Prescribing Eigenvalues with the Diagonal
Triangular matrices make it possible to build examples with chosen eigenvalues without first solving a characteristic polynomial. Their diagonal entries determine the characteristic polynomial, so repeating a diagonal value prescribes its algebraic multiplicity. But the diagonal does not, by itself, determine the dimension of the eigenspace: off-diagonal entries can impose extra equations on eigenvectors.
Throughout, matrices and vectors are over \(\mathbb{R}\). An eigenvalue \(\lambda\) of a matrix \(\mathbf{A}\) is a scalar for which there is a nonzero vector \(\mathbf{v}\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Its eigenspace is \(E_\lambda=\{\mathbf{v}:(\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}\}\). The algebraic multiplicity counts how many times \(\lambda\) occurs as a root of the characteristic polynomial \(p_{\mathbf{A}}(s)=\det(\mathbf{A}-s\mathbf{I})\); the geometric multiplicity is \(\dim E_\lambda\).
Proof. The matrix \(\mathbf{A}-s\mathbf{I}\) is upper triangular, with diagonal entries \(d_i-s\). In the determinant expansion, a term indexed by a permutation \(\pi\) can be nonzero only if the entry in row \(i\), column \(\pi(i)\) is not below the diagonal, so \(\pi(i)\geq i\) for every \(i\). Since \(\sum_i\pi(i)=\sum_i i\), these inequalities can all hold only when \(\pi(i)=i\) for every \(i\). The only potentially nonzero term is therefore the product of the diagonal entries, \(\prod_i(d_i-s)\). Its roots, with repetitions, are exactly the diagonal values with their counts. \(\square\)
This gives a direct construction rule: put each desired eigenvalue on the diagonal as many times as its desired algebraic multiplicity. The sum of those multiplicities must equal the matrix size. Once the diagonal is fixed, choose off-diagonal entries and solve \((\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}\) to find the geometric multiplicity. The following examples show why that second calculation matters.
A Repeated Eigenvalue in Two Dimensions
Worked Example: Algebraic Multiplicity Two, Two Possible Eigenspace Dimensions
Construct two upper-triangular matrices with the repeated eigenvalue \(2\), one whose eigenspace has dimension two and one whose eigenspace has dimension one.
Use \(2\) in both diagonal positions. First choose zero off-diagonal entry:
For an eigenvector \(\mathbf{v}=(x,y)^{\top}\), the equation \((\mathbf{A}_0-2\mathbf{I})\mathbf{v}=\mathbf{0}\) becomes \((0,0)^{\top}=\mathbf{0}\). Both \(x\) and \(y\) are free, so \(E_2=\mathbb{R}^2\) and the geometric multiplicity is two. For example, \(\mathbf{A}_0\mathbf{e}_1=2\mathbf{e}_1\) and \(\mathbf{A}_0\mathbf{e}_2=2\mathbf{e}_2\).
Now change only the off-diagonal entry:
The eigenvector equation is \(\begin{bmatrix}0&3\\0&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}0\\0\end{bmatrix}\), which gives \(3y=0\), hence \(y=0\), while \(x\) is free. Thus \(E_2=\operatorname{span}\{\mathbf{e}_1\}\), with geometric multiplicity one. Check: \(\mathbf{A}_1\mathbf{e}_1=(2,0)^{\top}=2\mathbf{e}_1\). Both matrices have algebraic multiplicity two, but the change in the off-diagonal entry changes the eigenspace dimension.
Proof. The characteristic polynomial is \((\lambda-s)^2\), so the algebraic multiplicity is two. Also, \((\mathbf{A}-\lambda\mathbf{I})(x,y)^{\top}=(ay,0)^{\top}\). If \(a=0\), every pair \((x,y)\) solves the eigenvector equation, so the eigenspace has dimension two. If \(a\ne0\), the equation \(ay=0\) forces \(y=0\), while \(x\) is free, so the eigenspace has dimension one. \(\square\)
Three Repeated Diagonal Entries
With three copies of the same diagonal value, more off-diagonal entries can interact. Write a general upper-triangular matrix with diagonal \(\lambda,\lambda,\lambda\) as \(\begin{bmatrix}\lambda&a&b\\0&\lambda&c\\0&0&\lambda\end{bmatrix}\). Its characteristic polynomial is \((\lambda-s)^3\), regardless of \(a,b,c\). To find the geometric multiplicity, subtract \(\lambda\mathbf{I}\) and solve for a vector \((x,y,z)^{\top}\).
Proof. The eigenvector equation reduces to \(ay+bz=0\) and \(cz=0\); \(x\) is free in every case. If \(c\ne0\), then \(z=0\). If also \(a\ne0\), then \(y=0\), leaving only \(x\) free, so the dimension is one. If \(c\ne0\) but \(a=0\), then \(z=0\) and the first equation imposes no condition on \(y\); \(x,y\) are free, giving dimension two. If \(c=0\), the only possible constraint on \(y,z\) is \(ay+bz=0\). When at least one of \(a,b\) is nonzero, this is one nontrivial equation in two variables, leaving one free variable in addition to \(x\), so the dimension is two. When \(a=b=0\) as well, there is no constraint on \(y,z\), and all three variables are free. This proves all cases. \(\square\)
Worked Example: Constructing Three Geometric Multiplicities
Keep the algebraic multiplicity equal to three and construct examples with geometric multiplicities one, two, and three.
For geometric multiplicity one, set \(\lambda=1\), \(a=2\), \(b=-1\), and \(c=3\):
The eigenvector equations are \(2y-z=0\) and \(3z=0\). They give \(z=0\), then \(y=0\), with \(x\) free. Thus \(E_1=\operatorname{span}\{\mathbf{e}_1\}\). Directly, \(\mathbf{A}_1\mathbf{e}_1=(1,0,0)^{\top}=\mathbf{e}_1\), confirming that the nonzero vector spanning this eigenspace is an eigenvector.
For geometric multiplicity two, set \(a=0\), \(b=4\), and \(c=0\):
The only eigenvector equation is \(4z=0\), so \(z=0\), while \(x\) and \(y\) are free. Hence \(E_1=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\}\). The checks are \(\mathbf{A}_2\mathbf{e}_1=(1,0,0)^{\top}\) and \(\mathbf{A}_2\mathbf{e}_2=(0,1,0)^{\top}\), each equal to the corresponding basis vector.
Finally, use zero off-diagonal entries:
Every vector solves \((\mathbf{A}_3-\mathbf{I}_3)\mathbf{v}=\mathbf{0}\), so \(E_1=\mathbb{R}^3\) has dimension three. In particular, \(\mathbf{A}_3\mathbf{e}_i=\mathbf{e}_i\) for \(i=1,2,3\). In all three constructions, the repeated diagonal sets the same algebraic multiplicity; the off-diagonal entries determine how many independent eigenvector directions remain.
Prescribing a Repeated Eigenvalue and a Distinct One
The diagonal can prescribe more than one eigenvalue at once. For example, placing \(\lambda,\lambda,\mu\), with \(\lambda\ne\mu\), on a three-by-three triangular matrix gives algebraic multiplicity two for \(\lambda\) and one for \(\mu\). Entries joining positions with different diagonal values do not necessarily affect the eigenspace in the same way as entries between repeated values; solving the eigenvector equation identifies which entries matter.
Worked Example: Algebraic Multiplicities Two and One
Construct matrices with algebraic multiplicities two for \(2\) and one for \(-1\), with geometric multiplicity for \(2\) equal to two in one case and one in another.
Consider the family \(\mathbf{B}_a=\begin{bmatrix}2&a&7\\0&2&-3\\0&0&-1\end{bmatrix}\). For every \(a\), the characteristic polynomial is \(p_{\mathbf{B}_a}(s)=(2-s)^2(-1-s)\). Thus \(2\) has algebraic multiplicity two and \(-1\) has algebraic multiplicity one.
For eigenvalue \(2\), subtract \(2\mathbf{I}_3\). The system is \(ay+7z=0\), \(-3z=0\), and \(-3z=0\). The second equation gives \(z=0\), so the first becomes \(ay=0\). If \(a=0\), both \(x,y\) are free and \(E_2=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\}\). If \(a=5\), then \(y=0\) and \(x\) is free, so \(E_2=\operatorname{span}\{\mathbf{e}_1\}\).
Check both cases: \(\mathbf{B}_0\mathbf{e}_1=2\mathbf{e}_1\) and \(\mathbf{B}_0\mathbf{e}_2=2\mathbf{e}_2\); also \(\mathbf{B}_5\mathbf{e}_1=2\mathbf{e}_1\). The entries \(7\) and \(-3\) do not change these eigenspace calculations because the equations force \(z=0\). Changing \(a\), which lies between the two repeated diagonal positions, does change the dimension.
Common Pitfall: Repeated Diagonal Values Do Not Count Eigenvectors
A tempting but incorrect calculation is to see three copies of \(1\) on the diagonal and conclude that there must be three independent eigenvectors. For instance, take \(\mathbf{C}=\begin{bmatrix}1&5&0\\0&1&0\\0&0&1\end{bmatrix}\). Its characteristic polynomial is \((1-s)^3\), so the algebraic multiplicity is three. But \((\mathbf{C}-\mathbf{I}_3)(x,y,z)^{\top}=(5y,0,0)^{\top}\), so an eigenvector must satisfy \(y=0\). The eigenspace has dimension two, not three, and \(\mathbf{C}\mathbf{e}_2=(5,1,0)^{\top}\ne\mathbf{e}_2\). The diagonal prescribes algebraic multiplicity; it does not count independent eigenvectors.
A reliable construction-and-check procedure is:
- Write the desired eigenvalues on the diagonal, repeating each according to its desired algebraic multiplicity.
- Choose off-diagonal entries, keeping the matrix triangular so its characteristic polynomial is immediate.
- For each repeated eigenvalue, solve \((\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}\) and count the free variables.
- Substitute a basis for the resulting eigenspace into \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) to check the eigenvector calculations.
This separates two design choices that are easy to confuse: the diagonal determines the roots and their algebraic multiplicities, while the eigenspace equations reveal the geometric multiplicities. Small triangular matrices make both choices visible entry by entry.
Check Your Understanding
For each construction, first identify the characteristic polynomial, then solve the eigenvector equations to verify the geometric multiplicity.
- For \(\begin{bmatrix}4&0\\0&4\end{bmatrix}\), state the algebraic and geometric multiplicities of \(4\).
- For \(\begin{bmatrix}3&2\\0&3\end{bmatrix}\), find a basis for \(E_3\) and state its dimension.
- Choose \(a,b,c\) so that \(\begin{bmatrix}0&a&b\\0&0&c\\0&0&0\end{bmatrix}\) has algebraic multiplicity three and geometric multiplicity two for \(0\).
- For \(\begin{bmatrix}2&0&6\\0&2&1\\0&0&-1\end{bmatrix}\), determine the algebraic multiplicities of \(2\) and \(-1\), and find a basis for \(E_2\).
- Explain why the diagonal entries alone are not enough to determine the dimension of an eigenspace.
