From Diagonal Entries to Eigenspaces
An upper triangular matrix makes its eigenvalues easy to identify, but its eigenvectors are not generally just standard basis vectors. The entries above the diagonal can link coordinates together, forcing some coordinates of an eigenvector to depend on others. To find the eigenspace, subtract the candidate eigenvalue from the diagonal and solve the resulting homogeneous system.
Throughout, matrices and vectors are over \(\mathbb{R}\). A matrix \(\mathbf{A}=(a_{ij})\in\mathbb{R}^{n\times n}\) is upper triangular if \(a_{ij}=0\) whenever \(i>j\). Thus every entry below its main diagonal is zero. An eigenvalue \(\lambda\) of \(\mathbf{A}\) is a scalar for which there is a nonzero vector \(\mathbf{x}\) satisfying \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\); such a vector is an eigenvector.
The definition gives the calculation directly: for each diagonal value \(\lambda\), form \(\mathbf{A}-\lambda\mathbf{I}_n\) and solve \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\). A diagonal entry may occur more than once, but it is still one eigenvalue; its eigenspace is found by solving the system once for that value. The diagonal identifies which values to test, while the full matrix determines the solutions.
Why the Homogeneous System Gives the Eigenspace
Proof. By definition, \(\mathbf{x}\) belongs to \(E_{\lambda}\) exactly when \((\mathbf{A}-\lambda\mathbf{I})\mathbf{x}=\mathbf{0}\). By definition, the null space of a matrix \(\mathbf{B}\) is the set of vectors \(\mathbf{x}\) satisfying \(\mathbf{B}\mathbf{x}=\mathbf{0}\). Taking \(\mathbf{B}=\mathbf{A}-\lambda\mathbf{I}\) proves the equality of the two sets. The zero vector is a solution, but it is not an eigenvector; every nonzero solution satisfies \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) and is an eigenvector. \(\square\)
The triangular structure also gives a useful restriction on the coordinates of any eigenvector. If a nonzero vector has some zero coordinates at its end, its last nonzero coordinate must correspond to a diagonal entry equal to the eigenvalue.
Proof. In row \(j\) of \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\), the left side is \(\sum_{k=j}^{n}a_{jk}x_k\), because entries below the diagonal are zero. By the choice of \(j\), each \(x_k=0\) for \(k>j\). The row equation therefore reduces to \(a_{jj}x_j=\lambda x_j\). Since \(x_j\ne0\), division by \(x_j\) gives \(a_{jj}=\lambda\). \(\square\)
This result does not say that every eigenvector is supported only at coordinates whose diagonal entries equal \(\lambda\). Earlier coordinates can be nonzero as well, and off-diagonal entries can determine their values. It says specifically that the final nonzero coordinate must occur at a diagonal position equal to \(\lambda\).
Solving for Each Eigenspace
Worked Example: Two Eigenvalues, Two Different Constraints
Find the eigenspaces of \(\mathbf{A}=\begin{bmatrix}2&1\\0&3\end{bmatrix}\). Its diagonal entries show that the eigenvalues are \(2\) and \(3\).
For \(\lambda=2\), subtract \(2\mathbf{I}_2\) and solve:
Writing \(\mathbf{x}=(x,y)^{\top}\), the remaining equation is \(y=0\), while \(x\) is free. Hence \(E_2=\{(x,0)^{\top}:x\in\mathbb{R}\}=\operatorname{span}\{(1,0)^{\top}\}\). Check: \(\mathbf{A}(1,0)^{\top}=(2,0)^{\top}=2(1,0)^{\top}\).
For \(\lambda=3\),
The equation is \(-x+y=0\), so \(y=x\). Therefore \(E_3=\{(x,x)^{\top}:x\in\mathbb{R}\}=\operatorname{span}\{(1,1)^{\top}\}\). Check: \(\mathbf{A}(1,1)^{\top}=(3,3)^{\top}=3(1,1)^{\top}\). The off-diagonal \(1\) links the coordinates for \(\lambda=3\), rather than leaving the second standard basis vector as an eigenvector.
In a larger system, the equations still have a triangular pattern: the last row involves only the last coordinate, the row above it involves the last two, and so on. Solving the equations from the bottom upward can make the constraints especially visible. Row operations are also valid because they preserve the solutions of a homogeneous system.
Worked Example: A Repeated Diagonal Entry
Let \(\mathbf{B}=\begin{bmatrix}4&1&0\\0&4&1\\0&0&2\end{bmatrix}\). Find the eigenspace for \(\lambda=4\). Subtracting \(4\mathbf{I}_3\) gives
For the row reduction, first scale \(R_3\) by \(-\tfrac12\), then replace \(R_2\) by \(R_2-R_3\). In the resulting system, \(y=0\) and \(z=0\); \(x\) is free. Thus \(E_4=\operatorname{span}\{(1,0,0)^{\top}\}\). Although \(4\) appears twice on the diagonal, the equations constrain the second and third coordinates. Check: \(\mathbf{B}(1,0,0)^{\top}=(4,0,0)^{\top}=4(1,0,0)^{\top}\).
The other eigenvalue is \(\lambda=2\). Here
The equations are \(2x+y=0\) and \(2y+z=0\). Set the free variable \(z=4\). Then \(2y+4=0\), so \(y=-2\); next \(2x-2=0\), so \(x=1\). A nonzero solution is \((1,-2,4)^{\top}\), and \(E_2=\operatorname{span}\{(1,-2,4)^{\top}\}\). Check by multiplication:
Free Coordinates and the Shape of an Eigenspace
When solving \((\mathbf{A}-\lambda\mathbf{I})\mathbf{x}=\mathbf{0}\), a free variable can be chosen arbitrarily, and the pivot variables are then determined by the equations. Each choice gives a solution, so the resulting eigenspace includes all linear combinations of the vectors obtained by assigning one free variable the value \(1\) and the others \(0\). In simple cases there is one free variable and the eigenspace is a span of one vector; there may also be more than one.
Worked Example: Two Free Coordinates
Let \(\mathbf{C}=\begin{bmatrix}5&0&2\\0&5&0\\0&0&1\end{bmatrix}\). For \(\lambda=5\),
Scale \(R_1\) by \(\tfrac12\), then replace \(R_3\) by \(R_3+4R_1\). The only constraint is \(z=0\), leaving \(x\) and \(y\) free. Thus
Check the spanning vectors separately: \(\mathbf{C}(1,0,0)^{\top}=(5,0,0)^{\top}\) and \(\mathbf{C}(0,1,0)^{\top}=(0,5,0)^{\top}\). Both have eigenvalue \(5\). The entry \(2\) in position \((1,3)\) forces \(z=0\), while the other two coordinates remain unrestricted.
These examples illustrate a reliable sequence: identify a diagonal value, form \(\mathbf{A}-\lambda\mathbf{I}\), solve the homogeneous equations, express all solutions using free variables, and verify a spanning vector by multiplying it by \(\mathbf{A}\). The multiplication check confirms the eigenvector equation, while the system solution describes the entire eigenspace, not just one eigenvector.
Common Mistake: Treating an Upper Triangular Matrix as Diagonal
A common error is to assume that a diagonal entry \(\lambda\) automatically makes the corresponding standard basis vector an eigenvector. That shortcut works for a diagonal matrix, but an upper triangular matrix may have nonzero entries above its diagonal.
For \(\mathbf{A}=\begin{bmatrix}2&1\\0&3\end{bmatrix}\), it is wrong to claim that \((0,1)^{\top}\) is an eigenvector for \(\lambda=3\), just because \(3\) is the second diagonal entry. Direct multiplication gives \(\mathbf{A}(0,1)^{\top}=(1,3)^{\top}\), which is not \(3(0,1)^{\top}=(0,3)^{\top}\). Solving \((\mathbf{A}-3\mathbf{I})\mathbf{x}=\mathbf{0}\) instead gives \(-x+y=0\), so \((1,1)^{\top}\) is an eigenvector. Indeed, \(\mathbf{A}(1,1)^{\top}=(3,3)^{\top}=3(1,1)^{\top}\).
Keep the distinction precise: diagonal entries identify the eigenvalues of an upper triangular matrix; the full system identifies the eigenspaces. A correct eigenvector must be nonzero and must pass the substitution check \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\).
Check Your Understanding
For each problem, write the homogeneous system and check any proposed eigenvector by multiplication.
- Find the eigenspace for \(\lambda=1\) of \(\begin{bmatrix}1&2\\0&4\end{bmatrix}\).
- For \(\begin{bmatrix}3&1&0\\0&3&0\\0&0&2\end{bmatrix}\), solve for the eigenspace corresponding to \(\lambda=3\).
- Find one eigenvector for \(\lambda=2\) of \(\begin{bmatrix}4&1&0\\0&2&1\\0&0&2\end{bmatrix}\).
- Explain why \((0,1)^{\top}\) is not an eigenvector for \(\lambda=3\) of \(\begin{bmatrix}2&1\\0&3\end{bmatrix}\).
- For an upper triangular matrix, what must be true of the diagonal entry at the largest index where an eigenvector has a nonzero coordinate?
