Equal Row Sums Give an Eigenpair
When finding eigenvalues by solving a characteristic equation, it is useful to recognize an eigenvalue before doing any polynomial calculation. Equal row sums provide such a shortcut: the same number appears when the matrix acts on the vector whose entries are all \(1\). This gives one eigenvalue and a particular eigenvector, though it does not by itself determine the rest of the spectrum.
Throughout, matrices are real \(n\times n\) matrices acting on column vectors in \(\mathbb{R}^n\). For a matrix \(\mathbf{A}=(a_{ij})\), the sum of row \(i\) is \(\sum_{j=1}^{n}a_{ij}\). Write \(\mathbf{1}=(1,\dots,1)^{\top}\). An eigenvalue of \(\mathbf{A}\) means a real scalar \(\lambda\) for which there is a nonzero real vector \(\mathbf{v}\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The corresponding nonzero vector is an eigenvector.
The word “constant” here means that all rows have one and the same sum. It does not mean merely that the entries within each individual row are equal. The vector \(\mathbf{1}\) is nonzero, so if matrix multiplication gives \(\mathbf{A}\mathbf{1}=r\mathbf{1}\), it meets the definition of an eigenpair directly.
Proof. Assume each row sum of \(\mathbf{A}\) is \(r\). The \(i\)-th entry of \(\mathbf{A}\mathbf{1}\) is the dot product of row \(i\) with \(\mathbf{1}\), so it equals \(\sum_{j=1}^{n}a_{ij}\cdot 1=r\). This holds for every \(i\), hence \(\mathbf{A}\mathbf{1}=(r,\dots,r)^{\top}=r\mathbf{1}\). Since \(\mathbf{1}\ne\mathbf{0}\), the eigenvalue definition shows that \(r\) is an eigenvalue and \(\mathbf{1}\) is an eigenvector. \(\square\)
The same computation gives an exact equivalence: \(\mathbf{A}\) has constant row sum \(r\) if and only if \(\mathbf{A}\mathbf{1}=r\mathbf{1}\). In the reverse direction, the \(i\)-th coordinate of that matrix equation says precisely that row \(i\) sums to \(r\). This equivalence is often the fastest way to check whether a proposed constant-row-sum claim is correct.
Using the Shortcut in Examples
Worked Example: A Three-by-Three Matrix
Find an eigenvalue and eigenvector of
The row sums are \(1+1+1=3\), \(2+0+1=3\), and \(0+3+0=3\). Thus the common row sum is \(3\). The shortcut gives the eigenpair \((3,\mathbf{1})\). Check it by multiplication:
Check: the displayed multiplication verifies \(\mathbf{A}\mathbf{1}=3\mathbf{1}\), and \(\mathbf{1}\) is nonzero. Therefore \(3\) really is an eigenvalue. This establishes one eigenpair; it does not claim that \(3\) is the only eigenvalue.
For a small matrix, the known eigenvalue can also make it easier to find other eigenvalues. The following example uses the characteristic polynomial as a follow-up calculation. Under the convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I})\), a real root gives a real eigenvalue; in this example, the eigenvectors can also be checked directly.
Worked Example: A Two-by-Two Matrix and Its Other Eigenvalue
Consider \(\mathbf{B}=\begin{bmatrix}2&1\\4&-1\end{bmatrix}\). Both rows sum to \(3\), so \(\mathbf{1}\) is an eigenvector for \(3\). Find and verify another eigenpair.
The trace is \(2+(-1)=1\). Calculate the characteristic polynomial:
The other root is \(-2\). To verify it directly, take \(\mathbf{v}=(1,-4)^{\top}\), which is nonzero:
Check: the known eigenpair also satisfies \(\mathbf{B}(1,1)^{\top}=(3,3)^{\top}=3(1,1)^{\top}\). The two verified eigenpairs are \((3,(1,1)^{\top})\) and \((-2,(1,-4)^{\top})\).
The shortcut supplies a starting point, not a complete eigenvalue list. In the two-by-two example, the characteristic polynomial provided an efficient way to locate the other candidate, and direct multiplication verified its eigenvector. In larger matrices, finding one eigenvalue may simplify a characteristic-polynomial calculation, but further work is still needed to determine the remaining eigenvalues.
Worked Example: A Parameterized Family
For real \(t\), consider
The first row sums to \(t+(4-t)=4\), and the second to \((t+1)+(3-t)=4\). Therefore \(4\) is an eigenvalue for every real \(t\), with eigenvector \((1,1)^{\top}\). A second eigenpair can be verified without dividing by an expression that might vanish. Set \(\mathbf{w}=(t-4,t+1)^{\top}\). Its entries cannot both be zero: that would require \(t=4\) and \(t=-1\) at once. Thus \(\mathbf{w}\ne\mathbf{0}\). Multiplication gives
So \(-1\) is an eigenvalue for every real \(t\), with eigenvector \(\mathbf{w}\). As a cross-check, the determinant calculation is \(\det(\mathbf{C}(t)-\lambda\mathbf{I})=\lambda^2-3\lambda-4=(\lambda-4)(\lambda+1)\). Check: the direct matrix products above verify both eigenpairs, including the parameter values where an entry of \(\mathbf{w}\) is zero.
What the Shortcut Can and Cannot Establish
An eigenvalue is defined by the existence of at least one nonzero eigenvector. The constant-row-sum theorem provides that vector explicitly, so there is no need first to solve a homogeneous system. But it does not show that the eigenspace has dimension one, that the eigenvalue occurs only once in the characteristic polynomial, or that there are no other eigenvalues. Those are separate questions. For instance, in the two-by-two example the known eigenvalue \(3\) was accompanied by \(-2\), while the parameterized example also had a second eigenvalue.
Proof. By the constant-row-sum equivalence, \(\mathbf{B}\mathbf{1}=s\mathbf{1}\) and \(\mathbf{A}\mathbf{1}=r\mathbf{1}\). Associativity of matrix multiplication gives \(\mathbf{A}\mathbf{B}\mathbf{1}=\mathbf{A}(s\mathbf{1})=s(\mathbf{A}\mathbf{1})=sr\mathbf{1}\). Since \(sr=rs\), this is \(\mathbf{A}\mathbf{B}\mathbf{1}=rs\mathbf{1}\). Applying the reverse direction of the equivalence shows that every row of \(\mathbf{A}\mathbf{B}\) sums to \(rs\). \(\square\)
Worked Example: Row Sums of a Product
Let \(\mathbf{D}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\), whose row sums are \(3\), and \(\mathbf{E}=\begin{bmatrix}1&1\\2&0\end{bmatrix}\), whose row sums are \(2\). The theorem predicts that \(\mathbf{D}\mathbf{E}\) has row sums \(6\). Multiply to check:
The product's row sums are \(5+1=6\) and \(6+0=6\). Equivalently, its action on \(\mathbf{1}\) is \(\begin{bmatrix}5&1\\6&0\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}6\\6\end{bmatrix}=6\mathbf{1}\), confirming the result.
Common Pitfall: Checking Column Sums Instead
The shortcut concerns row sums because the entries of \(\mathbf{A}\mathbf{1}\) are row sums. Constant column sums instead describe \(\mathbf{A}^{\top}\mathbf{1}\); they do not generally give the same conclusion for \(\mathbf{A}\). For a concrete counterexample, take \(\mathbf{F}=\begin{bmatrix}0&1\\3&2\end{bmatrix}\). Its column sums are \(0+3=3\) and \(1+2=3\), but its row sums are \(1\) and \(5\), not a common value. The incorrect column-sum calculation would claim \(\mathbf{F}\mathbf{1}=3\mathbf{1}\). In fact,
Thus \(\mathbf{1}\) is not an eigenvector of \(\mathbf{F}\) for \(3\). Before using the shortcut, add the entries across each row and verify that every row has the same sum. Also keep the conclusion appropriately limited: equal row sums establish one eigenpair, not by themselves the complete set of eigenvalues.
Check Your Understanding
Use the row-sum computation and, where requested, verify the resulting matrix equation.
- Find an eigenvalue and a corresponding eigenvector of \(\begin{bmatrix}2&0&1\\1&1&1\\0&2&1\end{bmatrix}\).
- For which real \(t\) does \(\begin{bmatrix}t&5-t\\2&3\end{bmatrix}\) have constant row sums? For those values, state the eigenvalue given by the shortcut.
- Give an example of a matrix with equal column sums but unequal row sums, and compute its product with \(\mathbf{1}\).
- If \(\mathbf{A}\) and \(\mathbf{B}\) have row sums \(4\) and \(-2\), respectively, what is the common row sum of \(\mathbf{A}\mathbf{B}\)?
- Explain why the constant-row-sum result does not, by itself, prove that its eigenvalue is the only eigenvalue.
