Why an Eigenvector Stays in Its Eigenspace
An eigenspace is built from vectors that a linear map sends to scalar multiples of themselves. That description already suggests why the eigenspace should be preserved by the map: if \(\mathbf{v}\) belongs to the eigenspace for \(\lambda\), then applying the map gives \(\lambda\mathbf{v}\), which is still in that eigenspace. Making this observation precise connects eigenvectors with the earlier idea of an invariant subspace and gives a useful way to understand what can happen when an eigenvalue is repeated or when several eigenspaces are combined.
Let \(V\) be a vector space over \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(T:V\to V\) be linear. Write \(I_V\) for the identity map on \(V\), so \(I_V(\mathbf{v})=\mathbf{v}\). A scalar \(\lambda\in\mathbb{F}\) is an eigenvalue of \(T\) if there is a nonzero vector \(\mathbf{v}\in V\) such that \(T(\mathbf{v})=\lambda\mathbf{v}\). Such a vector is an eigenvector for \(\lambda\).
A subspace \(W\) of \(V\) is invariant under \(T\) if \(T(\mathbf{w})\in W\) for every \(\mathbf{w}\in W\). In other words, \(T(W)\subseteq W\); equality is not part of the definition. Before using the word “subspace” for \(E_{\lambda}\), note that it is a kernel: if \(\mathbf{u},\mathbf{v}\in E_{\lambda}\) and \(a,b\in\mathbb{F}\), linearity gives \(T(a\mathbf{u}+b\mathbf{v})-\lambda(a\mathbf{u}+b\mathbf{v})=a(T(\mathbf{u})-\lambda\mathbf{u})+b(T(\mathbf{v})-\lambda\mathbf{v})=\mathbf{0}\). Also \(\mathbf{0}\in E_{\lambda}\), so \(E_{\lambda}\) is a subspace.
Proof. Let \(\mathbf{v}\in E_{\lambda}\). By the definition of \(E_{\lambda}\), \(T(\mathbf{v})=\lambda\mathbf{v}\). Since \(E_{\lambda}\) is a subspace, it contains the scalar multiple \(\lambda\mathbf{v}\). Thus \(T(\mathbf{v})\in E_{\lambda}\) for every \(\mathbf{v}\in E_{\lambda}\), which is exactly \(T(E_{\lambda})\subseteq E_{\lambda}\). This also covers \(\mathbf{v}=\mathbf{0}\), since \(T(\mathbf{0})=\mathbf{0}\). If \(\lambda\) is not an eigenvalue, then \(E_{\lambda}=\{\mathbf{0}\}\), and the same argument shows that the zero subspace is invariant. \(\square\)
Repeated Eigenvalues: The Eigenspace Can Vary in Size
The invariance proof uses only the equation \(T(\mathbf{v})=\lambda\mathbf{v}\); it does not depend on whether \(\lambda\) occurs once or repeatedly in a characteristic polynomial. A repeated root does not by itself tell us how many independent eigenvectors there are. The next examples compute the eigenspaces directly, then check the action of the matrix on their vectors.
Worked Example: A Repeated Eigenvalue with a Two-Dimensional Eigenspace
Let \(T:\mathbb{R}^3\to\mathbb{R}^3\) have matrix \(\mathbf{A}=\begin{bmatrix}3&0&1\\0&3&0\\0&0&3\end{bmatrix}\) in the standard basis. Find \(E_3\) and verify that it is invariant.
By definition, \(\mathbf{x}=(x,y,z)^{\top}\) belongs to \(E_3\) exactly when \((\mathbf{A}-3\mathbf{I}_3)\mathbf{x}=\mathbf{0}\). The matrix and the resulting equations are
Thus \(z=0\), while \(x\) and \(y\) are free. Therefore \(E_3=\{(x,y,0)^{\top}:x,y\in\mathbb{R}\}=\operatorname{span}\{(1,0,0)^{\top},(0,1,0)^{\top}\}\). The characteristic polynomial, using \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_3)\), is \((3-\lambda)^3\), so \(3\) is a repeated root. Here the eigenspace has dimension two, not three.
Check: For every \(\mathbf{x}=(x,y,0)^{\top}\in E_3\), \(\mathbf{A}\mathbf{x}=(3x,3y,0)^{\top}=3\mathbf{x}\in E_3\). On the spanning vectors, \(\mathbf{A}(1,0,0)^{\top}=(3,0,0)^{\top}\) and \(\mathbf{A}(0,1,0)^{\top}=(0,3,0)^{\top}\). Both images remain in \(E_3\), as required.
Worked Example: A Repeated Eigenvalue with a One-Dimensional Eigenspace
Let \(\mathbf{B}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\) define \(S:\mathbb{R}^2\to\mathbb{R}^2\). Find \(E_2\), then check invariance.
For \(\mathbf{x}=(x,y)^{\top}\), the kernel equation is \((\mathbf{B}-2\mathbf{I}_2)\mathbf{x}=\mathbf{0}\). Computing its left side gives
Thus \(y=0\) and \(x\) is free, so \(E_2=\operatorname{span}\{(1,0)^{\top}\}\). The characteristic polynomial is \((2-\lambda)^2\), but this eigenspace is only one-dimensional. For any \((x,0)^{\top}\in E_2\), \(\mathbf{B}(x,0)^{\top}=(2x,0)^{\top}=2(x,0)^{\top}\in E_2\).
Check: With \(\mathbf{v}=(1,0)^{\top}\), direct multiplication gives \(\mathbf{B}\mathbf{v}=(2,0)^{\top}=2\mathbf{v}\). The computed spanning vector is indeed an eigenvector, and its scalar multiples stay in the eigenspace.
Sums of Eigenspaces
If \(U\) and \(W\) are subspaces of \(V\), their sum is \(U+W=\{\mathbf{u}+\mathbf{w}:\mathbf{u}\in U,\mathbf{w}\in W\}\). For a finite list of subspaces, the sum consists of all sums formed by choosing one vector from each subspace. The sum need not be a direct sum: different choices of summands can sometimes represent the same vector. Regardless of whether representations are unique, a sum of invariant subspaces is invariant.
Proof. Take any \(\mathbf{x}\in U_1+\cdots+U_q\). By the definition of the sum, there are vectors \(\mathbf{u}_i\in U_i\) such that \(\mathbf{x}=\mathbf{u}_1+\cdots+\mathbf{u}_q\). Linearity gives \(T(\mathbf{x})=T(\mathbf{u}_1)+\cdots+T(\mathbf{u}_q)\). Since each \(U_i\) is invariant, \(T(\mathbf{u}_i)\in U_i\). Hence \(T(\mathbf{x})\in U_1+\cdots+U_q\). Each eigenspace is invariant by the preceding theorem, so the conclusion also applies when the \(U_i\) are eigenspaces. \(\square\)
Worked Example: The Sum of Two Eigenspaces
Consider the linear map on \(\mathbb{R}^3\) with matrix \(\mathbf{C}=\begin{bmatrix}2&0&0\\0&2&0\\1&0&-1\end{bmatrix}\). Find the eigenspaces for \(2\) and \(-1\), and show directly that their sum is invariant.
For \(\lambda=2\), the equation \((\mathbf{C}-2\mathbf{I}_3)(x,y,z)^{\top}=\mathbf{0}\) becomes \((x-3z,0,-3z)^{\top}=\mathbf{0}\) only if we make an incorrect subtraction, so compute the matrix carefully: \(\mathbf{C}-2\mathbf{I}_3=\begin{bmatrix}0&0&0\\0&0&0\\1&0&-3\end{bmatrix}\). Its kernel condition is \(x-3z=0\), with \(y,z\) free. Consequently, \(E_2=\operatorname{span}\{(3,0,1)^{\top},(0,1,0)^{\top}\}\).
For \(\lambda=-1\), \(\mathbf{C}+\mathbf{I}_3=\begin{bmatrix}3&0&0\\0&3&0\\1&0&0\end{bmatrix}\). The equations are \(3x=0\), \(3y=0\), and \(x=0\); thus \(x=y=0\), \(z\) is free, and \(E_{-1}=\operatorname{span}\{(0,0,1)^{\top}\}\).
To check the eigenspace descriptions by multiplication,
Every \(\mathbf{x}\in E_2+E_{-1}\) can be written as \(\mathbf{x}=\mathbf{u}+\mathbf{v}\), where \(\mathbf{u}\in E_2\) and \(\mathbf{v}\in E_{-1}\). The displayed multiplications show \(T(\mathbf{u})=2\mathbf{u}\in E_2\) and \(T(\mathbf{v})=-\mathbf{v}\in E_{-1}\), so \(T(\mathbf{x})=2\mathbf{u}-\mathbf{v}\in E_2+E_{-1}\). In fact, the three spanning vectors displayed above span \(\mathbb{R}^3\), so this sum is all of \(\mathbb{R}^3\), which is invariant as well.
Distinct Eigenspaces Have No Overlap Except Zero
For distinct eigenvalues, a stronger fact holds: their eigenspaces form a direct sum. A sum is called direct when each vector in it has only one representation as a sum of vectors from the specified subspaces. Equivalently, if a sum of one vector from each subspace is zero, then every one of those vectors must be zero. We prove this finite-dimensional-independent statement for any vector space \(V\).
Proof. Suppose \(\mathbf{v}_1+\cdots+\mathbf{v}_q=\mathbf{0}\), where \(\mathbf{v}_i\in E_{\lambda_i}\). If \(q=1\), the equation itself gives \(\mathbf{v}_1=\mathbf{0}\). Now let \(q\ge2\), and fix an index \(i\). Apply to the equation the composition of the maps \(T-\lambda_j I_V\) for all \(j\ne i\). These maps commute because each is a scalar shift of the same map \(T\). For a term \(\mathbf{v}_k\) with \(k\ne i\), one factor is \(T-\lambda_k I_V\), and \((T-\lambda_k I_V)(\mathbf{v}_k)=\mathbf{0}\), so that term is sent to zero. For the remaining term, each factor acts by a scalar: \((T-\lambda_j I_V)(\mathbf{v}_i)=(\lambda_i-\lambda_j)\mathbf{v}_i\). Therefore the resulting equation is \(\bigl(\prod_{j\ne i}(\lambda_i-\lambda_j)\bigr)\mathbf{v}_i=\mathbf{0}\). All the eigenvalues are distinct, so every factor in this product is nonzero. The product is a nonzero scalar in the field \(\mathbb{F}\), and division by it gives \(\mathbf{v}_i=\mathbf{0}\). This holds for each index \(i\), proving the direct-sum condition. \(\square\)
The theorem does not say that eigenspaces for different eigenvalues exhaust \(V\); their sum may be a proper subspace. It says that whenever a vector belongs to their sum, its components in those eigenspaces are unique. The earlier invariance result says that the sum is preserved by \(T\), whether or not it fills the whole space.
Common Pitfall: Multiplicity Does Not Determine the Eigenspace
A common incorrect calculation is to see the characteristic polynomial \((2-\lambda)^2\) for \(\mathbf{B}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\) and conclude that every vector in \(\mathbb{R}^2\) belongs to \(E_2\). This confuses the repeated root of the characteristic polynomial with the eigenspace itself. For example, the proposed vector \((0,1)^{\top}\) fails the eigenvector equation: \(\mathbf{B}(0,1)^{\top}=(1,2)^{\top}\ne(0,2)^{\top}=2(0,1)^{\top}\).
The correction is to solve the defining kernel equation, not infer the eigenspace from the root’s repetition. As computed above, \((\mathbf{B}-2\mathbf{I}_2)(x,y)^{\top}=(y,0)^{\top}=\mathbf{0}\) forces \(y=0\), so \(E_2=\operatorname{span}\{(1,0)^{\top}\}\), a one-dimensional subspace. It is invariant because the map sends \((x,0)^{\top}\) to \((2x,0)^{\top}\), not because the characteristic root is repeated.
For a real matrix and a nonreal eigenvalue, the field also matters. A real vector space does not contain eigenvectors with genuinely complex entries. To use a complex eigenvalue, regard the matrix as defining a linear map on the corresponding complex vector space; then the definition and invariance proof apply over \(\mathbb{C}\) with \(\lambda\in\mathbb{C}\).
Check Your Understanding
Use the kernel definition and the invariance condition to justify each answer.
- For \(\mathbf{D}=\begin{bmatrix}4&0\\0&4\end{bmatrix}\), find \(E_4\) and verify directly that it is invariant.
- For \(\mathbf{A}=\begin{bmatrix}3&0&1\\0&3&0\\0&0&3\end{bmatrix}\), check whether \((1,1,1)^{\top}\) belongs to \(E_3\). Show the kernel equation used.
- Explain why the zero eigenspace \(\{\mathbf{0}\}\), when a scalar is not an eigenvalue, is still invariant.
- If \(U\) and \(W\) are invariant subspaces of the same linear map, prove that \(U+W\) is invariant.
- For two distinct eigenvalues \(\lambda_1,\lambda_2\), suppose \(\mathbf{v}_1+\mathbf{v}_2=\mathbf{0}\) with \(\mathbf{v}_i\in E_{\lambda_i}\). Use \(T-\lambda_2 I_V\) to show \(\mathbf{v}_1=\mathbf{0}\), then conclude \(\mathbf{v}_2=\mathbf{0}\).
