Reading Eigenvalues from the Diagonal
For a general matrix, finding an eigenvalue may require solving a characteristic equation, and finding its eigenvectors may require solving a system of equations. A diagonal matrix makes both tasks more direct: each coordinate of the eigenvalue equation can be examined separately. Repeated entries have a specific effect: they allow more than one coordinate to vary in the corresponding eigenspace.
Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\) be diagonal, with diagonal entries \(d_1,\dots,d_n\). Thus the entry in row \(i\), column \(i\) is \(d_i\), and all off-diagonal entries are zero. Recall that an eigenvalue \(\lambda\in\mathbb{R}\) of \(\mathbf{A}\) is a scalar for which there is a nonzero vector \(\mathbf{x}\in\mathbb{R}^n\) satisfying \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). The eigenspace \(E_\lambda\) is the set of all solutions of this equation, including \(\mathbf{0}\); the eigenvectors for \(\lambda\) are its nonzero vectors.
Proof. Write \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\). Since \(\mathbf{A}\) is diagonal, the equation \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) is equivalent, coordinate by coordinate, to \(d_i x_i=\lambda x_i\), or \((d_i-\lambda)x_i=0\), for every \(i\). If \(d_i\ne\lambda\), then \(d_i-\lambda\ne0\), so this equation forces \(x_i=0\). If \(d_i=\lambda\), the equation becomes \(0\cdot x_i=0\), which places no restriction on \(x_i\). Therefore all solutions have zero coordinates outside the positions where \(d_i=\lambda\), and every vector with that coordinate pattern solves the equation. Such vectors are precisely the linear combinations of the standard basis vectors \(\mathbf{e}_i\) for which \(d_i=\lambda\). If \(\lambda\) occurs on the diagonal, at least one such basis vector is nonzero and belongs to \(E_\lambda\), so \(\lambda\) is an eigenvalue. If it does not occur, every coordinate is forced to be zero, so there is no nonzero solution and \(\lambda\) is not an eigenvalue. \(\square\)
The theorem gives two useful ways to describe an eigenspace. The coordinate description specifies which entries of \(\mathbf{x}\) must be zero. The span description identifies the standard basis directions along which \(\mathbf{x}\) may vary. The eigenspace is a subspace, so it contains the zero vector; to obtain eigenvectors, exclude that vector.
Distinct Entries and Repeated Entries
Worked Example: Distinct Diagonal Entries
Find all eigenvalues and eigenspaces of \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&-1&0\\0&0&5\end{bmatrix}\).
The diagonal entries are \(2,-1,5\), all distinct, so these are exactly the eigenvalues. For \(\lambda=2\), the coordinate equations are \((2-2)x_1=0\), \((-1-2)x_2=0\), and \((5-2)x_3=0\). The last two equations give \(x_2=x_3=0\), while \(x_1\) is free. Hence \(E_2=\{(x_1,0,0)^{\top}:x_1\in\mathbb{R}\}=\operatorname{span}\{\mathbf{e}_1\}\).
For \(\lambda=-1\), the first and third coordinates must be zero, giving \(E_{-1}=\operatorname{span}\{\mathbf{e}_2\}\). For \(\lambda=5\), the first and second coordinates must be zero, giving \(E_5=\operatorname{span}\{\mathbf{e}_3\}\).
Check: \(\mathbf{A}\mathbf{e}_1=(2,0,0)^{\top}=2\mathbf{e}_1\), \(\mathbf{A}\mathbf{e}_2=(0,-1,0)^{\top}=-1\mathbf{e}_2\), and \(\mathbf{A}\mathbf{e}_3=(0,0,5)^{\top}=5\mathbf{e}_3\). Each listed eigenvalue has a nonzero eigenvector in its claimed eigenspace.
When an entry is repeated, each coordinate at a matching diagonal position is free. Consequently, the eigenspace can have dimension greater than one. It is not a single eigendirection: all combinations of the matching standard basis vectors are included.
Worked Example: A Repeated Diagonal Entry
Find the eigenvalues and eigenspaces of \(\mathbf{B}=\begin{bmatrix}3&0&0&0\\0&-2&0&0\\0&0&3&0\\0&0&0&-2\end{bmatrix}\).
The distinct diagonal values, and therefore the eigenvalues, are \(3\) and \(-2\). For \(\lambda=3\), the coordinate equations are \(0x_1=0\), \(-5x_2=0\), \(0x_3=0\), and \(-5x_4=0\). Thus \(x_2=x_4=0\), while \(x_1\) and \(x_3\) are free: \(E_3=\{(a,0,b,0)^{\top}:a,b\in\mathbb{R}\}=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_3\}\).
For \(\lambda=-2\), the equations force \(x_1=x_3=0\), while \(x_2\) and \(x_4\) are free. Therefore \(E_{-2}=\{(0,a,0,b)^{\top}:a,b\in\mathbb{R}\}=\operatorname{span}\{\mathbf{e}_2,\mathbf{e}_4\}\).
Check: for \(\mathbf{x}=(a,0,b,0)^{\top}\), direct multiplication gives \(\mathbf{B}\mathbf{x}=(3a,0,3b,0)^{\top}=3\mathbf{x}\). For \(\mathbf{y}=(0,a,0,b)^{\top}\), it gives \(\mathbf{B}\mathbf{y}=(0,-2a,0,-2b)^{\top}=-2\mathbf{y}\). Each eigenspace contains nonzero vectors, for example \(\mathbf{e}_1\in E_3\) and \(\mathbf{e}_2\in E_{-2}\).
Dimension and Multiplicity
The number of times an eigenvalue occurs on the diagonal determines the number of coordinate directions available in its eigenspace. In particular, the dimension of the eigenspace is the number of matching diagonal positions. For a diagonal matrix, this also agrees with the algebraic multiplicity: the number of times the eigenvalue is a root of the characteristic polynomial.
Proof. Let \(k\) be the number of indices \(i\) for which \(d_i=\lambda\). By the eigenspace description proved above, \(E_\lambda\) is spanned by the \(k\) standard basis vectors with those indices. These vectors are linearly independent: if a linear combination of distinct standard basis vectors is zero, its coordinate at each included index is the corresponding coefficient, so every coefficient is zero. They therefore form a basis of \(E_\lambda\), and \(\dim E_\lambda=k\). For the algebraic multiplicity, the characteristic polynomial convention is \(p_{\mathbf{A}}(t)=\det(\mathbf{A}-t\mathbf{I}_n)\). Since \(\mathbf{A}-t\mathbf{I}_n\) is diagonal, \(p_{\mathbf{A}}(t)=\prod_{i=1}^n(d_i-t)\). Each index with \(d_i=\lambda\) contributes one factor \(\lambda-t=-(t-\lambda)\), and every other index contributes a factor nonzero at \(t=\lambda\). Thus \(t=\lambda\) is a root of order exactly \(k\), so \(\operatorname{am}(\lambda)=k\). Combining the two equalities gives \(\dim E_\lambda=\operatorname{am}(\lambda)\). \(\square\)
This equality is a conclusion about diagonal matrices. The key point for determining the eigenspace directly is still the coordinate equation: matching positions are free, and all other positions must be zero. The polynomial calculation provides a second count, through the repeated factor in the characteristic polynomial.
Worked Example: Zero as a Repeated Eigenvalue
For \(\mathbf{C}=\begin{bmatrix}0&0&0&0\\0&5&0&0\\0&0&0&0\\0&0&0&-2\end{bmatrix}\), determine the eigenvalues, the eigenspace for \(0\), and its dimension.
The diagonal values are \(0,5,0,-2\), so the distinct eigenvalues are \(0,5,-2\). For \(\lambda=0\), the coordinate equations are \(0x_1=0\), \(5x_2=0\), \(0x_3=0\), and \(-2x_4=0\). Hence \(x_2=x_4=0\), with \(x_1,x_3\) free. Therefore \(E_0=\{(a,0,b,0)^{\top}:a,b\in\mathbb{R}\}=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_3\}\), and its dimension is \(2\).
Check: for every \(\mathbf{x}=(a,0,b,0)^{\top}\), multiplication gives \(\mathbf{C}\mathbf{x}=(0,0,0,0)^{\top}=0\mathbf{x}\). The nonzero choice \((a,b)=(1,0)\) confirms that \(0\) is an eigenvalue. The characteristic polynomial is \((-t)(5-t)(-t)(-2-t)\), which has \(t=0\) as a root of order \(2\), agreeing with the eigenspace dimension.
When a Candidate Is Not an Eigenvalue
A scalar not appearing on the diagonal cannot be an eigenvalue of a diagonal matrix. The coordinate equations explain why: if every \(d_i\ne\lambda\), each coefficient \(d_i-\lambda\) is nonzero, so every coordinate \(x_i\) must be zero. The only solution is then \(\mathbf{x}=\mathbf{0}\), which is not an eigenvector. This distinction prevents a common error: solving the eigenvalue equation is not enough; an eigenvalue requires a nonzero solution.
Worked Example: Testing a Value Absent from the Diagonal
Determine whether \(\lambda=4\) is an eigenvalue of \(\mathbf{D}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&6\end{bmatrix}\).
The equations \(\mathbf{D}\mathbf{x}=4\mathbf{x}\), for \(\mathbf{x}=(x_1,x_2,x_3)^{\top}\), are \((1-4)x_1=0\), \((2-4)x_2=0\), and \((6-4)x_3=0\). They become \(-3x_1=0\), \(-2x_2=0\), and \(2x_3=0\), so \(x_1=x_2=x_3=0\). There is no nonzero solution; therefore \(4\) is not an eigenvalue.
Check: the only vector satisfying the equation is \(\mathbf{0}\), and \(\mathbf{D}\mathbf{0}=4\mathbf{0}=\mathbf{0}\). This verifies that the solution set is \(E_4=\{\mathbf{0}\}\), but it does not make \(4\) an eigenvalue, because no nonzero eigenvector exists.
Common Mistake: Ignoring a Repeated Coordinate
Consider again \(\mathbf{B}=\operatorname{diag}(3,-2,3,-2)\). A mistaken calculation for \(E_3\) might claim that \(x_1\) is free but \(x_3=0\), and therefore \(E_3=\operatorname{span}\{\mathbf{e}_1\}\). The correction is to write every coordinate equation. The third equation is \((3-3)x_3=0\), or \(0x_3=0\), which does not force \(x_3=0\). It is free, just like \(x_1\). For instance, \(\mathbf{x}=(0,0,1,0)^{\top}\) satisfies \(\mathbf{B}\mathbf{x}=(0,0,3,0)^{\top}=3\mathbf{x}\). The correct eigenspace is \(\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_3\}\).
When reading a diagonal matrix, use the following checks:
- List distinct diagonal values to identify the eigenvalues; do not count a repeated value as several different eigenvalues.
- For a chosen eigenvalue, keep every coordinate position where it occurs; those coordinates are free.
- Set every other coordinate to zero, then state the eigenspace as a coordinate subspace or as a span.
- To verify an eigenvalue, exhibit a nonzero vector in the eigenspace and check the eigenvalue equation.
Check Your Understanding
For each question, use the coordinate equations and distinguish the eigenspace from its nonzero eigenvectors.
- Find the eigenvalues and eigenspaces of \(\operatorname{diag}(4,1,4)\).
- For \(\operatorname{diag}(-3,0,-3,2)\), describe \(E_{-3}\) and state its dimension.
- Is \(5\) an eigenvalue of \(\operatorname{diag}(1,3,7)\)? Explain using the eigenvalue equation.
- For \(\operatorname{diag}(0,2,0)\), give one nonzero eigenvector for the eigenvalue \(0\), and state the full eigenspace.
- Explain why the dimension of an eigenspace of a diagonal matrix equals the number of occurrences of its eigenvalue on the diagonal.
