Opening the book…
A fundamental question in linear algebra is the following: If \(T : V \to V\) is a linear operator, how can a basis \(B\) of \(V\) be chosen so the matrix \(M_{B}(T)\) is as simple as possible? A basic technique for answering such questions will be explained in this section. If \(U\) is a subspace of \(V\), write its image under \(T\) as
\[T(U) = \{T(\mathbf{u}) \mid \mathbf{u} \mbox{ in } U\} \nonumber \]
Let \(T : V \to V\) be an operator. A subspace \(U \subseteq V\) is called \(\boldsymbol{T}\)-invariant if \(T(U) \subseteq U\), that is, \(T(\mathbf{u}) \in U\) for every vector \(\mathbf{u} \in U\). Hence \(T\) is a linear operator on the vector space \(U\).
This is illustrated in the diagram, and the fact that \(T : U \to U\) is an operator on \(U\) is the primary reason for our interest in \(T\)-invariant subspaces.
Let \(T : V \to V\) be any linear operator. Then:
- \(\{\mathbf{0}\}\) and \(V\) are \(T\)-invariant subspaces.
- Both \(\text{ker }T\) and \(im \;T = T(V)\) are \(T\)-invariant subspaces.
- If \(U\) and \(W\) are \(T\)-invariant subspaces, so are \(T(U)\), \(U \cap W\), and \(U + W\).
Solution
Item 1 is clear, and the rest is left as Exercises 9.3.1 and 9.3.2.
Define \(T : \mathbb{R}^3 \to \mathbb{R}^3\) by \(T(a, b, c) = (3a + 2b, b - c, 4a + 2b - c)\). Then \(U = \{(a, b, a) \mid a, b \mbox{ in } \mathbb{R}\}\) is \(T\)-invariant because
\[T(a, b, a) = (3a + 2b, b - a, 3a + 2b) \nonumber \]
is in \(U\) for all \(a\) and \(b\) (the first and last entries are equal).
If a spanning set for a subspace \(U\) is known, it is easy to check whether \(U\) is \(T\)-invariant.
Let \(T : V \to V\) be a linear operator, and suppose that \(U = span \;\{\mathbf{u}_{1}, \mathbf{u}_{2}, \dots, \mathbf{u}_{k}\}\) is a subspace of \(V\). Show that \(U\) is \(T\)-invariant if and only if \(T(\mathbf{u}_{i})\) lies in \(U\) for each \(i = 1, 2, \dots, k\).
Solution
Given \(\mathbf{u}\) in \(U\), write it as \(\mathbf{u} = r_{1}\mathbf{u}_{1} + \cdots + r_{k}\mathbf{u}_{k}\), \(r_{i}\) in \(\mathbb{R}\). Then
\[T(\mathbf{u}) = r_1T(\mathbf{u}_1) + \cdots + r_kT(\mathbf{u}_k) \nonumber \]
and this lies in \(U\) if each \(T(\mathbf{u}_{i})\) lies in \(U\). This shows that \(U\) is \(T\)-invariant if each \(T(\mathbf{u}_{i})\) lies in \(U\); the converse is clear.
Define \(T : \mathbb{R}^2 \to \mathbb{R}^2\) by \(T(a, b) = (b, -a)\). Show that \(\mathbb{R}^2\) contains no \(T\)-invariant subspace except \(0\) and \(\mathbb{R}^2\).
Solution
Suppose, if possible, that \(U\) is \(T\)-invariant, but \(U \neq 0\), \(U \neq \mathbb{R}^2\). Then \(U\) has dimension \(1\) so \(U = \mathbb{R}\mathbf{x}\) where \(\mathbf{x} \neq \mathbf{0}\). Now \(T(\mathbf{x})\) lies in \(U\)—say \(T(\mathbf{x}) = r\mathbf{x}\), \(r\) in \(\mathbb{R}\). If we write \(\mathbf{x} = (a, b)\), this is \((b, -a) = r(a, b)\), which gives \(b = ra\) and \(-a = rb\). Eliminating \(b\) gives \(r^{2}a = rb = -a\), so \((r^{2} + 1)a = 0\). Hence \(a = 0\). Then \(b = ra = 0\) too, contrary to the assumption that \(\mathbf{x} \neq \mathbf{0}\). Hence no one-dimensional \(T\)-invariant subspace exists.
Let \(T : V \to V\) be a linear operator. If \(U\) is any \(T\)-invariant subspace of \(V\), then
\[T : U \to U \nonumber \]
is a linear operator on the subspace \(U\), called the restriction of \(T\) to \(U\).
This is the reason for the importance of \(T\)-invariant subspaces and is the first step toward finding a basis that simplifies the matrix of \(T\).
Let \(T : V \to V\) be a linear operator where \(V\) has dimension \(n\) and suppose that \(U\) is any \(T\)-invariant subspace of \(V\). Let \(B_{1} = \{\mathbf{b}_{1}, \dots, \mathbf{b}_{k}\}\) be any basis of \(U\) and extend it to a basis \(B = \{\mathbf{b}_{1}, \dots, \mathbf{b}_{k}, \mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\}\) of \(V\) in any way. Then \(M_{B}(T)\) has the block triangular form
\[M_B(T) = \left[ \begin{array}{cc} M_{B_1}(T) & Y \\ 0 & Z \end{array} \right] \nonumber \]
where \(Z\) is \((n - k) \times (n - k)\) and \(M_{B_1}(T)\) is the matrix of the restriction of \(T\) to \(U\).
The matrix of (the restriction) \(T : U \to U\) with respect to the basis \(B_{1}\) is the \(k \times k\) matrix
\[M_{B_1}(T) = \left[ \begin{array}{cccc} C_{B_1}[T(\mathbf{b}_1)] & C_{B_1}[T(\mathbf{b}_2)] & \cdots & C_{B_1}[T(\mathbf{b}_k)] \end{array} \right] \nonumber \]
Now compare the first column \(C_{B_1}[T(\mathbf{b}_1)]\) here with the first column \(C_{B}[T(\mathbf{b}_{1})]\) of \(M_{B}(T)\). The fact that \(T(\mathbf{b}_{1})\) lies in \(U\) (because \(U\) is \(T\)-invariant) means that \(T(\mathbf{b}_{1})\) has the form
\[T(\mathbf{b}_1) = t_1\mathbf{b}_1 + t_2\mathbf{b}_2 + \cdots + t_k\mathbf{b}_k + 0\mathbf{b}_{k+1} + \cdots + 0\mathbf{b}_n \nonumber \]
Consequently,
\[C_{B_1}[T(\mathbf{b}_1)] = \left[ \begin{array}{c} t_1 \\ t_2 \\ \vdots \\ t_k \end{array} \right] \mbox{ in } \mathbb{R}^k \quad \mbox{ whereas } \quad C_B[T(\mathbf{b}_1)] = \left[ \begin{array}{c} t_1 \\ t_2 \\ \vdots \\ t_k \\ 0 \\ \vdots \\ 0 \end{array} \right] \mbox{ in } \mathbb{R}^n \nonumber \]
This shows that the matrices \(M_{B}(T)\) and \(\left[ \begin{array}{cc} M_{B_1}(T) & Y \\ 0 & Z \end{array} \right]\) have identical first columns.
Similar statements apply to columns \(2, 3, \dots, k\), and this proves the theorem.
\(\square\)
The block upper triangular form for the matrix \(M_{B}(T)\) in Theorem 9.3.1 is very useful because the determinant of such a matrix equals the product of the determinants of each of the diagonal blocks. This is recorded in Theorem 9.3.2 for reference, together with an important application to characteristic polynomials.
Let \(A\) be a block upper triangular matrix, say
\[A = \left[ \begin{array}{ccccc} A_{11} & A_{12} & A_{13} & \cdots & A_{1n} \\ 0 & A_{22} & A_{23} & \cdots & A_{2n} \\ 0 & 0 & A_{33} & \cdots & A_{3n} \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & A_{nn} \\ \end{array} \right] \nonumber \]
where the diagonal blocks are square. Then:
- \(\det A = (\det A_{11})(\det A_{22})(\det A_{33})\cdots(\det A_{nn})\).
- \(c_A(x) = c_{A_{11}}(x)c_{A_{22}}(x)c_{A_{33}}(x){\cdots}c_{A_{nn}}(x)\).
If \(n = 2\), (1) is Theorem 3.1.5; the general case (by induction on \(n\)) is left to the reader. Then (2) follows from (1) because
\[xI - A = \left[ \begin{array}{ccccc} xI - A_{11} & -A_{12} & -A_{13} & \cdots & -A_{1n} \\ 0 & xI - A_{22} & -A_{23} & \cdots & -A_{2n} \\ 0 & 0 & xI - A_{33} & \cdots & -A_{3n} \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & xI - A_{nn} \\ \end{array} \right] \nonumber \]
where, in each diagonal block, the symbol \(I\) stands for the identity matrix of the appropriate size.
\(square\)
Consider the linear operator \(T :\|{P}_{2} \to\|{P}_{2}\) given by
\[T(a + bx + cx^2) = (-2a - b + 2c) + (a + b)x + (-6a -2b + 5c)x^2 \nonumber \]
Show that \(U = span \;\{x, 1 + 2x^{2}\}\) is \(T\)-invariant, use it to find a block upper triangular matrix for \(T\), and use that to compute \(c_{T}(x)\).
Solution
\(U\) is \(T\)-invariant by Example 9.3.3 because \(U = span \;\{x, 1 + 2x^{2}\}\) and both \(T(x)\) and \(T(1 + 2x^{2})\) lie in \(U\):
\[\begin{aligned} T(x) & = -1 + x - 2x^2 = x - (1 + 2x^2) \\ T(1 + 2x^2) & = 2 + x + 4x^2 = x + 2(1 + 2x^2)\end{aligned} \nonumber \]
Extend the basis \(B_{1} = \{x, 1 + 2x^{2}\}\) of \(U\) to a basis \(B\) of \(\mathbf{P}_{2}\) in any way at all—say, \(B = \{x, 1 + 2x^{2}, x^{2}\}\). Then
\[\begin{aligned} M_B(T) & = \left[ \begin{array}{ccc} C_B[T(x)] & C_B[T(1 + 2x^2)] & C_B[T(x^2)] \end{array} \right] \\ & = \left[ \begin{array}{ccc} C_B(-1 + x - 2x^2) & C_B(2 + x + 4x^2) & C_B(2 + 5x^2) \end{array} \right] \\ & = \left[ \begin{array}{rr|r} 1 & 1 & 0 \\ -1 & 2 & 2 \\ \hline 0 & 0 & 1 \end{array} \right]\end{aligned} \nonumber \]
is in block upper triangular form as expected. Finally,
\[c_T(x) = \det \left[ \begin{array}{cc|c} x - 1 & -1 & 0 \\ 1 & x - 2 & -2 \\ \hline 0 & 0 & x - 1 \end{array} \right] = (x^2 - 3x + 3)(x - 1) \nonumber \]
Eigenvalues
Let \(T : V \to V\) be a linear operator. A one-dimensional subspace \(\mathbb{R}\mathbf{v}\), \(\mathbf{v} \neq \mathbf{0}\), is \(T\)-invariant if and only if \(T(r\mathbf{v}) = rT(\mathbf{v})\) lies in \(\mathbb{R}\mathbf{v}\) for all \(r\) in \(\mathbb{R}\). This holds if and only if \(T(\mathbf{v})\) lies in \(\mathbb{R}\mathbf{v}\); that is, \(T(\mathbf{v}) = \lambda\mathbf{v}\) for some \(\lambda\) in \(\mathbb{R}\). A real number \(\lambda\) is called an eigenvalue of an operator \(T : V \to V\) if
\[T(\mathbf{v}) = \lambda\mathbf{v} \nonumber \]
holds for some nonzero vector \(\mathbf{v}\) in \(V\). In this case, \(\mathbf{v}\) is called an eigenvector of \(T\) corresponding to \(\lambda\). The subspace
\[E_{\lambda}(T) = \{\mathbf{v} \mbox{ in } V \mid T(\mathbf{v}) = \lambda\mathbf{v}\} \nonumber \]
is called the eigenspace of \(T\) corresponding to \(\lambda\). These terms are consistent with those used in Section 5.5 for matrices. If \(A\) is an \(n \times n\) matrix, a real number \(\lambda\) is an eigenvalue of the matrix operator \(T_{A} : \mathbb{R}^n \to \mathbb{R}^n\) if and only if \(\lambda\) is an eigenvalue of the matrix \(A\). Moreover, the eigenspaces agree:
\[E_{\lambda}(T_A) = \{\mathbf{x} \mbox{ in } \mathbb{R}^n \mid A\mathbf{x} = \lambda\mathbf{x} \} = E_{\lambda}(A) \nonumber \]
The following theorem reveals the connection between the eigenspaces of an operator \(T\) and those of the matrices representing \(T\).
Let \(T : V \to V\) be a linear operator where \(dim \;V = n\), let \(B\) denote any ordered basis of \(V\), and let \(C_{B} : V \to \mathbb{R}^n\) denote the coordinate isomorphism. Then:
- The eigenvalues \(\lambda\) of \(T\) are precisely the eigenvalues of the matrix \(M_{B}(T)\) and thus are the roots of the characteristic polynomial \(c_{T}(x)\).
- In this case the eigenspaces \(E_{\lambda}(T)\) and \(E_{\lambda}[M_{B}(T)]\) are isomorphic via the restriction \(C_{B} : E_{\lambda}(T) \to E_{\lambda}[M_{B}(T)]\).
Write \(A = M_{B}(T)\) for convenience. If \(T(\mathbf{v}) = \lambda\mathbf{v}\), then \({\lambda}C_{B}(\mathbf{v}) = C_{B}[T(\mathbf{v})] = AC_{B}(\mathbf{v})\) because \(C_{B}\) is linear. Hence \(C_{B}(\mathbf{v})\) lies in \(E_{\lambda}(A)\), so we do have a function \(C_{B} : E_{\lambda}(T) \to E_{\lambda}(A)\). It is clearly linear and one-to-one; we claim it is onto. If \(\mathbf{x}\) is in \(E_{\lambda}(A)\), write \(\mathbf{x} = C_{B}(\mathbf{v})\) for some \(\mathbf{v}\) in \(V\) (\(C_{B}\) is onto). This \(\mathbf{v}\) actually lies in \(E_{\lambda}(T)\). To see why, observe that
\[C_B[T(\mathbf{v})] = AC_B(\mathbf{v}) = A\mathbf{x} = \lambda\mathbf{x} = {\lambda}C_B(\mathbf{v}) = C_B(\lambda\mathbf{v}) \nonumber \]
Hence \(T(\mathbf{v}) = \lambda\mathbf{v}\) because \(C_{B}\) is one-to-one, and this proves (2). As to (1), we have already shown that eigenvalues of \(T\) are eigenvalues of \(A\). The converse follows, as in the foregoing proof that \(C_{B}\) is onto.
\(square\)
Theorem 9.3.3 shows how to pass back and forth between the eigenvectors of an operator \(T\) and the eigenvectors of any matrix \(M_{B}(T)\) of \(T\):
\[\mathbf{v} \mbox{ lies in } E_{\lambda}(T) \quad \mbox{ if and only if } \quad C_B(\mathbf{v}) \mbox{ lies in } E_{\lambda}[M_B(T)] \nonumber \]
Find the eigenvalues and eigenspaces for \(T :\|{P}_{2} \to\|{P}_{2}\) given by
\[T(a + bx + cx^2) = (2a + b + c) + (2a + b - 2c)x - (a + 2c)x^2 \nonumber \]
Solution
If \(B = \{1, x, x^{2}\}\), then
\[M_B(T) = \left[ \begin{array}{ccc} C_B[T(1)] & C_B[T(x)] & C_B[T(x^2)] \end{array} \right] = \left[ \begin{array}{rrr} 2 & 1 & 1 \\ 2 & 1 & -2 \\ -1 & 0 & -2 \end{array} \right] \nonumber \]
Hence \(c_{T}(x) = \det [xI - M_{B}(T)] = (x + 1)^{2}(x - 3)\) as the reader can verify.
Moreover, \(E_{-1}[M_B(T)] = \mathbb{R}\left[ \begin{array}{r} -1 \\ 2 \\ 1 \end{array} \right]\) and \(E_3[M_B(T)] = \mathbb{R}\left[ \begin{array}{r} 5 \\ 6 \\ -1 \end{array} \right]\), so Theorem 9.3.3 gives \(E_{-1}(T) = \mathbb{R}(-1 + 2x + x^{2})\) and \(E_{3}(T) = \mathbb{R}(5 + 6x - x^{2})\).
Each eigenspace of a linear operator \(T : V \to V\) is a \(T\)-invariant subspace of \(V\).
If \(\mathbf{v}\) lies in the eigenspace \(E_{\lambda}(T)\), then \(T(\mathbf{v}) = \lambda\mathbf{v}\), so \(T[T(\mathbf{v})] = T(\lambda\mathbf{v}) = {\lambda}T(\mathbf{v})\). This shows that \(T(\mathbf{v})\) lies in \(E_{\lambda}(T)\) too.
\(\square\)
Direct Sums
Sometimes vectors in a space \(V\) can be written naturally as a sum of vectors in two subspaces. For example, in the space \(\mathbf{M}_{nn}\) of all \(n \times n\) matrices, we have subspaces
\[U = \{P \mbox{ in }\|{M}_{nn} \mid P \mbox{ is symmetric } \} \quad \mbox{ and } \quad W = \{Q \mbox{ in }\|{M}_{nn} \mid Q \mbox{ is skew symmetric} \} \nonumber \]
where a matrix \(Q\) is called skew-symmetric if \(Q^{T} = -Q\). Then every matrix \(A\) in \(\mathbf{M}_{nn}\) can be written as the sum of a matrix in \(U\) and a matrix in \(W\); indeed,
\[A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T) \nonumber \]
where \(\frac{1}{2}(A + A^T)\) is symmetric and \(\frac{1}{2}(A - A^T)\) is skew symmetric. Remarkably, this representation is unique: If \(A = P + Q\) where \(P^{T} = P\) and \(Q^{T} = -Q\), then \(A^{T} = P^{T} + Q^{T} = P - Q\); adding this to \(A = P + Q\) gives \(P = \frac{1}{2}(A + A^T)\), and subtracting gives \(Q = \frac{1}{2}(A - A^{T})\). In addition, this uniqueness turns out to be closely related to the fact that the only matrix in both \(U\) and \(W\) is \(0\). This is a useful way to view matrices, and the idea generalizes to the important notion of a direct sum of subspaces.
If \(U\) and \(W\) are subspaces of \(V\), their sum \(U + W\) and their intersection \(U \cap W\) were defined in Section 6.4 as follows:
\[\begin{aligned} U + W & = \{ \mathbf{u} + \mathbf{w} \mid \mathbf{u} \mbox{ in } U \mbox{ and } \mathbf{w} \mbox{ in } W \} \\ U \cap W & = \{ \mathbf{v} \mid \mathbf{v} \mbox{ lies in both } U \mbox{ and } W \}\end{aligned} \nonumber \]
These are subspaces of \(V\), the sum containing both \(U\) and \(W\) and the intersection contained in both \(U\) and \(W\). It turns out that the most interesting pairs \(U\) and \(W\) are those for which \(U \cap W\) is as small as possible and \(U + W\) is as large as possible.
A vector space \(V\) is said to be the direct sum of subspaces \(U\) and \(W\) if
\[U \cap W = \{ \mathbf{0} \} \quad \mbox{ and } \quad U + W = V \nonumber \]
In this case we write \(V = U \oplus W\). Given a subspace \(U\), any subspace \(W\) such that \(V = U \oplus W\) is called a complement of \(U\) in \(V\).
In the space \(\mathbb{R}^5\), consider the subspaces \(U = \{(a, b, c, 0, 0) \mid a, b, \mbox{ and } c \mbox{ in } \mathbb{R}\}\) and \(W = \{(0, 0, 0, d, e) \mid d \mbox{ and } e \mbox{ in } \mathbb{R}\}\). Show that \(\mathbb{R}^5 = U \oplus W\).
Solution
If \(\mathbf{x} = (a, b, c, d, e)\) is any vector in \(\mathbb{R}^5\), then \(\mathbf{x} = (a, b, c, 0, 0) + (0, 0, 0, d, e)\), so \(\mathbf{x}\) lies in \(U + W\). Hence \(\mathbb{R}^5 = U + W\). To show that \(U \cap W = \{\mathbf{0}\}\), let \(\mathbf{x} = (a, b, c, d, e)\) lie in \(U \cap W\). Then \(d = e = 0\) because \(\mathbf{x}\) lies in \(U\), and \(a = b = c = 0\) because \(\mathbf{x}\) lies in \(W\). Thus \(\mathbf{x} = (0, 0, 0, 0, 0) = \mathbf{0}\), so \(\mathbf{0}\) is the only vector in \(U \cap W\). Hence \(U \cap W = \{\mathbf{0}\}\).
If \(U\) is a subspace of \(\mathbb{R}^n\), show that \(\mathbb{R}^n = U \oplus U^{\perp}\).
Solution
The equation \(\mathbb{R}^n = U + U^{\perp}\) holds because, given \(\mathbf{x}\) in \(\mathbb{R}^n\), the vector \(\text{proj}{U}{\mathbf{x}}\) lies in \(U\) and \(\mathbf{x} - \text{proj}{U}{\mathbf{x}}\) lies in \(U^{\perp}\). To see that \(U \cap U^{\perp} = \{\mathbf{0}\}\), observe that any vector in \(U \cap U^{\perp}\) is orthogonal to itself and hence must be zero.
Let \(\{\mathbf{e}_{1}, \mathbf{e}_{2}, \dots, \mathbf{e}_{n}\}\) be a basis of a vector space \(V\), and partition it into two parts: \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{k}\}\) and \(\{\mathbf{e}_{k+1}, \dots, \mathbf{e}_{n}\}\). If \(U = span \;\{\mathbf{e}_{1}, \dots, \mathbf{e}_{k}\}\) and \(W = span \;\{\mathbf{e}_{k+1}, \dots, \mathbf{e}_{n}\}\), show that \(V = U \oplus W\).
Solution
If \(\mathbf{v}\) lies in \(U \cap W\), then \(\mathbf{v} = a_{1}\mathbf{e}_{1} + \cdots + a_{k}\mathbf{e}_{k}\) and \(\mathbf{v} = b_{k+1}\mathbf{e}_{k+1} + \cdots + b_{n}\mathbf{e}_{n}\) hold for some \(a_{i}\) and \(b_{j}\) in \(\mathbb{R}\). The fact that the \(\mathbf{e}_{i}\) are linearly independent forces all \(a_{i} = b_{j} = 0\), so \(\mathbf{v} = \mathbf{0}\). Hence \(U \cap W = \{\mathbf{0}\}\). Now, given \(\mathbf{v}\) in \(V\), write \(\mathbf{v} = v_{1}\mathbf{e}_{1} + \cdots + v_{n}\mathbf{e}_{n}\) where the \(v_{i}\) are in \(\mathbb{R}\). Then \(\mathbf{v} = \mathbf{u} + \mathbf{w}\), where \(\mathbf{u} = v_{1}\mathbf{e}_{1} + \cdots + v_{k}\mathbf{e}_{k}\) lies in \(U\) and \(\mathbf{w} = v_{k+1}\mathbf{e}_{k+1} + \cdots + v_{n}\mathbf{e}_{n}\) lies in \(W\). This proves that \(V = U + W\).
Example 9.3.9 is typical of all direct sum decompositions.
Let \(U\) and \(W\) be subspaces of a finite dimensional vector space \(V\). The following three conditions are equivalent:
- \(V = U \oplus W\).
- Each vector \(v\) in \(V\) can be written uniquely in the form \[\boldsymbol{v}=\boldsymbol{u}+\boldsymbol{w} \quad \boldsymbol{u} \text { in } U, \boldsymbol{w} \text { in } W \nonumber\]
- If \(\{\mathbf{u}_{1}, \dots, \mathbf{u}_{k}\}\) and \(\{\mathbf{w}_{1}, \dots, \mathbf{w}_{m}\}\) are bases of \(U\) and \(W\), respectively, then \(B = \{\mathbf{u}_{1}, \dots, \mathbf{u}_{k}, \mathbf{w}_{1}, \dots, \mathbf{w}_{m}\}\) is a basis of \(V\).
(The uniqueness in (2) means that if \(\mathbf{v} = \mathbf{u}_{1} + \mathbf{w}_{1}\) is another such representation, then \(\mathbf{u}_{1} = \mathbf{u}\) and \(\mathbf{w}_{1} = \mathbf{w}\).)
Example 9.3.9 shows that (3) \(\Rightarrow\) (1).
(1) \(\Rightarrow\) (2). Given \(\mathbf{v}\) in \(V\), we have \(\mathbf{v} = \mathbf{u} + \mathbf{w}\), \(\mathbf{u}\) in \(U\), \(\mathbf{w}\) in \(W\), because \(V = U + W\).
If also \(\mathbf{v} = \mathbf{u}_{1} + \mathbf{w}_{1}\), then \(\mathbf{u} - \mathbf{u}_{1} = \mathbf{w}_{1} - \mathbf{w}\) lies in \(U \cap W = \{\mathbf{0}\}\), so \(\mathbf{u} = \mathbf{u}_{1}\) and \(\mathbf{w} = \mathbf{w}_{1}\).
(2) \(\Rightarrow\) (3). Given \(\mathbf{v}\) in \(V\), we have \(\mathbf{v} = \mathbf{u} + \mathbf{w}\), \(\mathbf{u}\) in \(U\), \(\mathbf{w}\) in \(W\). Hence \(\mathbf{v}\) lies in span \(B\); that is, \(V = span \;B\). To see that \(B\) is independent, let \(a_{1}\mathbf{u}_{1} + \cdots + a_{k}\mathbf{u}_{k} + b_{1}\mathbf{w}_{1} + \cdots + b_{m}\mathbf{w}_{m} = \mathbf{0}\). Write \(\mathbf{u} = a_{1}\mathbf{u}_{1} + \cdots + a_{k}\mathbf{u}_{k}\) and \(\mathbf{w} = b_{1}\mathbf{w}_{1} + \cdots + b_{m}\mathbf{w}_{m}\). Then \(\mathbf{u} + \mathbf{w} = \mathbf{0}\), and so \(\mathbf{u} = \mathbf{0}\) and \(\mathbf{w} = \mathbf{0}\) by the uniqueness in (2). Hence \(a_{i} = 0\) for all \(i\) and \(b_{j} = 0\) for all \(j\).
\(\square\)
Condition (3) in Theorem 9.3.5 gives the following useful result.
If a finite dimensional vector space \(V\) is the direct sum \(V = U \oplus W\) of subspaces \(U\) and \(W\), then
\[dim \; V = dim \; U + dim \; W \nonumber \]
These direct sum decompositions of \(V\) play an important role in any discussion of invariant subspaces. If \(T : V \to V\) is a linear operator and if \(U_{1}\) is a \(T\)-invariant subspace, the block upper triangular matrix
\[\label{eqn:9_3_1} M_B(T) = \left[ \begin{array}{cc} M_{B_1}(T) & Y \\ 0 & Z \end{array} \right] \]
in Theorem 9.3.1 is achieved by choosing any basis \(B_{1} = \{\mathbf{b}_{1}, \dots, \mathbf{b}_{k}\}\) of \(U_{1}\) and completing it to a basis \(B = \{\mathbf{b}_{1}, \dots, \mathbf{b}_{k}, \mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\}\) of \(V\) in any way at all. The fact that \(U_{1}\) is \(T\)-invariant ensures that the first \(k\) columns of \(M_{B}(T)\) have the form in (\ref{eqn:9_3_1}) (that is, the last \(n - k\) entries are zero), and the question arises whether the additional basis vectors \(\mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\) can be chosen such that
\[U_2 = span \;\{\mathbf{b}_{k+1}, \dots, \mathbf{b}_n\} \nonumber \]
is also \(T\)-invariant. In other words, does each \(T\)-invariant subspace of \(V\) have a \(T\)-invariant complement? Unfortunately the answer in general is no (see Example 9.3.11 below); but when it is possible, the matrix \(M_{B}(T)\) simplifies further. The assumption that the complement \(U_{2} = span \;\{\mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\}\) is \(T\)-invariant too means that \(Y = 0\) in equation \ref{eqn:9_3_1} above, and that \(Z = M_{B_{2}}(T)\) is the matrix of the restriction of \(T\) to \(U_{2}\) (where \(B_{2} = \{\mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\}\)). The verification is the same as in the proof of Theorem 9.3.1.
Let \(T : V \to V\) be a linear operator where \(V\) has dimension \(n\). Suppose \(V = U_{1} \oplus U_{2}\) where both \(U_{1}\) and \(U_{2}\) are \(T\)-invariant. If \(B_{1} = \{\mathbf{b}_{1}, \dots, \mathbf{b}_{k}\}\) and \(B_{2} = \{\mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\}\) are bases of \(U_{1}\) and \(U_{2}\) respectively, then
\[B = \{\mathbf{b}_1, \dots, \mathbf{b}_k, \mathbf{b}_{k+1}, \dots, \mathbf{b}_n\} \nonumber \]
is a basis of \(V\), and \(M_{B}(T)\) has the block diagonal form
\[M_B(T) = \left[ \begin{array}{cc} M_{B_1}(T) & 0 \\ 0 & M_{B_2}(T) \end{array} \right] \nonumber \]
where \(M_{B_{1}}(T)\) and \(M_{B_{2}}(T)\) are the matrices of the restrictions of \(T\) to \(U_{1}\) and to \(U_{2}\) respectively.
The linear operator \(T : V \to V\) is said to be reducible if nonzero \(T\)-invariant subspaces \(U_{1}\) and \(U_{2}\) can be found such that \(V = U_{1} \oplus U_{2}\).
Then \(T\) has a matrix in block diagonal form as in Theorem 9.3.7, and the study of \(T\) is reduced to studying its restrictions to the lower-dimensional spaces \(U_{1}\) and \(U_{2}\). If these can be determined, so can \(T\). Here is an example in which the action of \(T\) on the invariant subspaces \(U_{1}\) and \(U_{2}\) is very simple indeed. The result for operators is used to derive the corresponding similarity theorem for matrices.
Let \(T : V \to V\) be a linear operator satisfying \(T^{2} = 1_{V}\) (such operators are called involutions). Define
\[U_1 = \{ \mathbf{v} \mid T(\mathbf{v}) = \mathbf{v} \} \quad \mbox{ and } \quad U_2 = \{ \mathbf{v} \mid T(\mathbf{v}) = -\mathbf{v} \} \nonumber \]
- Show that \(V = U_{1} \oplus U_{2}\).
- If \(dim \;V = n\), find a basis \(B\) of \(V\) such that \(M_B(T) = \left[ \begin{array}{cc} I_k & 0 \\ 0 & -I_{n-k} \end{array} \right]\) for some \(k\).
- Conclude that, if \(A\) is an \(n \times n\) matrix such that \(A^{2} = I\), then \(A\) is similar to \(\left[ \begin{array}{cc} I_k & 0 \\ 0 & -I_{n-k} \end{array} \right]\) for some \(k\).
Solution a
Then \(\mathbf{v} + T(\mathbf{v})\) lies in \(U_{1}\), because \(T[\mathbf{v} + T(\mathbf{v})] = T(\mathbf{v}) + T^{2}(\mathbf{v}) = \mathbf{v} + T(\mathbf{v})\). Similarly, \(\mathbf{v} - T(\mathbf{v})\) lies in \(U_{2}\), and it follows that \(V = U_{1} + U_{2}\). This proves part (a).
Solution b
A similar argument shows that \(M_{B_2}(T) = -I_{n-k}\), so part (b) follows with \(B = \{\mathbf{b}_{1}, \mathbf{b}_{2}, \dots, \mathbf{b}_{n}\}\).
Solution c
But Theorem 9.2.4 shows that \(M_{B}(T_{A}) = P^{-1}AP\) for some invertible matrix \(P\), and this proves part (c).
Note that the passage from the result for operators to the analogous result for matrices is routine and can be carried out in any situation, as in the verification of part (c) of Example 9.3.10. The key is the analysis of the operators. In this case, the involutions are just the operators satisfying \(T^{2} = 1_{V}\), and the simplicity of this condition means that the invariant subspaces \(U_{1}\) and \(U_{2}\) are easy to find.
Unfortunately, not every linear operator \(T : V \to V\) is reducible. In fact, the linear operator in Example 9.3.4 has no invariant subspaces except \(0\) and \(V\). On the other hand, one might expect that this is the only type of nonreducible operator; that is, if the operator has an invariant subspace that is not \(0\) or \(V\), then some invariant complement must exist. The next example shows that even this is not valid.
Consider the operator \(T : \mathbb{R}^2 \to \mathbb{R}^2\) given by \(T\left[ \begin{array}{c} a \\ b \end{array} \right] = \left[ \begin{array}{c} a + b \\ b \end{array} \right]\). Show that \(U_1 = \mathbb{R}\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]\) is \(T\)-invariant but that \(U_1\) has not \(T\)-invariant complement in \(\mathbb{R}^2\).
Solution
Because \(U_1 = span \;\left\{\left[ \begin{array}{c} 1 \\ 0 \end{array} \right] \right\}\) and \(T\left[ \begin{array}{c} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 1 \\ 0 \end{array} \right]\), it follows (by Example 9.3.3) that \(U_{1}\) is \(T\)-invariant. Now assume, if possible, that \(U_{1}\) has a \(T\)-invariant complement \(U_{2}\) in \(\mathbb{R}^2\). Then \(U_{1} \oplus U_{2} = \mathbb{R}^2\) and \(T(U_{2}) \subseteq U_{2}\). Theorem 9.3.6 gives
\[2 = dim \; \mathbb{R}^2 = dim \; U_1 + dim \; U_2 = 1 + dim \; U_2 \nonumber \]
so \(dim \;U_{2} = 1\). Let \(U_{2} = \mathbb{R}\mathbf{u}_{2}\), and write \(\mathbf{u}_2 = \left[ \begin{array}{c} p \\ q \end{array} \right]\). We claim that \(\mathbf{u}_{2}\) is not in \(U_{1}\). For if \(\mathbf{u}_{2} \in U_{1}\), then \(\mathbf{u}_{2} \in U_{1} \cap U_{2} = \{\mathbf{0}\}\), so \(\mathbf{u}_{2} = \mathbf{0}\). But then \(U_{2} = \mathbb{R}\mathbf{u}_{2} = \{\mathbf{0}\}\), a contradiction, as \(dim \;U_{2} = 1\). So \(\mathbf{u}_{2} \notin U_{1}\), from which \(q \neq 0\). On the other hand, \(T(\mathbf{u}_{2}) \in U_{2} = \mathbb{R}\mathbf{u}_{2}\) (because \(U_{2}\) is \(T\)-invariant), say \(T(\mathbf{u}_2) = \lambda\mathbf{u}_2 = \lambda\left[ \begin{array}{c} p \\ q \end{array} \right]\).
Thus
\[\left[ \begin{array}{c} p + q \\ q \end{array} \right] = T\left[ \begin{array}{c} p \\ q \end{array} \right] = \lambda\left[ \begin{array}{c} p \\ q \end{array} \right] \mbox{ where } \lambda \in \mathbb{R} \nonumber \]
Hence \(p + q = \lambda p\) and \(q = \lambda q\). Because \(q \neq 0\), the second of these equations implies that \(\lambda = 1\), so the first equation implies \(q = 0\), a contradiction. So a \(T\)-invariant complement of \(U_{1}\) does not exist.
This is as far as we take the theory here, but in Chapter 11 the techniques introduced in this section will be refined to show that every matrix is similar to a very nice matrix indeed—its Jordan canonical form.
This page titled 9.3: Invariant Subspaces and Direct Sums was last modified on Sat, 17 Jan 2026 00:22:27 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by W. Keith Nicholson via source content that was edited to the style and standards of the LibreTexts platform.
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