How Scaling and Shifting Affect Eigenpairs
An eigenvalue describes the scalar action of a matrix on a particular nonzero vector. If a matrix is multiplied by a scalar, that action is scaled too. If a scalar multiple of the identity is added, the action on every vector gains the same extra scalar. These observations lead to simple transformation rules, but the zero-scalar case needs care: multiplying by zero makes every vector an eigenvector of the resulting zero matrix.
Fix a field \(\mathbb{F}\), either \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), where \(n\geq1\). A scalar \(c\) below belongs to \(\mathbb{F}\). An eigenpair of \(\mathbf{A}\) is a scalar \(\lambda\) and a nonzero vector \(\mathbf{v}\in\mathbb{F}^n\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The scalar \(\lambda\) is then an eigenvalue, and \(\mathbf{v}\) is an eigenvector. The eigenspace is \(E_{\lambda}=\{\mathbf{x}\in\mathbb{F}^n:\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\}\); it includes the zero vector, although the zero vector is not an eigenvector.
Proof. First let \(c\ne0\). If \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for a nonzero \(\mathbf{v}\), then \((c\mathbf{A})\mathbf{v}=c(\mathbf{A}\mathbf{v})=c\lambda\mathbf{v}\), so \(c\lambda\) is an eigenvalue of \(c\mathbf{A}\). Conversely, if \((c\mathbf{A})\mathbf{v}=\mu\mathbf{v}\) for a nonzero \(\mathbf{v}\), division by \(c\) gives \(\mathbf{A}\mathbf{v}=(\mu/c)\mathbf{v}\). Taking \(\mu=c\lambda\) proves both directions and shows that the vectors in the two stated eigenspaces are exactly the same. Now let \(c=0\). For every vector \(\mathbf{x}\), \(\mathbf{O}\mathbf{x}=\mathbf{0}\), so every nonzero vector satisfies the eigenvalue equation with eigenvalue \(0\). If a nonzero \(\mathbf{x}\) satisfied \(\mathbf{O}\mathbf{x}=\mu\mathbf{x}\), then \(\mu\mathbf{x}=\mathbf{0}\), which forces \(\mu=0\) over a field. Thus \(0\) is the only eigenvalue, and every vector belongs to its eigenspace. \(\square\)
Proof. For any nonzero \(\mathbf{v}\), the equation \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) gives \((\mathbf{A}+c\mathbf{I}_n)\mathbf{v}=\mathbf{A}\mathbf{v}+c\mathbf{v}=(\lambda+c)\mathbf{v}\). This proves one direction. Conversely, if \((\mathbf{A}+c\mathbf{I}_n)\mathbf{v}=\mu\mathbf{v}\) for a nonzero \(\mathbf{v}\), subtracting \(c\mathbf{v}\) gives \(\mathbf{A}\mathbf{v}=(\mu-c)\mathbf{v}\). Thus \(\mu\) is an eigenvalue exactly when \(\mu-c\) is an eigenvalue of \(\mathbf{A}\), and the equations show that the corresponding eigenspaces contain exactly the same vectors. \(\square\)
For nonzero scaling, the eigenvalue correspondence can be reversed by dividing by \(c\). For a shift, it can be reversed by subtracting \(c\). In either case, the rules describe all eigenvalues, not just some eigenpairs already known. The shift rule also applies when \(\lambda+c=0\): a shift can move an eigenvalue to zero without changing its eigenspace.
What Happens to the Characteristic Polynomial?
The characteristic polynomial convention used here is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\). Its roots are the eigenvalues. The same transformations can be read directly from this polynomial. In the calculations below, the determinant of an \(n\times n\) matrix multiplied by a scalar \(c\) is multiplied by \(c^n\); this follows by scaling each of its \(n\) rows by \(c\).
Proof. If \(c\ne0\), then \(c\mathbf{A}-\mu\mathbf{I}_n=c(\mathbf{A}-(\mu/c)\mathbf{I}_n)\). Taking determinants gives \(p_{c\mathbf{A}}(\mu)=c^n p_{\mathbf{A}}(\mu/c)\). When \(c=0\), the characteristic polynomial is \(\det(-\mu\mathbf{I}_n)=(-\mu)^n\). For a shift, \(\mathbf{A}+c\mathbf{I}_n-\mu\mathbf{I}_n=\mathbf{A}-(\mu-c)\mathbf{I}_n\), so taking determinants gives the stated identity. \(\square\)
These identities also show what happens to algebraic multiplicity, the multiplicity of a root of the characteristic polynomial. If \(\lambda\) is a root of multiplicity \(k\) for \(\mathbf{A}\), then for \(c\ne0\), \(c\lambda\) is a root of the same multiplicity for \(c\mathbf{A}\); the nonzero factor \(c^n\) cannot change a root’s multiplicity, and the substitution \(\mu/c\) scales its location without changing the number of repeated factors. For a shift, substituting \(\mu-c\) moves the root from \(\lambda\) to \(\lambda+c\), again preserving its multiplicity. When \(c=0\), the polynomial \((-\mu)^n\) has only the root \(0\), with multiplicity \(n\).
Worked Applications
Worked Example: Scaling by a Nonzero Scalar
Let \(\mathbf{A}=\begin{bmatrix}2&1\\0&3\end{bmatrix}\). Find its eigenvalues and eigenspaces, then determine those of \(-2\mathbf{A}\).
The characteristic polynomial is
Thus the eigenvalues are \(2\) and \(3\). For \(\lambda=2\), the equation \((\mathbf{A}-2\mathbf{I}_2)\mathbf{x}=\mathbf{0}\) gives \(x_2=0\), so \(E_2(\mathbf{A})=\operatorname{span}\{(1,0)^{\top}\}\). For \(\lambda=3\), it gives \(x_1=x_2\), so \(E_3(\mathbf{A})=\operatorname{span}\{(1,1)^{\top}\}\).
Now \(-2\mathbf{A}=\begin{bmatrix}-4&-2\\0&-6\end{bmatrix}\). The scaling theorem predicts eigenvalues \(-4\) and \(-6\), with the same respective eigenspaces. Check the eigenpairs directly:
The characteristic polynomial confirms that these are all the eigenvalues: \(\det\begin{bmatrix}-4-\mu&-2\\0&-6-\mu\end{bmatrix}=(-4-\mu)(-6-\mu)\).
Worked Example: Scaling by Zero
Take \(\mathbf{A}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\) and scale by \(c=0\). This example makes the difference between scaling by a nonzero scalar and scaling by zero explicit.
The original characteristic polynomial is \(\det\begin{bmatrix}1-\lambda&2\\0&3-\lambda\end{bmatrix}=(1-\lambda)(3-\lambda)\), so its eigenvalues are \(1\) and \(3\). For \(\lambda=1\), the eigenvalue equation gives \(x_2=0\), so \(E_1(\mathbf{A})=\operatorname{span}\{(1,0)^{\top}\}\). For \(\lambda=3\), it gives \(x_1=x_2\), so \(E_3(\mathbf{A})=\operatorname{span}\{(1,1)^{\top}\}\).
Scaling gives \(0\mathbf{A}=\mathbf{O}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\). For an arbitrary vector \(\mathbf{x}=(x_1,x_2)^{\top}\),
So every nonzero vector is an eigenvector with eigenvalue \(0\), and \(E_0(\mathbf{O})=\mathbb{F}^2\). No nonzero vector can have another eigenvalue \(\mu\), since \(\mathbf{O}\mathbf{x}=\mu\mathbf{x}\) implies \(\mu\mathbf{x}=\mathbf{0}\) and hence \(\mu=0\). The check is consistent with \(p_{\mathbf{O}}(\mu)=\det(-\mu\mathbf{I}_2)=\mu^2\).
Worked Example: Shifting an Eigenvalue to Zero
Let \(\mathbf{A}=\begin{bmatrix}2&1\\0&3\end{bmatrix}\) and shift by \(c=-2\). Compute the eigenvalues and eigenspaces of \(\mathbf{A}-2\mathbf{I}_2\).
The shifted matrix is \(\mathbf{B}=\begin{bmatrix}0&1\\0&1\end{bmatrix}\). The shift rule predicts that the eigenvalues \(2,3\) of \(\mathbf{A}\) become \(0,1\), respectively. The original eigenspaces were \(E_2(\mathbf{A})=\operatorname{span}\{(1,0)^{\top}\}\) and \(E_3(\mathbf{A})=\operatorname{span}\{(1,1)^{\top}\}\), so these should be the corresponding eigenspaces for \(\mathbf{B}\).
Direct calculation verifies both eigenpairs:
To check the eigenspaces, write \(\mathbf{x}=(x_1,x_2)^{\top}\). The equation \(\mathbf{B}\mathbf{x}=\mathbf{0}\) says \((x_2,x_2)^{\top}=(0,0)^{\top}\), so \(x_2=0\) and \(E_0(\mathbf{B})=\operatorname{span}\{(1,0)^{\top}\}\). The equation \(\mathbf{B}\mathbf{x}=\mathbf{x}\) says \((x_2,x_2)^{\top}=(x_1,x_2)^{\top}\), so \(x_1=x_2\) and \(E_1(\mathbf{B})=\operatorname{span}\{(1,1)^{\top}\}\). Finally, \(p_{\mathbf{B}}(\mu)=\det\begin{bmatrix}-\mu&1\\0&1-\mu\end{bmatrix}=\mu(\mu-1)\), confirming that \(0\) and \(1\) are all its eigenvalues.
Common Pitfall: Treating Zero Scaling Like Nonzero Scaling
A tempting but incorrect rule is: “Scaling by \(c\) changes eigenvalues to \(c\lambda\) and leaves each eigenspace unchanged, even when \(c=0\).” For the zero matrix in the previous example, that would incorrectly give \(E_0(\mathbf{O})=E_1(\mathbf{A})\), which is only \(\operatorname{span}\{(1,0)^{\top}\}\). In fact, multiplying by zero makes \(\mathbf{O}\mathbf{x}=\mathbf{0}\) for every \(\mathbf{x}\), so \(E_0(\mathbf{O})=\mathbb{F}^2\). The correction is to use the eigenspace-preservation rule only when \(c\ne0\), and handle \(c=0\) separately.
For a shift, keep the sign straight: adding \(c\mathbf{I}_n\) changes \(\lambda\) to \(\lambda+c\), not \(\lambda-c\). This follows directly from \((\mathbf{A}+c\mathbf{I}_n)\mathbf{v}=\mathbf{A}\mathbf{v}+c\mathbf{v}\). The inverse calculation subtracts \(c\) when recovering the original eigenvalue from the shifted one.
Check Your Understanding
Use the eigenvalue equations to justify each answer, and check numerical eigenpairs by multiplication.
- If \(3\) is an eigenvalue of \(\mathbf{A}\) with eigenspace \(E_3(\mathbf{A})\), what eigenvalue and eigenspace does it give for \(5\mathbf{A}\)?
- For a nonzero \(n\times n\) matrix \(\mathbf{A}\), describe the eigenvalue and eigenspace of \(0\mathbf{A}\).
- If \(-2\) is an eigenvalue of \(\mathbf{A}\), what shift \(c\) makes the corresponding eigenvalue of \(\mathbf{A}+c\mathbf{I}_n\) equal to \(4\)?
- For \(\mathbf{A}=\begin{bmatrix}1&0\\0&4\end{bmatrix}\), find the eigenvalues of \(-\mathbf{A}+2\mathbf{I}_2\) and state their eigenspaces.
- Use the characteristic-polynomial identity for a shift to explain why algebraic multiplicity is preserved.
