When a Parameter Changes the Answer
A matrix family can be diagonalizable for some parameter values and fail to be diagonalizable for others. Often the change occurs when two eigenvalues meet, but a repeated eigenvalue does not by itself decide the question. What matters is whether the eigenspace has enough independent eigenvectors to match the eigenvalue’s algebraic multiplicity. The calculations in this tutorial track both features as the parameter varies.
Throughout, matrices and vectors are over \(\mathbb{R}\), and each parameter is real. For a matrix \(\mathbf{A}\), an eigenvalue \(\lambda\) is a scalar for which \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for some nonzero vector \(\mathbf{v}\). Its eigenspace is \(E_{\lambda}=\{\mathbf{x}:\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\}\). The algebraic multiplicity \(\operatorname{am}(\lambda)\) is the multiplicity of \(\lambda\) as a root of the characteristic polynomial; the geometric multiplicity is \(\operatorname{gm}(\lambda)=\dim E_{\lambda}\).
All the characteristic polynomials considered below split over \(\mathbb{R}\). The criterion gives a useful test, but each parameter value still requires an eigenspace calculation. For a two-by-two matrix, diagonalizability is also equivalent to having two linearly independent eigenvectors, as in the earlier characterization of diagonalization by an eigenbasis.
A Complete Classification for a Triangular Two-by-Two Family
Consider the real upper triangular matrix \(\mathbf{A}=\begin{bmatrix}a&b\\0&d\end{bmatrix}\), where \(a,b,d\in\mathbb{R}\). Its characteristic polynomial is \(\det(\mathbf{A}-\lambda\mathbf{I}_2)=(a-\lambda)(d-\lambda)\), so the diagonal entries give the eigenvalues, counting repetition. The parameter values of interest are those for which \(a=d\), or for which a parameter in \(b\) makes \(b=0\) or nonzero.
Proof. Suppose first that \(a\ne d\). The vector \(\mathbf{e}_1=(1,0)\) satisfies \(\mathbf{A}\mathbf{e}_1=a\mathbf{e}_1\). Set \(\mathbf{v}=(-b,a-d)\). This vector is nonzero because its second coordinate \(a-d\) is nonzero. Direct multiplication gives \(\mathbf{A}\mathbf{v}=(-ab+b(a-d),d(a-d))=(-bd,d(a-d))=d(-b,a-d)=d\mathbf{v}\). The two eigenvectors are independent: if \(c\mathbf{e}_1+k\mathbf{v}=\mathbf{0}\), the second coordinate gives \(k(a-d)=0\), hence \(k=0\), and then the first coordinate gives \(c=0\). Thus they form an eigenbasis of \(\mathbb{R}^2\), so \(\mathbf{A}\) is diagonalizable. Now suppose \(a=d\). If \(b=0\), then \(\mathbf{A}=a\mathbf{I}_2\), which is already diagonal. If \(b\ne0\), then \(\mathbf{A}-a\mathbf{I}_2=\begin{bmatrix}0&b\\0&0\end{bmatrix}\), and \((\mathbf{A}-a\mathbf{I}_2)(x,y)=(by,0)\). The eigenspace for the only eigenvalue \(a\) is therefore given by \(y=0\), so it has dimension \(1\), not \(2\). There cannot be an eigenbasis of \(\mathbb{R}^2\), and \(\mathbf{A}\) is not diagonalizable. These cases prove both directions. \(\square\)
This classification separates two different events. If \(a\ne d\), the eigenvalues are distinct and the explicit eigenvectors above always give a basis, regardless of \(b\). If \(a=d\), the eigenvalue is repeated: the matrix is diagonalizable exactly when it is scalar. A nonzero off-diagonal entry at the collision leaves only a one-dimensional eigenspace.
Worked Example: An Off-Diagonal Parameter
For \(\mathbf{A}(s)=\begin{bmatrix}1&s\\0&1\end{bmatrix}\), determine the real values of \(s\) for which the matrix is diagonalizable.
The characteristic polynomial is \((1-\lambda)^2\), so the sole eigenvalue is \(1\), with algebraic multiplicity \(2\), for every \(s\). To find its eigenspace, solve \((\mathbf{A}(s)-\mathbf{I}_2)(x,y)=(sy,0)=(0,0)\). If \(s=0\), every vector is in the eigenspace, and \(\mathbf{A}(0)=\mathbf{I}_2\) is diagonalizable. If \(s\ne0\), then \(y=0\), so \(E_1=\operatorname{span}\{(1,0)\}\) has dimension \(1\). The multiplicity criterion gives that \(\mathbf{A}(s)\) is not diagonalizable in this case.
Check: At \(s=0\), the identity matrix has two independent eigenvectors, for example \((1,0)\) and \((0,1)\). At \(s=2\), \(\mathbf{A}(2)(1,0)=(1,0)\), but every eigenvector for \(1\) has second coordinate zero, confirming that there cannot be two independent eigenvectors.
Worked Example: A Collision at One Parameter Value
Analyze \(\mathbf{C}(t)=\begin{bmatrix}2&1\\0&t\end{bmatrix}\) and give an explicit diagonalization when \(t=4\).
The characteristic polynomial is \((2-\lambda)(t-\lambda)\). The two-by-two classification shows that \(\mathbf{C}(t)\) is diagonalizable whenever \(t\ne2\); at \(t=2\), the repeated eigenvalue has only a one-dimensional eigenspace because the upper-right entry is \(1\).
At \(t=4\), the matrix is \(\mathbf{C}(4)=\begin{bmatrix}2&1\\0&4\end{bmatrix}\). For eigenvalue \(2\), \(\mathbf{e}_1=(1,0)\) is an eigenvector. For eigenvalue \(4\), direct multiplication gives \(\mathbf{C}(4)(1,2)=(4,8)=4(1,2)\). These vectors are independent because the matrix with these columns has nonzero determinant:
Taking the eigenvalues in the same order as the columns gives \(\mathbf{D}=\begin{bmatrix}2&0\\0&4\end{bmatrix}\), and \(\mathbf{P}^{-1}=\begin{bmatrix}1&-\tfrac12\\0&\tfrac12\end{bmatrix}\). Verify the diagonalization by multiplication:
At the collision \(t=2\), the equation \((\mathbf{C}(2)-2\mathbf{I}_2)(x,y)=(y,0)=(0,0)\) forces \(y=0\). Thus \(E_2=\operatorname{span}\{(1,0)\}\) and \(\mathbf{C}(2)\) is not diagonalizable. The parameter changes the answer at exactly the value where the two diagonal entries meet.
A Three-by-Three Family: A Repeated Eigenvalue Can Be Harmless
With three dimensions, a repeated eigenvalue may still have a sufficiently large eigenspace. The following family has one parameter value where a repeated eigenvalue causes failure and another where a repeated eigenvalue causes no failure.
Proof. The characteristic polynomial is \((1-\lambda)(t-\lambda)(3-\lambda)\). First suppose \(t\ne1\) and \(t\ne3\). The vectors \(\mathbf{e}_1=(1,0,0)\), \(\mathbf{v}=(1,t-1,0)\), and \(\mathbf{e}_3=(0,0,1)\) are eigenvectors for \(1,t,3\), respectively: multiplication gives \(\mathbf{B}(t)\mathbf{e}_1=\mathbf{e}_1\), \(\mathbf{B}(t)\mathbf{v}=(t,t(t-1),0)=t\mathbf{v}\), and \(\mathbf{B}(t)\mathbf{e}_3=3\mathbf{e}_3\). Their column matrix has determinant \(t-1\ne0\), so they form a basis. If \(t=3\), the eigenvalues are \(1\) and \(3\), with \(3\) repeated. The equation \((\mathbf{B}(3)-\mathbf{I}_3)(x,y,z)=(y,2y,2z)=(0,0,0)\) gives \(E_1=\operatorname{span}\{(1,0,0)\}\). The equation \((\mathbf{B}(3)-3\mathbf{I}_3)(x,y,z)=(-2x+y,0,0)=(0,0,0)\) gives \(E_3=\operatorname{span}\{(1,2,0),(0,0,1)\}\). The three displayed vectors together are independent, since their column matrix has determinant \(2\), so \(\mathbf{B}(3)\) is diagonalizable. Finally, if \(t=1\), the only eigenvalues are \(1\) and \(3\). Solving \((\mathbf{B}(1)-\mathbf{I}_3)(x,y,z)=(y,0,2z)=(0,0,0)\) gives \(E_1=\operatorname{span}\{(1,0,0)\}\). Solving \((\mathbf{B}(1)-3\mathbf{I}_3)(x,y,z)=(-2x+y,-2y,0)=(0,0,0)\) gives \(y=0\), then \(x=0\), so \(E_3=\operatorname{span}\{(0,0,1)\}\). These eigenspaces together span only a two-dimensional space. No eigenbasis of \(\mathbb{R}^3\) exists, so \(\mathbf{B}(1)\) is not diagonalizable. The cases cover every real \(t\), proving the claim. \(\square\)
Worked Example: The Repeated Eigenvalue at t = 3
The theorem says that \(\mathbf{B}(3)\) is diagonalizable even though \(3\) is a repeated eigenvalue. Construct and verify a diagonalization.
Here \(\mathbf{B}(3)=\begin{bmatrix}1&1&0\\0&3&0\\0&0&3\end{bmatrix}\). The eigenvectors found in the proof can be ordered as \((1,0,0)\), \((1,2,0)\), and \((0,0,1)\). Thus take
First multiply \(\mathbf{P}\mathbf{D}=\begin{bmatrix}1&3&0\\0&6&0\\0&0&3\end{bmatrix}\). Then
The repeated eigenvalue \(3\) has two independent eigenvectors, so the eigenspace dimension matches its algebraic multiplicity. By contrast, at \(t=1\) the repeated eigenvalue \(1\) has only one independent eigenvector. The multiplicities, not the mere presence of repetition, explain the different outcomes.
A Reliable Parameter Check
For a parameter-dependent matrix, organize the work so that eigenvalue changes and eigenspace changes are both checked. The characteristic polynomial identifies candidate values where eigenvalues coincide; the eigenspace equations decide whether a collision permits diagonalization.
Factor it as far as possible and record the eigenvalues as functions of the parameter, including any repeated roots.
Check where eigenvalue expressions become equal. Also inspect other parameter values that change the equations defining an eigenspace.
For each eigenvalue in each case, solve \((\mathbf{A}-\lambda\mathbf{I})\mathbf{x}=\mathbf{0}\) and determine the dimension of the solution space.
Use the multiplicity criterion when the characteristic polynomial splits, or exhibit a full set of independent eigenvectors. A proposed diagonalization should satisfy \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
A parameter value at which eigenvalues coincide deserves attention, but it is only a candidate for a change in diagonalizability. For example, at \(t=3\) in the three-by-three family, two eigenvalues coincide but the eigenspace is large enough. Conversely, a characteristic polynomial alone cannot establish that a repeated eigenvalue has enough eigenvectors.
Common Pitfall: Treating Every Repetition as Failure
A tempting but incorrect conclusion is: “The eigenvalue \(3\) is repeated in \(\mathbf{B}(3)\), so the matrix is not diagonalizable.” The calculation contradicts that conclusion. At \(t=3\), the eigenspace for \(3\) contains the independent vectors \((1,2,0)\) and \((0,0,1)\), while the eigenspace for \(1\) contains \((1,0,0)\). Their column matrix has determinant \(2\ne0\), and the explicit product above verifies the diagonalization.
The opposite shortcut is also unsafe: seeing two different eigenvalues does not settle a matrix of dimension greater than two, because there may not be enough independent eigenvectors across all eigenspaces. Compute the eigenspace dimensions, or produce and check a full eigenbasis. For the two-by-two triangular family, the classification theorem provides a shortcut because it has been proved for that entire form; do not extend that conclusion to unrelated matrix families without checking their hypotheses.
Check Your Understanding
For each question, identify the parameter cases first, then use the eigenspace equations to justify the conclusion.
- For \(\begin{bmatrix}4&s\\0&4\end{bmatrix}\), which real values of \(s\) give a diagonalizable matrix?
- For \(\begin{bmatrix}1&2\\0&t\end{bmatrix}\), find the parameter value at which the eigenvalues coincide and decide whether the matrix is diagonalizable there.
- For \(\mathbf{B}(t)\), state the algebraic and geometric multiplicities of the repeated eigenvalue at \(t=3\).
- At \(t=1\) for \(\mathbf{B}(t)\), find a basis for each eigenspace and explain why their union is not an eigenbasis of \(\mathbb{R}^3\).
- Explain why a repeated eigenvalue is not, by itself, enough to conclude that a matrix is not diagonalizable.
