## Announcements:

Continue reading Section 3.5. We aren't covering 3.4. Work through recommended homework questions.

Midterm: It is Saturday, March 1, 6:30pm-9:30pm, one week after reading week. If you have a conflict, you must let me know this week. It will cover the material up to and including the lecture on Monday, Feb 24.

Four practice midterms have been posted on the course web page.

Office hour and tutorials: None during reading week. The tutorial after reading week will be for midterm review; no quiz.

Help Centers: Monday-Friday 2:30-6:30 in MC 106, but not during reading week.

## Partial review of Section 3.3, Lectures 16 and 17:

Definition: An inverse of an $n \times n$ matrix $A$ is an $n \times n$ matrix $A'$ such that $$A A' = I \qtext{and} A' A = I .$$ If such an $A'$ exists, we say that $A$ is invertible.

Theorem 3.6: If $A$ is an invertible matrix, then its inverse is unique.

We write $A^{-1}$ for the inverse of $A$, when $A$ is invertible.

Theorem 3.8: The matrix $A = \bmat{cc} a & b \\ c & d \emat$ is invertible if and only if $ad - bc \neq 0$. When this is the case, $$A^{-1} = \frac{1}{ad-bc} \, \bmat{rr} \red{d} & \red{-}b \\ \red{-}c & \red{a} \emat .$$

We call $ad-bc$ the determinant of $A$, and write it $\det A$.

### Properties of Invertible Matrices

Theorem 3.9: Assume $A$ and $B$ are invertible matrices of the same size. Then:
1. $A^{-1}$ is invertible and $(A^{-1})^{-1} = {A}$
2. If $c$ is a non-zero scalar, then $cA$ is invertible and $(cA)^{-1} = {\frac{1}{c} A^{-1}}$
3. $AB$ is invertible and $(AB)^{-1} = {B^{-1} A^{-1}}$ (socks and shoes rule)
4. $A^T$ is invertible and $(A^T)^{-1} = {(A^{-1})^T}$
5. $A^n$ is invertible for all $n \geq 0$ and $(A^n)^{-1} = {(A^{-1})^n}$

Remark: There is no formula for $(A+B)^{-1}$. In fact, $A+B$ might not be invertible, even if $A$ and $B$ are.

### The fundamental theorem of invertible matrices:

Very important! Will be used repeatedly, and expanded later.

Theorem 3.12: Let $A$ be an $n \times n$ matrix. The following are equivalent:
a. $A$ is invertible.
b. $A \vx = \vb$ has a unique solution for every $\vb \in \R^n$.
c. $A \vx = \vec 0$ has only the trivial (zero) solution.
d. The reduced row echelon form of $A$ is $I_n$.

Theorem 3.13: Let $A$ be a square matrix. If $B$ is a square matrix such that either $AB=I$ or $BA=I$, then $A$ is invertible and $B = A^{-1}$.

### Gauss-Jordan method for computing the inverse

Theorem 3.14: Let $A$ be a square matrix. If a sequence of row operations reduces $A$ to $I$, then the same sequence of row operations transforms $I$ into $A^{-1}$.

This gives a general purpose method for determining whether a matrix $A$ is invertible, and finding the inverse:

1. Form the $n \times 2n$ matrix $[A \mid I\,]$.

2. Use row operations to get it into reduced row echelon form.

3. If a zero row appears in the left-hand portion, then $A$ is not invertible.

4. Otherwise, $A$ will turn into $I$, and the right hand portion is $A^{-1}$.

True/false: If $A$ and $B$ are square matrices such that $AB$ is not invertible, then at least one of $A$ and $B$ is not invertible.

True/false: If $A$ and $B$ are matrices such that $AB = I$, then $BA = I$.

Question: Find invertible matrices $A$ and $B$ such that $A+B$ is not invertible.

## New material: Section 3.5: Subspaces, basis, dimension and rank

This section contains some of the most important concepts of the course.

### Subspaces

A generalization of lines and planes through the origin.

Definition: A subspace of $\R^n$ is any collection $S$ of vectors in $\R^n$ such that:
1. The zero vector $\vec 0$ is in $S$.
2. $S$ is closed under addition: If $\vu$ and $\vv$ are in $S$, then $\vu + \vv$ is in $S$.
3. $S$ is closed under scalar multiplication: If $\vu$ is in $S$ and $c$ is any scalar, then $c \vu$ is in $S$.

Conditions (2) and (3) together are the same as saying that $S$ is closed under linear combinations.

Example: $\R^n$ is a subspace of $\R^n$. Also, $S = \{ \vec 0 \}$ is a subspace of $\R^n$.

Example: A plane $\cP$ through the origin in $\R^3$ is a subspace. Applet.

Here's an algebraic argument. Suppose $\vv_1$ and $\vv_2$ are direction vectors for $\cP$, so $\cP = \span(\vv_1, \vv_2)$.
(1) $\vec 0$ is in $\cP$, since $\vec 0 = 0 \vv_1 + 0 \vv_2$.
(2) If $\vu = c_1 \vv_1 + c_2 \vv_2$ and $\vv = d_1 \vv_1 + d_2 \vv_2$, then \begin{aligned} \vu + \vv &= (c_1 \vv_1 + c_2 \vv_2) + (d_1 \vv_1 + d_2 \vv_2) \\ &= (c_1 + d_1) \vv_1 + (c_2 + d_2) \vv_2 \end{aligned} which is in $\span(\vv_1, \vv_2)$ as well.
(3) For any scalar $c$, $$c \vu = c (c_1 \vv_1 + c_2 \vv_2) = (c c_1) \vv_1 + (c c_2) \vv_2$$ which is also in $\span(\vv_1, \vv_2)$.

On the other hand, a plane not through the origin is not a subspace. It of course fails (1), but the other conditions fail as well, as shown in the applet.

As another example, a line through the origin in $\R^3$ is also a subspace.

The same method as used above proves:

Theorem 3.19: Let $\vv_1, \vv_2, \ldots, \vv_k$ be vectors in $\R^n$. Then $\span(\vv_1, \ldots, \vv_k)$ is a subspace of $\R^n$.

See text. We call $\span(\vv_1, \ldots, \vv_k)$ the subspace spanned by $\vv_1, \ldots, \vv_k$. This generalizes the idea of a line or a plane through the origin.

See Example 3.38 in the text for a similar question.

### Subspaces associated with matrices

Theorem 3.21: Let $A$ be an $m \times n$ matrix and let $N$ be the set of solutions of the homogeneous system $A \vx = \vec 0$. Then $N$ is a subspace of $\R^n$.

Proof: (1) Since $A \, \vec 0_n = \vec 0_m$, the zero vector $\vec 0_n$ is in $N$.
(2) Let $\vu$ and $\vv$ be in $N$, so $A \vu = \vec 0$ and $A \vv = \vec 0$. Then $$A (\vu + \vv) = A \vu + A \vv = \vec 0 + \vec 0 = \vec 0$$ so $\vu + \vv$ is in $N$.
(3) If $c$ is a scalar and $\vu$ is in $N$, then $$A (c \vu) = c A \vu = c \, \vec 0 = \vec 0$$ so $c \vu$ is in $N$. $\qquad \Box$

Aside: At this point, the book states Theorem 3.22, which says that every linear system has no solution, one solution or infinitely many solutions, and gives a proof of this. We already know this is true, using Theorem 2.2 from Section 2.2 (see Lecture 9). The proof given here is in a sense better, since it doesn't rely on knowing anything about row echelon form, but I won't use class time to cover it.

Spans and null spaces are the two main sources of subspaces.

Definition: Let $A$ be an $m \times n$ matrix.

1. The row space of $A$ is the subspace $\row(A)$ of $\R^n$ spanned by the rows of $A$.
2. The column space of $A$ is the subspace $\col(A)$ of $\R^m$ spanned by the columns of $A$.
3. The null space of $A$ is the subspace $\null(A)$ of $\R^n$ consisting of the solutions to the system $A \vx = \vec 0$.

Example: The column space of $A = \bmat{rr} 1 & 2 \\ 3 & 4 \emat$ is $\span(\coll 1 3, \coll 2 4)$. A vector $\vb$ is a linear combination of these columns if and only if the system $A \vx = \vb$ has a solution. But since $A$ is invertible (its determinant is $4 - 6 = -2 \neq 0$), every such system has a (unique) solution. So $\col(A) = \R^2$.

The row space of $A$ is the same as the column space of $A^T$, so by a similar argument, this is all of $\R^2$ as well.

Example: The column space of $A = \bmat{rr} 1 & 2 \\ 3 & 4 \\ 5 & 6 \emat$ is the span of the two columns, which is a subspace of $\R^3$. Since the columns are linearly independent, this is a plane through the origin in $\R^3$.

Determine whether $\colll 2 0 1$ and $\colll 2 0 {-2}$ are in $\col(A)$. (On board.)

We will learn methods to describe the three subspaces associated to a matrix $A$.