The Structure of a Scalar Matrix Plus a Rank-One Update
A matrix of the form \(\mathbf{A}=c\mathbf{I}_n+\mathbf{u}\mathbf{v}^{\top}\) is built from a scalar matrix and an update whose columns are all multiples of \(\mathbf{u}\). Here \(c\in\mathbb{R}\), \(n\geq 1\), and \(\mathbf{u},\mathbf{v}\in\mathbb{R}^n\) are column vectors. The matrix \(\mathbf{u}\mathbf{v}^{\top}\) has rank at most one: if both vectors are nonzero, every column is a multiple of \(\mathbf{u}\), and at least one column is nonzero. The update may be zero if either vector is zero.
The key quantity is \(\delta=\mathbf{v}^{\top}\mathbf{u}\), a real number. For every \(\mathbf{x}\in\mathbb{R}^n\), multiplication gives
Thus the update changes \(\mathbf{x}\) only through the scalar \(\mathbf{v}^{\top}\mathbf{x}\), and the direction of that change is \(\mathbf{u}\). In particular, if \(\mathbf{v}^{\top}\mathbf{x}=0\), then \(\mathbf{A}\mathbf{x}=c\mathbf{x}\). The vector \(\mathbf{u}\), when nonzero, is also especially useful: \(\mathbf{v}^{\top}\mathbf{u}=\delta\) gives \(\mathbf{A}\mathbf{u}=(c+\delta)\mathbf{u}\).
Computing the Characteristic Polynomial
The characteristic polynomial can be derived directly from the columns of \(\mathbf{A}-\lambda\mathbf{I}_n\). This avoids expanding a large determinant by a general-purpose procedure and makes the role of \(\delta\) visible.
Proof. Put \(d=c-\lambda\). Column \(j\) of \(\mathbf{A}-\lambda\mathbf{I}_n=d\mathbf{I}_n+\mathbf{u}\mathbf{v}^{\top}\) is \(d\mathbf{e}_j+v_j\mathbf{u}\). By multilinearity of the determinant in its columns, expand by choosing either \(d\mathbf{e}_j\) or \(v_j\mathbf{u}\) in each column. Any term choosing \(v_j\mathbf{u}\) in two or more columns is zero because those columns are proportional. Choosing \(d\mathbf{e}_j\) in every column contributes \(d^n\). If only column \(j\) uses \(v_j\mathbf{u}\), the contribution is \(d^{n-1}v_j u_j\): the determinant of the identity matrix with column \(j\) replaced by \(\mathbf{u}\) is \(u_j\), by expansion in that column. Adding the terms gives \[ \det(\mathbf{A}-\lambda\mathbf{I}_n) =d^n+d^{n-1}\sum_{j=1}^n v_j u_j =d^{n-1}(d+\delta). \] Substituting \(d=c-\lambda\) proves the formula. The argument also applies when \(n=1\), with \((c-\lambda)^0=1\). \(\square\)
The formula identifies the possible roots as \(c\) and \(c+\delta\), but it does not by itself say that both are eigenvalues: a root can have a multiplicity that depends on \(n\), and a factor with exponent zero contributes no root. The eigenspace calculation makes these distinctions explicit.
Worked Example: A Two-Dimensional Update with Nonzero Dot Product
Let \(c=2\), \(\mathbf{u}=\begin{bmatrix}1\\2\end{bmatrix}\), and \(\mathbf{v}=\begin{bmatrix}3\\-1\end{bmatrix}\). Compute the characteristic polynomial and eigenspaces.
First, \(\delta=\mathbf{v}^{\top}\mathbf{u}=3(1)+(-1)(2)=1\). The outer product and resulting matrix are
Since \(n=2\), the formula gives \(p_{\mathbf{A}}(\lambda)=(2-\lambda)(3-\lambda)\). For \(\lambda=2\), the equation \(\mathbf{v}^{\top}\mathbf{x}=0\) is \(3x_1-x_2=0\), so \(E_2=\operatorname{span}\left\{\begin{bmatrix}1\\3\end{bmatrix}\right\}\). For \(\lambda=3\), the eigenspace is \(\operatorname{span}\{\mathbf{u}\}\).
Check: Direct multiplication gives \(\mathbf{A}\begin{bmatrix}1\\3\end{bmatrix}=\begin{bmatrix}2\\6\end{bmatrix}=2\begin{bmatrix}1\\3\end{bmatrix}\) and \(\mathbf{A}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}3\\6\end{bmatrix}=3\begin{bmatrix}1\\2\end{bmatrix}\). Both claimed eigenspaces contain the required eigenvectors.
How the Dot Product Determines the Eigenspaces
- If \(\mathbf{u}=\mathbf{0}\) or \(\mathbf{v}=\mathbf{0}\), then \(\mathbf{A}=c\mathbf{I}_n\), and \(c\) is the only eigenvalue, with \(E_c=\mathbb{R}^n\).
- If \(\delta=0\) and \(\mathbf{u},\mathbf{v}\ne\mathbf{0}\), then \(c\) is the only eigenvalue and \(E_c=\{\mathbf{x}:\mathbf{v}^{\top}\mathbf{x}=0\}\).
- If \(\delta\ne0\), then \(c+\delta\) is an eigenvalue with \(E_{c+\delta}=\operatorname{span}\{\mathbf{u}\}\). If \(n\geq2\), \(c\) is also an eigenvalue with \(E_c=\{\mathbf{x}:\mathbf{v}^{\top}\mathbf{x}=0\}\). If \(n=1\), \(c\) is not an eigenvalue; \(c+\delta\) is the only eigenvalue.
Proof. Suppose first that \(\mathbf{u}=\mathbf{0}\) or \(\mathbf{v}=\mathbf{0}\). Then \(\mathbf{u}\mathbf{v}^{\top}=\mathbf{O}\) and \(\mathbf{A}\mathbf{x}=c\mathbf{x}\) for every \(\mathbf{x}\). Every nonzero vector is therefore an eigenvector for \(c\), and if \(\lambda\ne c\), the equation \(c\mathbf{x}=\lambda\mathbf{x}\) forces \(\mathbf{x}=\mathbf{0}\). This proves the first case.
Now assume \(\mathbf{u},\mathbf{v}\ne\mathbf{0}\), and let \(\lambda\) be an eigenvalue with eigenvector \(\mathbf{x}\ne\mathbf{0}\). The eigenvector equation rearranges to \[ (\lambda-c)\mathbf{x}=\mathbf{u}(\mathbf{v}^{\top}\mathbf{x}). \] If \(\lambda\ne c\), this equation makes \(\mathbf{x}\) a nonzero scalar multiple of \(\mathbf{u}\). In particular, \(\mathbf{v}^{\top}\mathbf{x}\ne0\); otherwise the equation would give \(\mathbf{x}=\mathbf{0}\). Applying \(\mathbf{v}^{\top}\) to both sides yields \((\lambda-c)\mathbf{v}^{\top}\mathbf{x}=\delta\mathbf{v}^{\top}\mathbf{x}\), so \(\lambda=c+\delta\). Conversely, \(\mathbf{A}\mathbf{u}=(c+\delta)\mathbf{u}\), and \(\mathbf{u}\ne\mathbf{0}\), so \(c+\delta\) is indeed an eigenvalue. If \(\lambda=c+\delta\), the original equation becomes \(\delta\mathbf{x}=\mathbf{u}(\mathbf{v}^{\top}\mathbf{x})\); since \(\delta\ne0\), every eigenvector is in \(\operatorname{span}\{\mathbf{u}\}\), and \(\mathbf{u}\) itself is an eigenvector. This proves the claimed eigenspace for \(c+\delta\).
For \(\lambda=c\), the eigenvector equation is \(\mathbf{u}(\mathbf{v}^{\top}\mathbf{x})=\mathbf{0}\). Since \(\mathbf{u}\ne\mathbf{0}\), it holds exactly when \(\mathbf{v}^{\top}\mathbf{x}=0\). If \(\delta=0\), then \(\mathbf{u}\) is a nonzero vector satisfying this condition, so \(c\) is an eigenvalue, and the displayed set is its eigenspace. The only possible eigenvalue other than \(c\) would be \(c+\delta=c\), so there is no other one.
If \(\delta\ne0\), then \(\mathbf{v}\ne\mathbf{0}\). When \(n\geq2\), the equation \(\mathbf{v}^{\top}\mathbf{x}=0\) has a nonzero solution: choose an index \(j\) with \(v_j\ne0\), assign an arbitrary value to one coordinate other than \(x_j\), assign zero to any other unspecified coordinates, and solve uniquely for \(x_j\). Thus \(c\) is an eigenvalue in this case. When \(n=1\), the equation \(v_1x_1=0\) forces \(x_1=0\), because \(v_1\ne0\); hence \(c\) is not an eigenvalue. This proves all cases. \(\square\)
The characteristic polynomial and eigenspaces give compatible multiplicity information. When \(\delta=0\), the characteristic polynomial is \((c-\lambda)^n\), although the eigenspace need not be all of \(\mathbb{R}^n\). When \(\delta\ne0\), it is \((c-\lambda)^{n-1}(c+\delta-\lambda)\). For \(n\geq2\), the two distinct eigenvalues have algebraic multiplicities \(n-1\) and \(1\); for \(n=1\), only \(c+\delta\) occurs as a root.
Worked Example: Orthogonal Update Vectors in Three Dimensions
Take \(c=-1\), \(\mathbf{u}=\begin{bmatrix}1\\1\\0\end{bmatrix}\), and \(\mathbf{v}=\begin{bmatrix}1\\-1\\2\end{bmatrix}\). Here \(\delta=1-1+0=0\), even though neither update vector is zero.
Compute the update and the matrix:
The characteristic polynomial is \((-1-\lambda)^3\). The eigenspace is described by \(x_1-x_2+2x_3=0\). Two independent vectors satisfying this equation are \(\begin{bmatrix}1\\1\\0\end{bmatrix}\) and \(\begin{bmatrix}-2\\0\\1\end{bmatrix}\), so they span the plane \(E_{-1}\).
Check: Multiplication gives \(\mathbf{A}\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}-1\\-1\\0\end{bmatrix}\) and \(\mathbf{A}\begin{bmatrix}-2\\0\\1\end{bmatrix}=\begin{bmatrix}2\\0\\-1\end{bmatrix}\). These are \(-1\) times the respective vectors. The two vectors are independent because one has third coordinate \(0\) and the other has third coordinate \(1\); their span therefore describes the full two-dimensional plane given by the single nonzero linear equation.
Worked Example: The One-Dimensional Case
Let \(n=1\), \(c=4\), \(u=2\), and \(v=3\). Then \(\delta=vu=6\), and
The only eigenvalue is \(10=c+\delta\), and its eigenspace is all of \(\mathbb{R}\). There is no eigenvalue \(4\): in one dimension the condition \(v x=0\) becomes \(3x=0\), which has no nonzero solution.
Check: \(\mathbf{A}[1]=[10]=10[1]\), while \(\mathbf{A}[1]=[10]\ne4[1]=[4]\). This confirms both the stated eigenvalue and the absence of the scalar-part value \(c\).
Common Pitfall: Treating Both Factors as Eigenvalues
A frequent error is to read \((c-\lambda)^{n-1}(c+\delta-\lambda)\) and conclude that \(c\) is always an eigenvalue. A concrete counterexample is the one-dimensional matrix above: \(c=4\), \(\delta=6\), and \(\mathbf{A}=[10]\). The incorrect claim would make \(4\) an eigenvalue with eigenspace \(\{x:3x=0\}=\{0\}\). But an eigenspace must contain a nonzero eigenvector, and direct substitution gives \([10][1]=[10]\ne[4]\). The correction is that when \(n=1\), the factor \((c-\lambda)^{n-1}\) has exponent zero, so it contributes no root.
Another important distinction is between \(\delta=0\) and a zero update. Orthogonality of nonzero \(\mathbf{u}\) and \(\mathbf{v}\) makes the only eigenvalue equal to \(c\), but it does not make \(\mathbf{u}\mathbf{v}^{\top}\) the zero matrix. In the three-dimensional example, the update matrix has nonzero entries, and the eigenspace is a plane rather than all of \(\mathbb{R}^3\). The update vanishes only if at least one of the two vectors is zero.
What the Formula Is Useful For
For these matrices, finding the characteristic polynomial does not require a full determinant expansion, and finding the eigenspaces does not require row-reducing a different matrix for each candidate eigenvalue. First compute \(\delta=\mathbf{v}^{\top}\mathbf{u}\). If it is nonzero, the update direction \(\mathbf{u}\) gives the eigenvector for \(c+\delta\), while vectors orthogonal to \(\mathbf{v}\) give the eigenspace for \(c\) when \(n\geq2\). If it is zero and the update is nonzero, the only eigenspace is the set of vectors orthogonal to \(\mathbf{v}\). If either update vector is zero, the matrix is simply \(c\mathbf{I}_n\).
This structure is useful when a matrix is presented as a simple scalar action plus a change in one direction. It also illustrates why characteristic-polynomial roots and eigenspace dimensions answer different questions: a repeated root does not, by itself, mean that every vector is an eigenvector. The action formula \(\mathbf{A}\mathbf{x}=c\mathbf{x}+\mathbf{u}(\mathbf{v}^{\top}\mathbf{x})\) is the direct way to check which vectors actually satisfy the eigenvector equation.
Check Your Understanding
For each problem, identify the scalar \(c\), compute \(\delta\), and use the update structure before checking by substitution.
- For \(c=3\), \(\mathbf{u}=\begin{bmatrix}1\\-1\end{bmatrix}\), and \(\mathbf{v}=\begin{bmatrix}2\\1\end{bmatrix}\), find the characteristic polynomial and both eigenspaces.
- Let \(c=0\), \(\mathbf{u}=\begin{bmatrix}1\\2\\0\end{bmatrix}\), and \(\mathbf{v}=\begin{bmatrix}2\\-1\\1\end{bmatrix}\). What is the only eigenvalue, and what equation describes its eigenspace?
- For \(n=1\), \(c=5\), and \(u=v=1\), state the eigenvalue and explain why \(c\) is not an eigenvalue.
- Suppose \(\delta\ne0\) and \(n=4\). State the eigenspace for \(c+\delta\) and the dimension of the eigenspace for \(c\).
- Explain why \(\delta=0\) does not imply \(\mathbf{u}\mathbf{v}^{\top}=\mathbf{O}\).
