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 1 · pp. 17–31

Chapter 1: The Minimum Polynomial

This chapter shows how to break a vector space into smaller pieces that a linear map $f$ treats separately. The tools are direct sums, projections and invariant subspaces. The minimum polynomial of $f$ controls the breakdown. It tells us when $f$ is diagonalizable and how to put $f$ in block diagonal form.

Sums and direct sums

The sum of subspaces $A_1,\dots,A_n$ of $V$ is the smallest subspace containing all of them. It is the set of all vectors $a_1+\dots+a_n$ with $a_i\in A_i$ . A linear combination of vectors from the union can be regrouped into one term per subspace, because a scalar multiple of a vector in $A_i$ stays in $A_i$ . In $\mathbb{R}^2$ , the x-axis plus the y-axis is all of $\mathbb{R}^2$ , since $(x,y)=(x,0)+(0,y)$ . (p. 17)

A sum is direct, written $\bigoplus A_i$ , when every vector in it can be written in exactly one way as $a_1+\dots+a_n$ . A sum always gives at least one way, and directness adds uniqueness. In $\mathbb{R}^2$ take the x-axis $X$ and the line $D=\{(x,x)\}$ . Then $(x,y)=(x-y,0)+(y,y)$ is the only decomposition, so $\mathbb{R}^2=X\oplus D$ . (p. 18)

A sum can equal the whole space without being direct. Take the planes $A: x+y=0$ and $B: x-y+z=0$ in $\mathbb{R}^3$ . They are different planes, so $A+B=\mathbb{R}^3$ . They share the line through $(-1,1,2)$ . That vector can be written as itself plus $0$ , or as $0$ plus itself, so decompositions are not unique. Overlap in more than $\{0\}$ always lets you shift a common vector from one part to the other. (p. 18)

These statements are equivalent for subspaces $A_1,\dots,A_n$ :

The sum is direct.
If $a_1+\dots+a_n=0$ with $a_i\in A_i$ , then every $a_i=0$ .
Each $A_i$ meets the sum of the other $A_j$ only in $\{0\}$ .

From (1) to (2): zero already has the decomposition $0+\dots+0$ , so uniqueness forces it to be the only one. From (2) to (3): a vector in $A_i$ and in the sum of the others gives a relation $a_i-\sum_{j\ne i}a_j=0$ , so it is $0$ . From (3) to (1): subtract two decompositions of one vector. Each difference $a_i-b_i$ lies in $A_i$ and in the sum of the others, so it is $0$ . (pp. 18–19)

For two subspaces this becomes a quick test: $V=A\oplus B$ if and only if $V=A+B$ and $A\cap B=\{0\}$ . (p. 19)

Standard splittings

Even and odd functions. The space of functions $\mathbb{R}\to\mathbb{R}$ is the direct sum of the even and the odd functions. Write $f=f^++f^-$ with $f^+(x)=\tfrac12[f(x)+f(-x)]$ (even) and $f^-(x)=\tfrac12[f(x)-f(-x)]$ (odd). A function that is both satisfies $f(x)=-f(x)$ , so it is $0$ . For example, $e^x=\cosh x+\sinh x$ . (p. 19)
Symmetric and skew-symmetric matrices. Every square real matrix $X$ equals $Y+Z$ with $Y=\tfrac12(X+X^T)$ symmetric and $Z=\tfrac12(X-X^T)$ skew-symmetric. A matrix that is both equals its own negative, so it is zero. For the matrix with rows $(1,2)$ and $(0,3)$ , the symmetric part has rows $(1,1)$ and $(1,3)$ . The skew part has rows $(0,1)$ and $(-1,0)$ . (p. 19)

Dimension of a direct sum

If $V=V_1\oplus\dots\oplus V_n$ is finite-dimensional and $B_i$ is a basis of $V_i$ , then the union of the $B_i$ is a basis of $V$ . So $\dim V=\sum\dim V_i$ . The union spans $V$ because every vector is a sum of vectors from the $V_i$ . It is independent because a zero combination splits into pieces $z_i\in V_i$ that sum to $0$ . Directness forces each $z_i=0$ , and then all coefficients are $0$ . In $\mathbb{R}^2$ , one basis vector from each axis gives dimension $1+1=2$ . (pp. 19–20)

Projections

If $V=A\oplus B$ , the projection on $A$ parallel to $B$ is the map $p$ sending $x=a+b$ to $a$ . Uniqueness of the decomposition makes $p$ well defined and linear. In $\mathbb{R}^2=X\oplus D$ above, $p(x,y)=(x-y,0)$ . It slides a point along a line parallel to $D$ until it reaches the x-axis. (p. 20)

A map $f:V\to V$ is idempotent if $f\circ f=f$ . It is a projection if it is the projection on some $A$ parallel to some $B$ with $V=A\oplus B$ . These two ideas turn out to be the same. (p. 20)

For a projection $p$ on $A$ parallel to $B$ , we have $A=\operatorname{Im}p=\{x: p(x)=x\}$ , $B=\operatorname{Ker}p$ , and $p^2=p$ . Vectors of $A$ decompose as $a+0$ , so $p$ fixes them. A vector goes to $0$ exactly when its $A$ -part is $0$ , which means it lies in $B$ . Since $p(x)\in A$ and $p$ fixes $A$ , applying $p$ twice changes nothing. (p. 21)

Conversely, every idempotent $f$ is a projection, and $V=\operatorname{Im}f\oplus\operatorname{Ker}f$ . If $z=f(y)$ is also in $\operatorname{Ker}f$ , then $z=f(f(y))=f(z)=0$ , so the intersection is trivial. Every $x$ splits as $f(x)+(x-f(x))$ . The second part lies in the kernel because $f(x-f(x))=f(x)-f^2(x)=0$ . (p. 21)

If $f$ is a projection, so is $\mathrm{id}_V-f$ . Indeed $(\mathrm{id}-f)^2=\mathrm{id}-2f+f^2=\mathrm{id}-f$ . Also $\operatorname{Im}f=\operatorname{Ker}(\mathrm{id}_V-f)$ , because $x=f(x)$ means $(\mathrm{id}-f)(x)=0$ . The two projections use the two summands in swapped roles. (p. 22)

For several summands, $V=V_1\oplus\dots\oplus V_n$ with non-zero $V_i$ holds if and only if there are non-zero linear maps $p_1,\dots,p_n$ with $\sum p_i=\mathrm{id}_V$ and $p_ip_j=0$ for $i\ne j$ . Then each $p_i$ is a projection with $V_i=\operatorname{Im}p_i$ . Given a direct sum, let $p_i$ project onto $V_i$ along the other summands. Then $p_ip_j=0$ because $p_j$ lands in $V_j$ , which $p_i$ kills. Conversely, $p_i=p_i\sum p_j=p_i^2$ shows each $p_i$ is idempotent. Also $x=\sum p_i(x)$ shows the images span $V$ with trivial overlaps. (pp. 22–23)

Invariant subspaces and block form

A subspace $W$ is $f$ -invariant (or $f$ -stable) if $x\in W$ implies $f(x)\in W$ . Then $f$ restricts to a map $W\to W$ . Examples: $\operatorname{Im}f$ and $\operatorname{Ker}f$ are always invariant. Polynomials of degree at most $n$ are invariant under differentiation. (pp. 23–24)

For a non-zero $x$ , the line spanned by $x$ is $f$ -invariant if and only if $x$ is an eigenvector. If $f(x)=\mu x$ , then $f(\lambda x)=\lambda\mu x$ stays in the line. Conversely, invariance forces $f(x)$ to be a multiple of $x$ . (p. 24)

For any polynomial $p$ over $F$ , both $\operatorname{Im}p(f)$ and $\operatorname{Ker}p(f)$ are $f$ -invariant. The reason is that $f$ commutes with $p(f)$ . If $x=p(f)(y)$ , then $f(x)=p(f)(f(y))$ is still in the image. If $p(f)(x)=0$ , then $p(f)(f(x))=f(p(f)(x))=0$ . (p. 24)

Suppose $V=W_1\oplus\dots\oplus W_n$ with each $W_i$ invariant. Join bases of the $W_i$ into one basis. Then the matrix of $f$ is block diagonal. The $i$ -th block $A_i$ is $\dim W_i\times\dim W_i$ and represents the map induced on $W_i$ . Each basis vector of $W_i$ is sent back into $W_i$ , so all entries outside the diagonal blocks are $0$ . With a single invariant $W$ of dimension $r$ , extending its basis gives an $r\times r$ block $A$ in the top left and zeros below it. (pp. 24–25)

Primary decomposition

Primary Decomposition Theorem. Let $\chi_f=p_1^{d_1}\cdots p_k^{d_k}$ and $m_f=p_1^{e_1}\cdots p_k^{e_k}$ with distinct irreducible $p_i$ . Then $V=\bigoplus V_i$ , where $V_i=\operatorname{Ker}p_i(f)^{e_i}$ is $f$ -invariant. The induced map $f_i$ has minimum polynomial $p_i^{e_i}$ and characteristic polynomial $p_i^{d_i}$ . For the proof, set $q_i=m_f/p_i^{e_i}$ . These have no common irreducible factor, so some polynomials $a_i$ satisfy $\sum q_ia_i=1$ . Then $t_i=q_ia_i$ give projections with $\sum t_i(f)=\mathrm{id}$ and $t_i(f)t_j(f)=0$ . One then shows $\operatorname{Im}t_i(f)=\operatorname{Ker}p_i(f)^{e_i}$ . Comparing minimum polynomials and the block diagonal determinant fixes the powers. (pp. 25–27)

The dimension satisfies $\dim V_i=d_i\deg p_i$ , the degree of $p_i^{d_i}$ . For linear factors $X-\lambda_i$ , this is $\dim V_i=d_i$ , the algebraic multiplicity. (p. 28)

When all eigenvalues lie in $F$ , we get $V=\bigoplus_i\operatorname{Ker}(f-\lambda_i\,\mathrm{id})^{e_i}$ with $\dim V_i=d_i$ . For example, if $\chi_f=m_f=(X+1)(X-1)^2$ on $\mathbb{R}^3$ , then $\mathbb{R}^3=\operatorname{Ker}(f+\mathrm{id})\oplus\operatorname{Ker}(f-\mathrm{id})^2$ , with dimensions $1$ and $2$ . (pp. 28–29)

To get block diagonal form:

Factor $\chi_f$ and $m_f$ to find the eigenvalues and the exponents $d_i$ , $e_i$ .
Solve $(f-\lambda_i\,\mathrm{id})^{e_i}(x)=0$ for a basis of each $V_i$ . The size should match $d_i$ .
Join the bases and put them as columns of $P$ . Then $P^{-1}AP$ is block diagonal. (pp. 28–29)

Differential equations. The solutions of $(D-a\,\mathrm{id})^nf=0$ are $e^{ax}$ times a polynomial of degree at most $n-1$ . A basis is $e^{ax},xe^{ax},\dots,x^{n-1}e^{ax}$ . This follows from $(D-a\,\mathrm{id})^nf=e^{ax}D^n(e^{-ax}f)$ , which reduces the problem to $D^n(e^{-ax}f)=0$ . For a general equation with polynomial $\prod(X-a_i)^{n_i}$ , the solution space is the direct sum of these pieces, one per root. (p. 29)

Diagonalizability and commuting maps

A map on a non-zero finite-dimensional space is diagonalizable if and only if $m_f$ is a product of distinct linear factors. If $m_f=\prod(X-\lambda_i)$ with distinct $\lambda_i$ , each $V_i=\operatorname{Ker}(f-\lambda_i\,\mathrm{id})$ consists of eigenvectors, and joining their bases gives an eigenvector basis. Conversely, given an eigenbasis, the product of $(X-\lambda_i)$ over the distinct eigenvalues kills every basis vector, so $m_f$ divides it and equals it. (p. 30)

Common mistake: thinking a repeated eigenvalue prevents diagonalization. What matters is $m_f$ , not $\chi_f$ . With $\chi=(X-6)^2(X-12)$ and $m=(X-6)(X-12)$ , the map is diagonalizable. (p. 30)

Two diagonalizable maps $f,g$ are simultaneously diagonalizable (they share an eigenbasis) if and only if $f\circ g=g\circ f$ . With a shared eigenbasis, both products multiply each basis vector by $\lambda\mu$ . If they commute, $g$ maps each eigenspace $V_i$ of $f$ into itself. An eigenbasis of the induced map on each $V_i$ then consists of common eigenvectors. For matrices: diagonalizable $A,B$ are diagonalized by one invertible $P$ if and only if $AB=BA$ . (p. 31)

Nilpotent maps

A map $f$ is nilpotent if $f^m=0$ for some positive integer $m$ . Example: $f(x,y,z)=(0,x,y)$ on $\mathbb{R}^3$ has $f^2(x,y,z)=(0,0,x)$ and $f^3=0$ . On a complex space, if all eigenvalues are $0$ , then $\chi_f=X^n$ , so $f^n=0$ by Cayley–Hamilton. (p. 32)

On each primary component $V_i=\operatorname{Ker}(f-\lambda_i\,\mathrm{id})^{e_i}$ , the map $f_i-\lambda_i\,\mathrm{id}$ is nilpotent, since its $e_i$ -th power is zero there. So on each component $f$ is $\lambda_i$ times the identity plus a nilpotent piece. This idea leads to triangular form when $m_f$ splits into linear factors. (p. 32)

Check yourself

In $\mathbb{R}^3$ , two distinct planes through the origin add up to $\mathbb{R}^3$ . Why is the sum not direct?
Show that an idempotent map $f$ satisfies $\operatorname{Ker}f\cap\operatorname{Im}f=\{0\}$ .
If $\chi_f=(X-2)^3(X+1)$ and $m_f=(X-2)(X+1)$ , is $f$ diagonalizable? What are the dimensions of the primary components?
Split the function $x^2+x^3$ into its even and odd parts.
Why does $f\circ g=g\circ f$ let us choose an eigenbasis of $f$ made of eigenvectors of $g$ too?