Lecture 03

[[lecture-data]]

2024-08-30

Readings

0. Chapter 0

Inverse

Suppose $A, B \in M_{n} (F)$ . We say $A, B$ are inverses if $A B = B A = I$ . Note that if the inverse exists, then it is unique.

If a matrix is non-invertible, we call it singular.

(see singluar matrix)

Note

If $A$ and $B$ are inverses, then $A : F^{n} \to F^{n}$ and $B : F^{n} \to F^{n}$ are inverse functions of each other. That is,
$\forall x \in F^{n}$ , we have $A (B x) = x$ and $B (A x) = x$ .

Theorem

Let $A \in M_{n} (F)$ . The following are equivalent:

$\exists B \in M_{n} (F)$ such that $A B = B A = I$
$det (A) \neq 0$
$null (A) = {0}$
$rank (A) = n$

Recall that we showed last time that $(3) ⟺$ columns are linearly independent $⟺ A$ is one-to-one as a function.

And note that $(4) ⟺$ the columns of $A$ span $F^{n} ⟺ A$ is onto as a function.

Also $(2) ⟺ rref (A) = I$

(see equivalent conditions for invertibility of a matrix)

Proof(s)

(1) ⟹ (2)

Suppose there exists a $B$ such that $A B = I$ . Then $1 = det I = det A B = det A det B ⟹ det A \neq 0$

(2) ⟹ (1)

Suppose $det A \neq 0$ , thus $rref (A) = I$ . Then solve $A X = I$ , where $X \in M_{n} (F^{n})$ . Note that this is equivalent to solving $A x_{i} = e_{i}$ and we can row reduce $[A I] \to [I B] ⟹ B = X$ ie $A B = I$ .
Note that $B A = I$ also. $[I B] \to [A I]$ by row reduction by just doing it in reverse. But we can also do the same process to get $[B I] \to [I A]$ . So $B A = I$ .

(3) ⟺ (4)

Columns of $A$ are lin. indep. $⟺$ columns of $A$ are a basis for $F^{n} ⟺$ columns of $A$ span $F^{n}$ .

(1) ⟹ (3)

Suppose there exists some $B$ such that $B A = I$ . If $x \in F^{n}, A x = 0$ , then $x = I x = B A x = B 0 = 0 ⟹ x = 0 ⟹ null (A) = {0}$

(3), (4) ⟹ (1)

Suppose that $A$ is one-to-one as a function and is also onto (see the equivalent defs above). Then $A$ has an inverse function! $A$ is a bijection, and therefore has an inverse function, call it $B$ . $B$ is linear (why?)

$A$ is a linear bijection, so linearity is preserved between the domain and the image. Thus $B$ is also linear, and can be expressed as a matrix (recall from Lecture 01).

Note, $\forall x \in F^{n}$ , $A B x = I x$ and also $B A x = I x ⟹ A B = I = B A$ . $◼$

Note

If $A, B \in M_{n} (F)$ and we have $A B = I$ , then $B = A^{- 1}$

(see a left inverse is also a right inverse)

Proof

$1 = det I = det A B = det A det B ⟹ det A \neq 0$ . So $A$ has an inverse. So $B = A^{- 1} A B = A^{- 1}$ .

1. Chapter 1 (similarity and eigenvalues)

Similarity

Suppose $A, B \in M_{n} (F)$ . $A$ is similar to $B$ if there exists $S \in M_{n} (F)$ invertible such that $A = S B S^{- 1}$ .

Note

Note that similarity is an equivalence relation on square matrices:

$A = I A I^{- 1}$ (reflexivity)
$A = S B S^{- 1} ⟹ S^{- 1} A S = B$ (symmetry)
$A = S B S^{- 1}, B = T C T^{- 1} ⟹ A = S T C T^{- 1} S^{- 1}$ (transitivity)

We denote similarity with $A \sim B$
(see similar matrices)

What does similarity mean? It means that $A$ and $B$ represent the same transformation, up to a change of basis. Where $S$ is transforming the space of $A$ (then $S^{- 1}$ is undoing it!)

Illustration:
$A x = S B S^{- 1} x$

Note

if $A \sim B$ then $det A = det B$ . (The reverse is NOT true!)

Proof

Say $A = S B S^{- 1}$ . Then $$\begin{aligned}
\det A&=\det SBS^{-1}\
&= \det S\det B\det S^{-1}\
&= \det S \det S^{-1} \det B ;;;\text{since these are scalars}\
&= \det SS^{-1}\det B\
&=\det I\det B \
&=\det B
\end{aligned}$$
see similarity implies equal determinants

Row Equivalence

Suppose $E, F \in M_{m, n} (F)$ . $E \sim_{rref} F ⟺ \exists G \in M_{m}$ invertible such that $G E = F$ .

(row equivalence)

$G$ describes the row reduction of $E$ to $F$