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 9 · pp. 118–132

Chapter 9: Bilinear and Quadratic Forms

This chapter section studies linear maps on a real inner product space that commute with their adjoints (normal maps). It starts with skew-adjoint maps, the opposite-sign partner of self-adjoint maps. It then shows that every real normal map can be put into a block-diagonal form built from 1×1 and 2×2 blocks, using an orthonormal basis. The special case of orthogonal maps gives rotations.

Skew-adjoint maps and the splitting of a map

A linear map $f$ on a finite-dimensional real inner product space is skew-adjoint when $f^* = -f$ . Self-adjoint maps have $f^*=f$ , so this is the opposite-sign case. In an orthonormal basis, $f$ is skew-adjoint exactly when its matrix satisfies $A^T=-A$ , which is called skew-symmetric. For example, $\begin{pmatrix}0&-2\\2&0\end{pmatrix}$ is skew-symmetric. (p. 118)

Every map $f$ splits in exactly one way as $f=g+h$ with $g$ self-adjoint and $h$ skew-adjoint. Take $g=\tfrac12(f+f^*)$ and $h=\tfrac12(f-f^*)$ . The split is unique because $f=g+h$ forces $f^*=g-h$ , which determines both parts. For matrices this reads $A=\tfrac12(A+A^T)+\tfrac12(A-A^T)$ . (p. 118)

The map $f$ is normal if and only if $g$ and $h$ commute. Expanding $f\circ f^*$ and $f^*\circ f$ shows they are equal exactly when $g\circ h=h\circ g$ . Conversely, if $g$ and $h$ commute, then $f\circ f^*=g^2-h^2=f^*\circ f$ . (p. 118)

There is also a handy test. A map $f$ is skew-adjoint if and only if $(f(x)|x)=0$ for every $x$ . If $f^*=-f$ , then $(f(x)|x)=(x|-f(x))=-(f(x)|x)$ , so the value is 0. For the converse, expand $0=(f(x+y)|x+y)$ to get $(f(x)|y)=(x|-f(y))$ . Uniqueness of adjoints then gives $f^*=-f$ . A 90° rotation of the plane is an example, since it sends each vector to one perpendicular to it. (p. 119)

Why a new canonical form is needed over the reals

On a real inner product space, the maps that have an orthonormal basis of eigenvectors are the self-adjoint ones. On a complex space they are the normal ones. This leaves a question: what do normal maps look like over the reals? Self-adjoint maps are normal, but not the other way round. A 90° rotation of the plane is normal but has no real eigenvectors, so it cannot be diagonalized. We therefore look for a different canonical form, under orthogonal similarity, for real matrices that commute with their transposes. (p. 118)

Irreducible polynomials and minimum polynomials

Over the reals, a monic irreducible polynomial is either linear, $X-a$ , or quadratic, $X^2-2aX+a^2+b^2$ with $b\neq 0$ . The quadratic type has complex roots $a\pm bi$ and no real roots. Examples are $X^2+1$ and $X^2-2X+5$ . These are the only possible factors of a minimum polynomial. The Primary Decomposition Theorem therefore splits the space into pieces where $f$ has one of these two forms. (p. 119)

For a normal map on a non-zero space, the minimum polynomial is a product of distinct irreducibles, so no factor repeats. Suppose a factor $p_i$ had exponent $e_i\ge 2$ . Then $p_i(f)^{e_i-1}$ would send the primary component into both the image and the kernel of the normal map $p_i(f)$ . For a normal map these two sets meet only in 0, so $p_i(f)^{e_i-1}$ would be zero there. That contradicts the fact that the induced map has minimum polynomial $p_i^{e_i}$ . A Jordan block with a repeated factor is therefore never normal. (pp. 119–120)

Structure of skew-adjoint maps

Allowed factors. Every irreducible factor of the minimum polynomial of a skew-adjoint map is $X$ or $X^2+c^2$ with $c\neq 0$ . A linear factor $X-a$ makes $f=a\,\mathrm{id}$ on its component, and skew-adjointness forces $a=0$ . For a quadratic factor $X^2-2aX+a^2+b^2$ , applying skew-adjointness and comparing gives $4af=0$ , so $a=0$ . The factor is then $X^2+b^2$ , as for $\begin{pmatrix}0&-b\\b&0\end{pmatrix}$ . (p. 120)

Possible minimum polynomials. Combining this with distinct factors, the minimum polynomial of a skew-adjoint $f$ is one of three kinds, where the $c_i$ are distinct non-zero reals. (pp. 120–121)

$X$ , if $f=0$ .
$\prod (X^2+c_i^2)$ , if $f$ is invertible.
$X\prod (X^2+c_i^2)$ , if $f$ is neither zero nor invertible.

The map $f$ is invertible exactly when the constant term of its minimum polynomial is non-zero, which rules out the factor $X$ .

Orthogonal components. The primary components of a skew-adjoint map are pairwise orthogonal. Take $x_i$ and $x_j$ in components where $f^2=-c_i^2$ and $f^2=-c_j^2$ , with $c_i^2\ne c_j^2$ . Computing $(f^2(x_i)|x_j)$ in two ways, using $f^*=-f$ , gives $(c_j^2-c_i^2)(x_i|x_j)=0$ , so $(x_i|x_j)=0$ . The component with minimum polynomial $X$ is handled the same way. (p. 121)

Invariant subspaces. A subspace $W$ is $f$ -invariant if and only if $W^\perp$ is $f^*$ -invariant. If $W$ is $f$ -invariant, $y\in W^\perp$ and $x\in W$ , then $(x|f^*(y))=(f(x)|y)=0$ , so $f^*(y)\in W^\perp$ . Applying this again with $W^{\perp\perp}=W$ gives the converse. For skew-adjoint $f$ , we have $f^*=-f$ , so $W$ is $f$ -invariant exactly when $W^\perp$ is. (pp. 121–122)

The case $X^2+b^2$ . Suppose $f$ is skew-adjoint with minimum polynomial $X^2+b^2$ , $b\neq0$ . Then $\dim V$ is even, and $V$ has an orthonormal basis in which $f$ is block diagonal with 2×2 blocks $\begin{pmatrix}0&-b\\b&0\end{pmatrix}$ . The reasoning is as follows. (pp. 122–123)

Pick a non-zero $y$ . The smallest $f$ -invariant subspace containing it has basis $\{y,f(y)\}$ , so it has dimension 2.
Its orthogonal complement is also $f$ -invariant, so repeat the step there until nothing is left. A leftover dimension of 1 is impossible, so the dimension is even.
Since $\|f(y)\|=b\|y\|$ , Gram–Schmidt gives the orthonormal basis $\{y/\|y\|,\ f(y)/(b\|y\|)\}$ for each piece. Joining the bases gives the block form.

General skew-adjoint maps. Every skew-adjoint map has an orthonormal basis in which its matrix is block diagonal. Each block is either the 1×1 zero matrix or a 2×2 block $\begin{pmatrix}0&-b\\b&0\end{pmatrix}$ . Equivalently, a real matrix is skew-symmetric if and only if it is orthogonally similar to such a matrix. The kernel of $f$ gives the zero blocks, and each factor $X^2+c_i^2$ gives 2×2 blocks with $b=c_i$ . A 3×3 skew-symmetric matrix always has a zero block, because odd dimension forces a non-zero kernel. (p. 123)

Structure of normal maps

One quadratic factor. Let $f=g+h$ be normal with irreducible quadratic minimum polynomial $X^2-2aX+a^2+b^2$ , $b\ne0$ . Then $h$ is invertible, $g=a\,\mathrm{id}$ , and $h$ has minimum polynomial $X^2+b^2$ . (p. 124)

If $h$ had a non-zero kernel, $f$ restricted to it would be self-adjoint and so would have real linear factors, contradicting irreducibility.
The identity $f^2-(f^*)^2=4ah$ , together with commutation, gives $(g-a\,\mathrm{id})\circ h=0$ . Since $h$ is invertible, $g=a\,\mathrm{id}$ .
Substituting $f=h+a\,\mathrm{id}$ into the minimum polynomial equation gives $h^2+b^2\mathrm{id}=0$ .

Orthogonal components. The primary components of a normal map on a real space are pairwise orthogonal. The induced maps split compatibly as $f_i=g_i+h_i$ . For two quadratic components, either $b_i^2\ne b_j^2$ , and a computation with $h^2+b_j^2$ forces $(x_i|x_j)=0$ , or $a_i\neq a_j$ , and the same argument with $g$ works. A component for a real linear factor is also orthogonal to the rest, because there $h_0=0$ . (pp. 124–126)

Canonical form. Every normal map on a real inner product space has an orthonormal basis in which its matrix is block diagonal. Each block is 1×1, or 2×2 of the form $\begin{pmatrix}a&-b\\b&a\end{pmatrix}$ with $b\neq0$ . Equivalently, a real matrix is normal if and only if it is orthogonally similar to such a matrix. (pp. 126–127)

On a component with a real linear factor, $f$ is self-adjoint, so an orthonormal basis diagonalizes it. This gives the 1×1 blocks.
On a quadratic component, $f=a\,\mathrm{id}+h$ and $h$ has the 2×2 skew form, so the matrix is $aI+\begin{pmatrix}0&-b\\b&0\end{pmatrix}$ .
The components are orthogonal, so the bases can be joined.

For example, $\begin{pmatrix}1&-2\\2&1\end{pmatrix}$ is normal with $a=1$ and $b=2$ .

Orthogonal maps and rotations

An orthogonal map satisfies $f^{-1}=f^*$ , so it is normal. Its canonical form has diagonal entries $\pm1$ and 2×2 blocks $\begin{pmatrix}a&-b\\b&a\end{pmatrix}$ with $b\neq0$ and $a^2+b^2=1$ . Each 2×2 block must itself be orthogonal, so multiplying it by its transpose gives the identity, which yields $a^2+b^2=1$ . The 1×1 blocks square to 1, so they are $\pm1$ . Writing $a=\cos\theta$ and $b=\sin\theta$ shows each 2×2 block is a plane rotation. (pp. 127–128)

An orthogonal map on $\mathbb{R}^3$ with determinant 1 is called a rotation. In a suitable orthonormal basis its matrix is

$\begin{pmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{pmatrix}.$

In dimension 3 the canonical form has a 1×1 block $\pm1$ and a 2×2 rotation block, or further $\pm1$ entries. Determinant 1 forces the 1×1 entry to be 1. The map is then a rotation about an axis by the angle $\theta$ . For $\theta=\pi/2$ , the plane perpendicular to the axis is turned by a quarter turn. (p. 128)

Check yourself

Split $A=\begin{pmatrix}1&3\\1&2\end{pmatrix}$ into a symmetric part and a skew-symmetric part. Is $A$ normal?
Why does $(f(x)|x)=0$ for all $x$ imply that $f$ is skew-adjoint? Which tool turns this into $f^*=-f$ ?
Why must a skew-adjoint map on a space of odd dimension have a non-zero kernel?
For the normal matrix $\begin{pmatrix}3&-4\\4&3\end{pmatrix}$ , find $a$ and $b$ . What are the minimum polynomials of its self-adjoint and skew-adjoint parts?
Why must the 2×2 blocks in the canonical form of an orthogonal map satisfy $a^2+b^2=1$ ? What does a 3×3 orthogonal map with determinant 1 do geometrically?