Linear Algebra · Book or PDF

Essential Student Algebra: Volume Four, Linear Algebra

T. S. Blyth & E. F. Robertson · First-year university · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 4 · pp. 54–67

Chapter 4: Reduction to Jordan Form

A linear map may have eigenvalues that do not lie in the ground field, so the Jordan form is not always available. This chapter builds a canonical matrix that works over any field and reduces to the Jordan form when all eigenvalues are in $F$ . The main tools are quotient spaces, cyclic subspaces, and the Cyclic Decomposition Theorem.

Why Jordan form can fail

The minimum polynomial of a linear map is a product of powers of irreducible polynomials. Those irreducible factors need not be linear, so the eigenvalues may lie outside the ground field (p. 54). Over $\mathbb{R}$ , the factor $X^2+1$ is irreducible but has no real root. A map with this factor has no Jordan form over $\mathbb{R}$ (p. 54). We therefore need a canonical matrix that does not depend on roots being present.

Quotient spaces

Let $W$ be a subspace of $V$ . The quotient space $V/W$ is the set of cosets $x+W$ . Addition is $(x+W)+(y+W)=(x+y)+W$ and scalar multiplication is $\lambda(x+W)=\lambda x+W$ (p. 54). Two vectors give the same coset exactly when their difference lies in $W$ . With these operations $V/W$ is a vector space over $F$ . Example: take $V=\mathbb{R}^3$ and $W$ the $xy$ -plane. Each coset is a plane parallel to $W$ , and the $z$ -coordinate identifies it (p. 54).

Suppose $V$ is finite-dimensional, $\{v_1,\dots,v_m\}$ is a basis of $W$ , and $\{x_1+W,\dots,x_k+W\}$ is a basis of $V/W$ . Then $\{v_1,\dots,v_m,x_1,\dots,x_k\}$ is a basis of $V$ (p. 55). The map $x\mapsto x+W$ is linear, so independent cosets give independent representatives. For spanning, take any $x\in V$ and write $x+W=\sum\lambda_i(x_i+W)$ . Then $x-\sum\lambda_i x_i$ lies in $W$ , so it is a combination of the $v_j$ (p. 55).

Counting basis vectors gives

$\dim V=\dim W+\dim V/W.$

For example, if $V=\mathbb{R}^5$ and $\dim W=2$ , then $\dim V/W=3$ (p. 55).

If $V=W\oplus Z$ , then $Z$ is isomorphic to $V/W$ . Both have dimension $\dim V-\dim W$ , and finite-dimensional spaces over $F$ of equal dimension are isomorphic (p. 55).

The induced map on a quotient

Let $W$ be $f$ -invariant. The induced map $f':V/W\to V/W$ is defined by $f'(x+W)=f(x)+W$ (pp. 56–57). It is well defined: if $x+W=y+W$ , then $x-y\in W$ , so $f(x)-f(y)=f(x-y)\in W$ by invariance. It is linear because $f$ is. Without invariance, different representatives of one coset could give different answers (pp. 56–57).

The minimum polynomial of $f'$ divides that of $f$ . By induction $(f^n)'=(f')^n$ , so $[p(f)]'=p(f')$ for every polynomial $p$ . Taking $p=m_f$ gives $m_f(f')=0$ , hence $m_{f'}\mid m_f$ (p. 56).

Cyclic subspaces and annihilators

For a non-zero $x\in V$ , the $f$ -cyclic subspace $Z_x$ is the set of all vectors $p(f)(x)$ with $p\in F[X]$ . It is an $f$ -invariant subspace (p. 57). Example: let $f(x,y,z)=(-y+2z,\,x+z,\,2z)$ and $x=(1,0,0)$ . Then $f(x)=(0,1,0)$ and $f^2(x)=-x$ , so $Z_x=\{(a,b,0)\}$ (p. 57).

The $f$ -annihilator $m_x$ of a vector $x$ is the monic polynomial of least degree with $m_x(f)(x)=0$ . It comes from the least $k$ for which $f^k(x)$ is a combination of $x,f(x),\dots,f^{k-1}(x)$ . In the example above, $f^2(x)=-x$ gives $m_x=X^2+1$ (p. 57).

If $m_x=a_0+a_1X+\dots+a_{k-1}X^{k-1}+X^k$ , then $\{x,f(x),\dots,f^{k-1}(x)\}$ is a basis of $Z_x$ , so $\dim Z_x=\deg m_x$ (p. 58). These vectors are independent by minimality of $k$ . Every higher power is a combination of them, by induction using $f^k(x)=-a_0x-\dots-a_{k-1}f^{k-1}(x)$ (p. 58).

Companion matrices

Relative to the basis $\{x,f(x),\dots,f^{k-1}(x)\}$ , the induced map $f_x$ on $Z_x$ has the companion matrix of $m_x$ . It has ones on the subdiagonal, last column $(-a_0,-a_1,\dots,-a_{k-1})^T$ , and zeros elsewhere (pp. 58–59). Each basis vector except the last goes to the next one, which gives the subdiagonal. The last vector goes to $f^k(x)$ , which gives the last column. For $m_x=X^2+1$ the matrix is $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ (pp. 58–59).

The minimum polynomial of $f_x$ equals $m_x$ . It divides $m_x$ , and its degree is at least $k$ , since otherwise $f^k(x)$ would be a combination of earlier vectors. Because $f_x$ is a restriction of $f$ , its minimum polynomial divides that of $f$ . So the annihilator of any vector divides $m_f$ . The same argument works in a quotient $V/W$ , using $m_{f'}\mid m_f$ (pp. 58–59).

A cyclic vector is a vector $x$ whose cyclic basis $\{x,f(x),\dots,f^{k-1}(x)\}$ spans $Z_x$ . A subspace with such a basis is a cyclic subspace. On it, $f$ has the companion matrix of the annihilator of $x$ . Cyclic subspaces are the building blocks of the rational form (p. 59).

The Cyclic Decomposition Theorem

Suppose $f$ has minimum polynomial $p^t$ with $p$ irreducible. Then $V=Z_{z_1}\oplus\dots\oplus Z_{z_k}$ , where $z_i$ has annihilator $p^{n_i}$ and each $n_i\le t$ (pp. 59–61). So the matrix of $f$ is block diagonal, with each block the companion matrix of $p^{n_i}$ . Example: for $p=X^2+1$ , $\dim V=4$ and $t=1$ , $V$ splits into two 2-dimensional cyclic pieces (pp. 59–61).

The proof is by induction on $\dim V$ (pp. 60–61):

Pick $z_1$ with $p^{t-1}(f)(z_1)\ne0$ . Its annihilator is $p^t$ , and $W=Z_{z_1}$ is $f$ -invariant.
Apply the inductive hypothesis to the induced map $f'$ on $V/W$ , which has smaller dimension.
For each generator $y_i+W$ with annihilator $p^{n_i}$ , replace $y_i$ by $z_i=y_i-q(f)(z_1)$ . This has annihilator exactly $p^{n_i}$ .
Combine the bases using the basis result for quotient spaces.

Since $\dim Z_{z_i}=n_i\deg p$ , we get

$\dim V=(n_1+\dots+n_k)\deg p.$

For $p=X^2+1$ and $\dim V=4$ , this means $n_1+\dots+n_k=2$ (p. 61).

Uniqueness and elementary divisors

The integers $t=n_1\ge n_2\ge\dots\ge n_k\ge1$ are determined by $f$ (p. 62). Let $d=\deg p$ . The image of $Z_{z_i}$ under $p(f)^j$ has dimension $d(n_i-j)$ if $j<n_i$ , and $0$ otherwise. Summing,

$\dim\operatorname{Im}p(f)^{j-1}-\dim\operatorname{Im}p(f)^j=d\cdot\#\{i:n_i\ge j\}.$

The left side depends only on $f$ , so the number of $n_i\ge j$ is fixed for every $j$ . That fixes the whole sequence (p. 62).

The polynomials $p^{n_1},\dots,p^{n_k}$ are the elementary divisors of $f$ . The first, $p^{n_1}=p^t$ , is the minimum polynomial. Repeats are allowed, and a repeated divisor means its companion matrix appears more than once (p. 63).

The rational canonical matrix

To build it (p. 63):

Split $V$ by the Primary Decomposition Theorem into kernels $V_i=\operatorname{Ker}p_i(f)^{m_i}$ .
Apply cyclic decomposition to each $V_i$ .
Assemble the companion matrices of all elementary divisors into one block diagonal matrix.

Each $V_i$ gives a block $A_i$ , itself block diagonal in companion matrices $C_{ij}$ of the divisors $p_i^{n_{ij}}$ . The result is unique up to the order of the $A_i$ . Each $A_i$ has size $d_i\deg p_i$ , where $d_i$ is the multiplicity of $p_i$ in the characteristic polynomial (p. 63).

Example: if $f:\mathbb{R}^4\to\mathbb{R}^4$ has minimum polynomial $X^2+1$ , the characteristic polynomial is $(X^2+1)^2$ . Then $n_1+\dots+n_k=2$ with $n_1=1$ and all $n_i\ge1$ forces $k=2$ and $n_1=n_2=1$ . The form is $C_{X^2+1}\oplus C_{X^2+1}$ (p. 64).

Two common mistakes

Mistake 1: the characteristic and minimum polynomials determine the form. They do not. Let $f$ on $\mathbb{R}^6$ have minimum polynomial $(X^2+1)(X-2)^2$ and characteristic polynomial $(X^2+1)(X-2)^4$ . The $X-2$ part has dimension 4 with $n_1=2$ . Both $n=(2,2)$ and $n=(2,1,1)$ are possible, giving two different rational forms (p. 64).

Mistake 2: the rational form equals the Jordan form when both exist. Take characteristic and minimum polynomial $(X-2)^3$ . The rational form is the single companion matrix of $X^3-6X^2+12X-8$ , with last column $(8,-12,6)^T$ . This looks nothing like a Jordan block with $2$ on the diagonal (p. 65).

The classical canonical matrix

Suppose $f$ has minimum polynomial $p^n$ with $\deg p=k$ , and $z$ is a cyclic vector. The $kn$ vectors $p(f)^jf^i(z)$ , for $0\le i<k$ and $0\le j<n$ , form a basis of $V$ (pp. 65–66). They are independent because a dependence would give a polynomial of degree less than $kn$ that kills $z$ , contradicting minimality of $p^n$ . Relative to this basis, the matrix of $f$ is block diagonal with $n$ copies of $C_p$ on the diagonal. The blocks $M$ sit just above them, where $M$ is the $k\times k$ matrix with a single $1$ in its bottom-left corner. This $1$ arises because, for the first vector in a row, $f$ gives a combination of that row's vectors plus a vector from the previous row (pp. 65–66).

The classical canonical matrix is obtained from the rational one by replacing each group of companion matrices of $p_i^{n_{ij}}$ with the classical $p_i$ -matrix. Each such block uses the companion matrix of $p_i$ rather than of $p_i^n$ . For $p=X^2+1$ this gives $2\times2$ blocks $C_p$ linked by connecting $1$ -entries (pp. 66–67).

When $p=X-a$ , we have $k=1$ , so $C_p=[a]$ and $M=[1]$ . The classical $p$ -matrix is the $n\times n$ Jordan block with $a$ on the diagonal and $1$ on the superdiagonal. This is the case where $f-a\,\mathrm{id}_V$ is nilpotent of index $n$ . So the classical form reduces to the Jordan form whenever all eigenvalues lie in the ground field (p. 67).

Check yourself

Why can a map on $\mathbb{R}^2$ with minimum polynomial $X^2+1$ not have a Jordan form over $\mathbb{R}$ ?
If $\dim V=7$ and $\dim W=3$ , what is $\dim V/W$ , and why?
Why must $W$ be $f$ -invariant for the induced map $f'$ on $V/W$ to be well defined?
If $f^2(x)=-x$ and $f(x)\ne\lambda x$ for every scalar $\lambda$ , what are $m_x$ , $\dim Z_x$ and the companion matrix?
A map on $\mathbb{R}^6$ has minimum polynomial $(X^2+1)(X-2)^2$ and characteristic polynomial $(X^2+1)(X-2)^4$ . List the possible sequences $n_i$ for the $X-2$ part.