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 2 · pp. 32–40

Chapter 2: Direct Sums of Subspaces

This chapter shows how a linear map $f:V\to V$ can be broken into simpler pieces. First we study nilpotent maps, which are the simplest non-trivial case. Then we use them to put any map with all its eigenvalues in the field into triangular form. Finally we split such a map into a diagonalizable part plus a nilpotent part (the Jordan Decomposition).

Nilpotent maps

A linear map $f:V\to V$ is nilpotent if $f^m=0$ for some integer $m\ge 1$ . Applying $f$ often enough sends every vector to $0_V$ . Differentiation $D$ on real polynomials of degree at most $n$ is an example. Each application lowers the degree by one, so $D^{n+1}=0$ . (p. 33)

The index of a nilpotent map is the least positive integer $k$ with $f^k=0$ . For $D$ on polynomials of degree at most 3, $D^3\neq 0$ but $D^4=0$ , so the index is 4. (p. 41)

A special basis. If $f$ is nilpotent on a non-zero finite-dimensional space $V$ , there is a basis $\{v_1,\dots,v_n\}$ with $f(v_1)=0$ and $f(v_j)\in\langle v_1,\dots,v_{j-1}\rangle$ for $j\ge 2$ . Each basis vector is sent into the span of the earlier ones. For instance, $f(v_3)=av_1+bv_2$ for some scalars $a,b$ . (p. 33)

The basis is built one vector at a time:

Let $k$ be the index. Pick $v$ with $f^{k-1}(v)\neq 0$ and set $v_1=f^{k-1}(v)$ . Then $f(v_1)=0$ .
Let $W$ be the span of the vectors chosen so far. If $\operatorname{Im} f\subseteq W$ , any vector outside $W$ can be added.
Otherwise choose $i$ with $\operatorname{Im} f^i\not\subseteq W$ but $\operatorname{Im} f^{i+1}\subseteq W$ . Take the new vector from $\operatorname{Im} f^i\setminus W$ .

Such an $i$ exists because of the chain $\{0\}=\operatorname{Im} f^k\subseteq\cdots\subseteq\operatorname{Im} f$ . The new vector lies outside $W$ , so it is independent of $W$ , and $f$ sends it into $W$ . (p. 33)

Matrix form. In this basis column $j$ has non-zero entries only in rows above $j$ , and column 1 is zero. So the matrix is upper triangular with every diagonal entry 0. Its entries $a_{ij}$ can be non-zero only when $i<j$ . (p. 34)

Triangular form

Let $\lambda_i$ be an eigenvalue of $f$ and let $V_i=\operatorname{Ker}(f-\lambda_i\,\mathrm{id})^{m_i}$ be its primary component. The induced map is $f_i=g_i+\lambda_i\,\mathrm{id}$ , where $g_i=f_i-\lambda_i\,\mathrm{id}$ is nilpotent. Choose a basis in which $g_i$ has zeros on the diagonal and entries only above it. Adding $\lambda_i$ times the identity matrix puts $\lambda_i$ on the whole diagonal and leaves the entries above it unchanged. (p. 34)

Triangular Form Theorem. Suppose $\chi_f=\prod (X-\lambda_i)^{d_i}$ and $m_f=\prod (X-\lambda_i)^{e_i}$ with distinct $\lambda_i\in F$ . Then $V$ has an ordered basis in which the matrix of $f$ is block diagonal. The $i$ -th block is a $d_i\times d_i$ upper triangular matrix with $\lambda_i$ on its diagonal. Here $d_i=\dim V_i$ , where $V_i=\operatorname{Ker}(f-\lambda_i\,\mathrm{id})^{e_i}$ , and $e_i\le d_i$ . Joining the triangular bases of the $V_i$ gives a basis of $V$ . If $\chi_f=(X+1)(X-1)^2$ , the blocks have sizes 1 and 2, and the diagonal reads $-1$ , then $1,1$ . (pp. 34–35)

Consequence over $\mathbb{C}$ . Every complex polynomial splits into linear factors, so all eigenvalues lie in $\mathbb{C}$ . Hence every square complex matrix is similar to an upper triangular matrix. (p. 35)

Common mistake. It is false that every real square matrix is similar to a real upper triangular matrix. Triangularizing needs all eigenvalues in the field. A rotation of the plane through $90^\circ$ has $\chi=X^2+1$ , which has no real roots, so it is not similar to a real triangular matrix. Over $\mathbb{C}$ it is. (p. 35)

Triangularizing in practice

One eigenvalue. Suppose $\chi_f=m_f=(X-\lambda)^3$ on a 3-dimensional space. Write $h=f-\lambda\,\mathrm{id}$ and choose:

$v_1\in\operatorname{Ker}h$ with $v_1\neq 0$ ;
$v_2$ independent of $v_1$ with $h(v_2)\in\langle v_1\rangle$ ;
$v_3$ independent of $v_1,v_2$ with $h(v_3)\in\langle v_1,v_2\rangle$ .

Each condition is a small linear system. Put the basis vectors as the columns of $P$ . Then $P^{-1}AP$ is upper triangular with $\lambda$ on the diagonal. In the chapter's worked example $\lambda=2$ and the diagonal is $2,2,2$ . (pp. 35–36)

Several eigenvalues. First split $V$ into the subspaces $\operatorname{Ker}(f-\lambda_i\,\mathrm{id})^{e_i}$ . Then triangularize $f$ on each one and join the bases. In the example with $\chi_f=(X+1)(X-1)^2$ , $\mathbb{R}^3=\operatorname{Ker}(f+\mathrm{id})\oplus\operatorname{Ker}(f-\mathrm{id})^2$ , with dimensions 1 and 2. A basis of the first part is a single eigenvector. A basis of the second part comes from the nilpotent-basis method applied to $f-\mathrm{id}$ . The joined basis gives a matrix with diagonal $-1,1,1$ . (pp. 36–37)

Jordan Decomposition

Local splitting. Each induced map $f_i$ is the sum of $\lambda_i\,\mathrm{id}$ and $f_i-\lambda_i\,\mathrm{id}$ . The first is diagonalizable, since its minimum polynomial is $X-\lambda_i$ . The second is nilpotent by the definition of the component. (p. 37)

Theorem. If all eigenvalues of $f:V\to V$ lie in $F$ , then $f=\delta+\eta$ with $\delta$ diagonalizable, $\eta$ nilpotent and $\delta\circ\eta=\eta\circ\delta$ . The commuting is what makes the result useful. The pair is also unique. (p. 38)

Construction. Let $p_i$ be the projection onto $V_i$ parallel to the other components. Define $\delta=\sum_i \lambda_i p_i,\qquad \eta=f-\delta.$ On $V_i$ , $\delta$ is multiplication by $\lambda_i$ . So a basis of $V$ made of eigenvectors of $\delta$ exists, and $\delta$ is diagonalizable. On $V_i$ , $\eta=f-\lambda_i\,\mathrm{id}$ , and $(f-\lambda_i\,\mathrm{id})^{e_i}$ kills $V_i$ , so $\eta^r=0$ for large $r$ . Each $V_i$ is $f$ -invariant, so each $p_i$ commutes with $f$ . Hence $\delta f=f\delta$ , and so $\delta\eta=\eta\delta$ . (pp. 38–39)

Polynomials in $f$ . Each projection is a polynomial $p_i=t_i(f)$ . So $\delta=p(f)$ with $p=\sum\lambda_i t_i$ , and $\eta=f-p(f)$ . Both therefore commute with $f$ and with any map that commutes with $f$ . (p. 39)

Uniqueness. Suppose $f=\delta+\eta=\delta'+\eta'$ with each pair commuting. The primed maps commute with $f$ , so they commute with $\delta$ and $\eta$ . Then $\delta-\delta'=\eta'-\eta$ . The right side is nilpotent by the binomial theorem. The left side is diagonal in a common eigenbasis. A matrix that is both diagonal and nilpotent is 0. So $\delta=\delta'$ and $\eta=\eta'$ . (p. 39)

Computing it for a matrix. Triangularize to $T=P^{-1}AP$ . Write $T=D+N$ , where $D$ is the diagonal part and $N$ is the rest. In the chapter's example the diagonal is $-1,1,1$ , and $N$ has one non-zero entry above the diagonal, inside the $2\times 2$ block for eigenvalue 1. $D$ and $N$ commute, so this is the Jordan decomposition of $T$ . Convert back with $PDP^{-1}$ and $PNP^{-1}$ . (pp. 39–40)

Powers. Since $\delta$ and $\eta$ commute, the binomial theorem applies: $f^n=(\delta+\eta)^n=\sum_{j=0}^{n}\binom{n}{j}\delta^{\,n-j}\eta^{\,j}.$ Powers of $\delta$ come from powers of a diagonal matrix. Also $\eta^j=0$ for large $j$ , so few terms survive. If $\eta^2=0$ , then $f^n=\delta^n+n\delta^{n-1}\eta$ . (p. 40)

Kernels of powers of a nilpotent map

For any linear $f$ and positive integer $i$ , $\operatorname{Ker}f^i\subseteq\operatorname{Ker}f^{i+1}$ . If $f^i(x)=0$ , then $f^{i+1}(x)=f(f^i(x))=0$ . Also, if $x\in\operatorname{Ker}f^{i+1}$ , then $f^i(f(x))=0$ , so $f(x)\in\operatorname{Ker}f^i$ . (p. 41)

Now let $f$ be nilpotent of index $k$ on a non-zero space $V$ . The kernels form a strictly increasing chain: $\{0\}\subset\operatorname{Ker}f\subset\operatorname{Ker}f^2\subset\cdots\subset\operatorname{Ker}f^{k-1}\subset\operatorname{Ker}f^k=V.$ The first step is strict because $f$ is not invertible, so $\operatorname{Ker}f\neq\{0\}$ . Suppose two consecutive kernels were equal. Then every later kernel would equal them as well. In particular $\operatorname{Ker}f^{k-1}=V$ , so $f^{k-1}=0$ . This contradicts the index being $k$ . (p. 41)

Check yourself

Why is differentiation on real polynomials of degree at most 4 nilpotent, and what is its index?
In a basis built for a nilpotent map, what shape does the matrix have, and why are the diagonal entries all zero?
Why is a $90^\circ$ rotation of the real plane not similar to a real upper triangular matrix, while every complex square matrix is similar to a triangular one?
If $\chi_f=(X+1)(X-1)^2$ , what are the block sizes and the diagonal entries in the triangular form?
Why do the diagonalizable and nilpotent parts of the Jordan decomposition have to commute, and how does this help in computing $f^n$ ?