From a Characteristic Polynomial to Eigenpairs
For a three-by-three matrix, finding eigenpairs means doing two related tasks. First, identify the eigenvalues from the characteristic polynomial. Then, for each eigenvalue, find the nonzero vectors that satisfy the eigenvector equation and describe the resulting eigenspace. A factorization into real linear factors makes the first task direct, but each candidate still needs its own equation to determine the eigenspace.
Throughout, matrices and vectors are over \(\mathbb{R}\). For a matrix \(\mathbf{A}\in\mathbb{R}^{3\times3}\), the characteristic polynomial is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_3)\). A real number \(\lambda\) is an eigenvalue if there is a nonzero vector \(\mathbf{v}\in\mathbb{R}^3\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The corresponding eigenspace is \(E_\lambda=\{\mathbf{v}\in\mathbb{R}^3:(\mathbf{A}-\lambda\mathbf{I}_3)\mathbf{v}=\mathbf{0}\}\), which includes the zero vector. Its nonzero elements are the eigenvectors for \(\lambda\).
The practical sequence is: compute and factor the characteristic polynomial; take its real roots as candidate eigenvalues; for each candidate solve \((\mathbf{A}-\lambda\mathbf{I}_3)\mathbf{v}=\mathbf{0}\); then express the solution set as a span and read off its dimension. A basis must consist of vectors that span the whole eigenspace and are linearly independent. In the examples, the general solution will be a single free scalar times a nonzero vector, which directly gives a one-vector basis.
Proof. By the definition of eigenvalue, \(\lambda\) is an eigenvalue exactly when there is a vector \(\mathbf{v}\ne\mathbf{0}\) satisfying \((\mathbf{A}-\lambda\mathbf{I}_3)\mathbf{v}=\mathbf{0}\). A square matrix has a nonzero solution to its homogeneous system exactly when it is singular, and a square matrix is singular exactly when its determinant is zero. Thus such a vector exists exactly when \(\det(\mathbf{A}-\lambda\mathbf{I}_3)=0\), which is \(p_{\mathbf{A}}(\lambda)=0\). Finally, the eigenspace consists of all vectors satisfying that homogeneous equation, including \(\mathbf{0}\), so it is precisely the null space stated. \(\square\)
Proof. By the preceding theorem, a real number \(\lambda\) is an eigenvalue exactly when \(p_{\mathbf{A}}(\lambda)=0\). The displayed product is zero exactly when at least one factor \(\lambda-r_i\) is zero, so its real roots are exactly \(r_1,r_2,r_3\). Since those numbers are distinct, each corresponding linear factor occurs once, giving algebraic multiplicity \(1\) for each eigenvalue. \(\square\)
Worked Eigenpair Problems
Worked Example: Three Distinct Integer Eigenvalues
Find all eigenvalues, eigenspace bases, and multiplicities of
First compute the characteristic polynomial, using expansion along the first row of \(\mathbf{A}-\lambda\mathbf{I}_3\):
Thus the eigenvalues are \(1,2,3\), each with algebraic multiplicity \(1\). For an eigenvalue \(\lambda\), write \(\mathbf{v}=(x,y,z)^\top\). The equation \((\mathbf{A}-\lambda\mathbf{I}_3)\mathbf{v}=\mathbf{0}\) is
For \(\lambda=1\), the first equation gives \(x=6z\), and the third gives \(y=-5z\). The second then reads \(6z-(-5z)-11z=0\). Hence \(\mathbf{v}=z(6,-5,1)^\top\), and a basis is \(\{(6,-5,1)^\top\}\). For \(\lambda=2\), the first and third equations give \(x=3z\) and \(y=-4z\); the second becomes \(3z-2(-4z)-11z=0\). A basis is \(\{(3,-4,1)^\top\}\). For \(\lambda=3\), they give \(x=2z\) and \(y=-3z\); the second becomes \(2z-3(-3z)-11z=0\). A basis is \(\{(2,-3,1)^\top\}\). In each case \(z\) is free, and choosing \(z=1\) gives a nonzero basis vector, so each geometric multiplicity is \(1\).
Check: Direct multiplication gives \(\mathbf{A}(6,-5,1)^\top=(6,-5,1)^\top\), \(\mathbf{A}(3,-4,1)^\top=(6,-8,2)^\top=2(3,-4,1)^\top\), and \(\mathbf{A}(2,-3,1)^\top=(6,-9,3)^\top=3(2,-3,1)^\top\).
Worked Example: Including a Zero Eigenvalue
Find all eigenpairs and multiplicities for
The first row of \(\mathbf{B}-\lambda\mathbf{I}_3\) has only one nonzero entry, so expansion along that row gives
The eigenvalues are \(-1,0,4\); each has algebraic multiplicity \(1\). For \(\mathbf{v}=(x,y,z)^\top\), the eigenvector equation is equivalent to
When \(\lambda=0\), the last two equations give \(x=-4z\) and \(y=-3z\). Thus \(E_0=\operatorname{span}\{(-4,-3,1)^\top\}\). When \(\lambda=-1\), the first equation gives \(x=0\), while the last gives \(y=-4z\); substituting these values into the middle equation gives \(0-(-1)(-4z)+4z=0\). Thus \(E_{-1}=\operatorname{span}\{(0,-4,1)^\top\}\). When \(\lambda=4\), the first equation gives \(x=0\), and the last gives \(y=z\); the middle equation then gives \(-4z+4z=0\). Thus \(E_4=\operatorname{span}\{(0,1,1)^\top\}\). Each displayed spanning vector is nonzero, so each is a basis for its one-dimensional eigenspace and each geometric multiplicity is \(1\).
Check: \(\mathbf{B}(-4,-3,1)^\top=(0,0,0)^\top\); \(\mathbf{B}(0,-4,1)^\top=(0,4,-1)^\top=-1(0,-4,1)^\top\); and \(\mathbf{B}(0,1,1)^\top=(0,4,4)^\top=4(0,1,1)^\top\).
Worked Example: Irrational Real Eigenvalues
Find the real eigenvalues, bases for their eigenspaces, and both multiplicities for the symmetric matrix
Set \(d=2-\lambda\). Expanding the determinant gives
The roots are \(\lambda=2\) and \(\lambda=2+\sqrt{2},2-\sqrt{2}\). These are three distinct real numbers, so each has algebraic multiplicity \(1\). For \(\lambda=2\), the system is \(y=0,\ x+z=0,\ y=0\). Therefore \(E_2=\operatorname{span}\{(1,0,-1)^\top\}\). For either of the other eigenvalues, write \(\lambda=2+s\), where \(s=\sqrt{2}\) or \(s=-\sqrt{2}\). The system becomes \(-sx+y=0,\ x-sy+z=0,\ y-sz=0\). Taking \(x=z=1\), the first and third equations require \(y=s\), and the middle equation holds because \(1-s^2+1=0\), using \(s^2=2\). Conversely, the first equation forces \(y=sx\), and the third then gives \(s(x-z)=0\). Since \(s\ne0\), \(z=x\). Thus each eigenspace is exactly \(\operatorname{span}\{(1,s,1)^\top\}\), a one-dimensional space, so its geometric multiplicity is \(1\).
Check: \(\mathbf{C}(1,0,-1)^\top=(2,0,-2)^\top=2(1,0,-1)^\top\). For \(s=\pm\sqrt{2}\), \(\mathbf{C}(1,s,1)^\top=(2+s,2+2s,2+s)^\top\), while \((2+s)(1,s,1)^\top=(2+s,2s+s^2,2+s)^\top=(2+s,2+2s,2+s)^\top\). The eigenvector equation holds for both signs.
How to Read Multiplicity and Check the Work
In these examples, every characteristic polynomial factors into three distinct real linear factors. Each factor supplies one real eigenvalue, and each has algebraic multiplicity \(1\). Solving the homogeneous system separately for each root produces a basis of its eigenspace; the number of vectors in that basis gives the geometric multiplicity. Do not infer the basis from the factorization: the polynomial identifies the eigenvalue, while the system \((\mathbf{A}-\lambda\mathbf{I}_3)\mathbf{v}=\mathbf{0}\) identifies its eigenvectors.
A useful check is to substitute every proposed basis vector into \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Also verify that the polynomial and the matrix convention agree. With \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_3)\), a three-by-three characteristic polynomial has leading coefficient \(-1\). Changing to \(\det(\lambda\mathbf{I}_3-\mathbf{A})\) changes the sign in dimension three, although it does not change the roots.
Common pitfall: losing the characteristic-polynomial sign. For the first example, a tempting but incorrect calculation writes \(p_{\mathbf{A}}(\lambda)=(\lambda-1)(\lambda-2)(\lambda-3)\). That expression has leading coefficient \(+1\), but the stated convention gives the leading term \(\det(-\lambda\mathbf{I}_3)=-\lambda^3\). The correct calculation is \(p_{\mathbf{A}}(\lambda)=-(\lambda-1)(\lambda-2)(\lambda-3)\). The incorrect sign happens not to change the roots, but it means the polynomial itself was computed incorrectly. A second check is that an eigenvector must be nonzero: the zero vector solves every homogeneous system but is not an eigenvector.
The three examples also illustrate a useful distinction. A zero eigenvalue is handled by the same procedure as any other root: solve the corresponding homogeneous system and find its null space. An irrational root is also valid when the matrix is real; the eigenvalue and eigenvector entries need not be rational. In every case, substitute the proposed vectors back into the defining equation rather than relying only on the determinant calculation.
Check Your Understanding
For each problem, give enough work to support the eigenvalues and eigenspace claims, and verify each proposed eigenvector by substitution.
- For \(\mathbf{A}=\begin{bmatrix}0&0&2\\1&0&-5\\0&1&4\end{bmatrix}\), compute \(p_{\mathbf{A}}(\lambda)\) and find its real eigenvalues.
- For \(\mathbf{B}=\begin{bmatrix}3&0&0\\1&1&0\\0&1&2\end{bmatrix}\), determine the eigenspace basis for \(\lambda=1\).
- Explain why a root of \(p_{\mathbf{A}}\) is an eigenvalue, but does not by itself specify an eigenvector.
- If a characteristic polynomial factors into three distinct real linear factors, what is the algebraic multiplicity of each eigenvalue?
- For the matrix \(\mathbf{C}\) in the last worked example, verify directly that \((1,\sqrt{2},1)^\top\) is an eigenvector and name its eigenvalue.
