the basis just described, the matrix has $r$ diagonal entries equal to $1$ and the rest equal to $0$. Its trace is therefore $r$, its rank. Trace is unchanged by a change of basis, so the original matrix has the same trace. Consequently, for a real or complex idempotent matrix,

$$\operatorname{rank}(A)=\operatorname{tr}(A).$$

This is a quick way to find the rank when the entries are known. For example, let

$$A=\begin{pmatrix}1&1\\0&0\end{pmatrix}.$$

Then

$$A^2=\begin{pmatrix}1&1\\0&0\end{pmatrix}=A,$$

so $A$ is idempotent. Its trace is $1+0=1$, and its rank is $1$; indeed, its two columns are $(1,0)^T$ and $(1,0)^T$, so they span a one-dimensional space.

Finding the rank from a specific matrix

If the problem provides the entries of $A$, either count the nonzero diagonal entries only after establishing a suitable diagonal form, or use standard row reduction:

1. Form the matrix $A$ from the given entries.
2. Reduce it to row-echelon form.
3. Count the pivot positions. That count is $\operatorname{rank}(A)$.
4. As a check over $\mathbb{R}$ or $\mathbb{C}$, add the diagonal entries of the original matrix. The sum should equal the rank.

The trace shortcut is convenient, but row reduction also works and does not rely on first using the idempotence property. If the matrix is already diagonal with entries only $0$ and $1$, its rank is simply the number of $1$s.

Common mistakes

• Claiming the rank must be 1. Idempotence does not force rank 1; the rank may be any value from $0$ to $n$.
• Treating $A^2=A$ as enough to calculate a number. You need the entries of $A$ or some other information, such as its trace or the dimension of its image.
• Using the diagonal entries as the rank in any basis. Rank is not generally the number of nonzero diagonal entries of the matrix as written. The trace shortcut works because trace is invariant under change of basis and an idempotent map has a basis in which its matrix is $\operatorname{diag}(I_r,0)$.
• Ignoring the number system. The identity $\operatorname{rank}(A)=\operatorname{tr}(A)$ is stated here over the real or complex numbers. In positive characteristic, the trace is interpreted in that field and can fail to distinguish ranks as ordinary integers.