Linear algebra

Example 2.1 (Dissimilar matrices with same characteristic polynomial)  

$$A= \left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}\right]$$    and $$B= \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}\right]$$    and the characteristic polynomials are $$f_A(x)=f_B(x)=(x-1)^2.$$

Definition 2.2   ${\mathbf{x}}$ is an eigenvector to the geometric eigenvalue $\lambda$ for matrix $A$ if ${\mathbf{x}}\neq0$ and $A{\mathbf{x}}=\lambda{\mathbf{x}}$. An algebraic eigenvalue of $A$ is a root of the characteristic polynomial $f_A(x)$.

Exercise 2.3   $A{\mathbf{x}}={\mathbf{b}}$ is an inhomogeneous system of equations. Show it has a solution iff ${\mathbf{b}}\in$ span $\{{\mathbf{a}}_1,\ldots,{\mathbf{a}}_n\}$ iff rk $A=$rk $(A\mid{\mathbf{b}})$.

Exercise 2.4   $A{\mathbf{x}}={\mathbf{0}}$ is a homogeneous system of equations. Show it has a nontrivial solution iff columns are linearly dependent.

Determinants - quick and dirty¹.
Given an $n\times n$ matrix $A$, let $A_{ij}$ denote the matrix formed by deleting the $i^{th}$ row and $j^{th}$ column. The determinat-expansion by the $j$-th column is

   det $$A=\sum_{i=1}^n (-1)^{i+j}a_{ij}$$det$$(A_{ij})$$ - the choice of $$j$$ doesn't matter,$$$$

   where det$$\left[\begin{array}{cc} a & b\\ c & d \end{array}\right] =ad-bc.$$

We could also reverse the roles of $i$ and $j$ in the above equation. Switching two rows or two columns changes the sign of the determinant. Adding some linear combination of rows (or columns) to another row (or column respectively) does not affect the determinant.

Note that the above rules are rules of calculation, not definition. It is far from obvious that these rules actually result in a unique number regardless of the choices made along the way. (What is the definition?)

Theorem 2.5   $\det A=0$ iff the columns are linearly dependent.

Exercise 2.6   $\lambda$ is an eigenvalue of $A$ iff $f_A(\lambda)=0$. (In other words the two definitions of eigenvalue are equivalent).

Exercise 2.7   Let $U=\{{\mathbf{x}}:A{\mathbf{x}}={\mathbf{0}}\}$. This is a subspace of $F^n$. Show dim $U=n-$rk $A$, where by rk $A$ we mean the row rank.

Exercise 2.8   Show that for any matrix the row rank equals the column rank.

Definition 2.9   The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial. The geometric multiplicity of $\lambda$ is dim $U_\lambda=\{{\mathbf{x}}:A{\mathbf{x}}=\lambda{\mathbf{x}}\}=n-\mathop{\rm rk}\nolimits (\lambda I-A)$.

Example 2.10  

$$\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}\right]$$    has eigenvalue 1 with algebraic multiplicity 2 and geometric multiplicity 1$$.$$

Example 2.11  

$$R_\alpha=\left[\begin{array}{cc} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{array}\right]$$    is a matrix with no eigenvalues over $$\mathbb{R}$$$$\mbox{ (unless $\pi {\,\mid\,}a$)}$$$$.$$

However if we extend our base field to $\mathbb{C}$ then $R_\alpha$ has eigenvectors

$$\left[\begin{array}{c} i\\ 1 \end{array}\right]$$ and $$\left[\begin{array}{c} -i\\ 1 \end{array}\right]$$    and eigenvalues $$e^{\pm i\alpha}=\cos\alpha\pm i\sin\alpha.$$

Exercise 2.12   Show that over $\mathbb{C}$ a matrix $A$ is diagonalizable (i.e., it has an eigenbasis) iff the geometric and algebraic multiplicities of all its eigenvalues coincide.