es

Before formulating the Primary Decomposition Theorem, we need to recall some definitions and facts that were explained in other sections. We remind that the minimal polynomial of a square matrix A (or corresponding lineat transformation) is the (unique) monic polynomial ψ(λ) of least degree that annihilates the matrix A, that is ψ(A) = 0. The minimal polynomial \( \psi_u (\lambda ) \) of a vector \( {\bf u} \in V \ \mbox{ or } \ {\bf u} \in \mathbb{R}^n \) relative to A is the monic polynomial of least degree such that \( \psi_u ({\bf A}) {\bf u} = {\bf 0} . \) It follows that \( \psi_u (\lambda ) \) divides the minimal polynomial ψ(λ) of the matrix A. There exists a vector \( {\bf u} \in V (\mathbb{R}^n ) \) such that \( \psi_u (\lambda ) = \psi (\lambda ) . \) This result can be proved by representing the minimal polynomial as the product of simple terms to each of which corresponds a subspace. Then the original vector space (or \( \mathbb{R}^n \) ) is the direct sum of these subspaces.

A subspace U of a vector space V is said to be T-cyclic with respect to a linear transformation T : V ⇾ V if there exists a vector \( {\bf u} \in U \) and a nonnegative integer r such that \( {\bf u}, T\,{\bf u} , \ldots , T^r {\bf u} \) form a basis for U. Thus, for the vector u if the degree of the minimal polynomial \( \psi_u (\lambda ) \) is k, then \( {\bf u}, T\,{\bf u} , \ldots , T^{k-1} {\bf u} \) are linearly independent and the space U is spanned by these k vectors is T-cyclic.

Theorem (Primary Decomposition Theorem): Let V be an n-dimensional vector space (n is finite) and T is a linear transformation on V. Then V is the direct sum of T-cyclic subspaces. ■

Let k be the degree of the minimal polynomial ψ(λ) of transformation T (or corresponding matrix written is specified basis), and let u be a vector in V with \( \psi_u (\lambda ) = \psi (\lambda ) . \) Then the space U spanned by \( {\bf u}, T{\bf u} , \ldots , T^{k-1} {\bf u} \) is T-cyclic. We shall prove that if \( U \ne V \quad (k \ne n), \) then there exists a T-invariant subspace W such that \( V = U\oplus W . \) Clearly, by induction on the dimension, W will then be the direct sum of T-cyclic subspaces and the proof is complete.

To show the existence of W enlarge the basis \( {\bf e}_1 = {\bf u}, {\bf e}_2 = T{\bf u} , \ldots , {\bf e}_k = T^{k-1} {\bf u} \) of U to a basis \( {\bf e}_1 , {\bf e}_2 , \ldots , {\bf e}_k , \ldots , {\bf e}_n \) of V and let \( {\bf e}_1^{\ast} , {\bf e}_2^{\ast} , \ldots , {\bf e}_k^{\ast} , \ldots , {\bf e}_n^{\ast} \) be the dual basis in the dual space. Recall that the dual space consists of all linear forms on V or, equivalently, of all functionals on V. To simplify notation, let z = ek*. Then

\[ \langle {\bf z} , {\bf e}_i \rangle =0 \quad\mbox{for } 1 \le i \le k-1 \quad\mbox{and}\quad \langle {\bf z} , {\bf e}_k \rangle =1 . \]
Consider the dual space U* spanned by \( {\bf z}, T^{\ast} {\bf z} , \ldots , T^{\ast\, k-1} {\bf z} . \) Since ψ(λ) is also the minimal polynomial of T*, the space U* is T*-invariant. Now observe that if \( U^{\ast} \cap U^{\perp} = \{ 0 \} \) and dim U* = k, then \( V^{\ast} = U^{\ast} \oplus U^{\perp} , \) where U* and \( U^{\perp} \) are T*-invariant (since dim \( U^{\perp} \) = n-k). This in turn implies the desired decomposition \( V = U^{\perp\perp} \oplus U^{\ast\perp} = U \oplus W , \) where \( U^{\perp\perp} = U \) and \( U^{\ast\perp} = W \) are T-invariant.

Finally, we shall prove that \( U^{\ast} \cap U^{\perp} = \{ 0 \} \) and dim U* = k simultaneously as follows. Suppose that \( a_0 {\bf z} + a_1 T^{\ast} {\bf z} + \cdots + a_s T^{\ast s} {\bf z} \in U^{\perp} , \) where \( a_s \ne 0 \) and \( 0 \le s \le k-1 . \) Then

\[ T^{\ast k-1-s} \left( a_0 {\bf z} + a_1 T^{\ast} {\bf z} + \cdots + a_s T^{\ast s} {\bf z} \right) = a_0 T^{\ast k-1-s} {\bf z} + a_1 T^{\ast k-s} {\bf z} + \cdots + a_s T^{\ast k-1} {\bf z} \]
is in \( U^{\perp} \) since \( U^{\perp} \) is T*-invariant. Therefore,

\[ \left\langle T^{\ast k-1-s} \left( a_0 {\bf z} + a_1 T^{\ast} {\bf z} + \cdots + a_s T^{\ast s} {\bf z} \right) , {\bf u} \right\rangle =0 . \]
This implies
\[ \left\langle {\bf z} , \left( a_0 T^{k-1-s} + a_1 T^{k-s} + \cdots + a_s T^{k-1} \right) {\bf u} \right\rangle = a_0 \left\langle {\bf z} , {\bf e}_{k-s} \right\rangle + a_1 \left\langle {\bf z} , {\bf e}_{k-s-1} \right\rangle + \cdots + a_s \left\langle {\bf z} , {\bf e}_k \right\rangle = a_s = 0 , \]
which is a contradiction. ■

Example 5: Consider the defective matrix (not diagonalizable)
\[ {\bf A} = \begin{bmatrix} 15&-6&2 \\ 35&-14&5 \\ 7&-3&2 \end{bmatrix} . \]
This matrix has the characteristic polynomial \( \chi (\lambda ) = \left( \lambda -1 \right)^3 \) while its minimal polynomial is \( \psi (\lambda ) = \left( \lambda -1 \right)^2 . \) The matrix A has two linearly independent eigenvectors
\[ {\bf u} = \left[ -1, 0, 7 \right]^{\mathrm T} \qquad\mbox{and} \qquad {\bf v} = \left[ 3, 7, 0 \right]^{\mathrm T} . \]
Let U and V be one-dimensional subspaces generated by spans of vectors u and v, respectively. The minimal polynomials of these vectors are the same: \( \psi_u (\lambda ) = \psi_v (\lambda ) = \lambda -1 \) because
\[ {\bf A}\,{\bf u} = {\bf u} \qquad\mbox{and} \qquad {\bf A}\,{\bf v} = {\bf v} . \]
Each of these one-dimensional subspaces U and V are A-cyclic and they cannot form the direct sum of \( \mathbb{R}^3 . \) We choose a vector \( {\bf z} = \left[ 7, -3, 1 \right]^{\mathrm T} , \) which is perpendicular to each u and v. matrix A transfers z into the vector \( {\bf A}\,{\bf z} = \left[ 125, 192, 60 \right]^{\mathrm T} , \) which is perpendicular to neither u nor v. Next application of A yields
\[ {\bf A}^2 \,{\bf z} = \begin{bmatrix} 243 \\ 587 \\ 119 \end{bmatrix} \qquad\Longrightarrow \qquad \det \begin{bmatrix} 7&125&243 \\ -3&192&587 \\ 1&60&119 \end{bmatrix} \ne 0 . \]
Hence, vectors z, Az, and A2z are linearly independent and \( \mathbb{R}^3 \) is the direct sum of A-cyclic. ■
End of Example 4

 


  1. Beezer, R.A., A First Course in Linear Algebra, 2017.