Math 1600B Lecture 20, Section 2, 26 Feb 2014

Announcements:

Read Section 3.6 for next class. Work through recommended homework questions.

Extra Midterm Review: Friday, 4:30-6:00pm, MC110. Bring questions.

Midterm location is based on first letter of last name: HSB35 A-H, HSB236 I-Q, HSB240 R-Z. Be sure to write in the correct room!

Midterm: Saturday, March 1, 6:30pm-9:30pm. It will cover the material up to and including last lecture (not today's lecture), but not electrical networks. Practice midterms are on the website.

Tutorials: Midterm review this week. No quiz.

Help Centers: Monday-Friday 2:30-6:30 in MC106.

Intent to Register Open House for Actuarial & Statistical Sciences, Applied Math, Mathematics, Computer Science: Thursday, Feb 27, 3:30-5pm, MC320. Free pizza!

Partial review of Lectures 18 and 19:

Subspaces

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 $S$.

Basis

Definition: A basis for a subspace $S$ of $\R^n$ is a set of vectors $\vv_1, \ldots, \vv_k$ such that:
1. $S = \span(\vv_1, \ldots, \vv_k)$, and
2. $\vv_1, \ldots, \vv_k$ are linearly independent.

Subspaces associated with matrices

Theorem 3.20: Let $A$ and $R$ be row equivalent matrices. Then $\row(A) = \row(R)$.

Also, $\null(A) = \null(R)$. But elementary row operations change the column space! So $\col(A) \neq \col(R)$.

Theorem: If $R$ is a matrix in row echelon form, then the nonzero rows of $R$ form a basis for $\row(R)$.

So if $R$ is a row echelon form of $A$, then a basis for $\row(A)$ is given by the nonzero rows of $R$.

Now, since $\null(A) = \null(R)$, the columns of $R$ have the same dependency relationships as the columns of $A$.

It is easy to see that the pivot columns of $R$ form a basis for $\col(R)$, so the corresponding columns of $A$ form a basis for $\col(A)$.

We learned in Chapter 2 how to use $R$ to find a basis for the null space of a matrix $A$, even though we didn't use this terminology.

Summary

Row echelon form is in fact enough. Then you look at the columns with leading nonzero entries (the pivot columns).

These methods can be used to compute a basis for a subspace $S$ spanned by some vectors $\vv_1, \ldots, \vv_k$.

The row method:

1. Form the matrix $A$ whose rows are $\vv_1, \ldots, \vv_k$, so $S = \row(A)$.
2. Reduce $A$ to row echelon form $R$.
3. The nonzero rows of $R$ will be a basis of $S = \row(A) = \row(R)$.

The column method:

1. Form the matrix $A$ whose columns are $\vv_1, \ldots, \vv_k$, so $S = \col(A)$.
2. Reduce $A$ to row echelon form $R$.
3. The columns of $A$ that correspond to the columns of $R$ with leading entries form a basis for $S = \col(A)$.

New material

Dimension and Rank

Idea of proof: Suppose that $\{ \vu_1, \vu_2 \}$ and $\{ \vv_1, \vv_2, \vv_3 \}$ were both bases for $S$. We'll show that this is impossible, by showing that $\{ \vv_1, \vv_2, \vv_3 \}$ is linearly dependent. Since $\{ \vu_1, \vu_2 \}$ is a basis, we can express each $\vv_i$ in terms of the $\vu_j$'s: $$ \begin{aligned} \vv_1 &= a_{11} \vu_1 + a_{21} \vu_2 \\ \vv_2 &= a_{12} \vu_1 + a_{22} \vu_2 \\ \vv_3 &= a_{13} \vu_1 + a_{23} \vu_2 \end{aligned} $$ Then $$ \kern-8ex \begin{aligned} &c_1 \vv_1 + c_2 \vv_2 + c_3 \vv_3 \\ = &c_1 ( a_{11} \vu_1 + a_{21} \vu_2 ) + c_2 ( a_{12} \vu_1 + a_{22} \vu_2 ) + c_3 ( a_{13} \vu_1 + a_{23} \vu_2 ) \\ = &(c_1 a_{11} + c_2 a_{12} + c_3 a_{13}) \vu_1 + (c_1 a_{21} + c_2 a_{22} + c_3 a_{23}) \vu_2 \end{aligned} $$ But the homogenous system $$ \begin{aligned} c_1 a_{11} + c_2 a_{12} + c_3 a_{13} &= 0 \\ c_1 a_{21} + c_2 a_{22} + c_3 a_{23} &= 0 \end{aligned} $$ has nontrivial solutions! (Why?) Therefore, we can find nontrivial $c_1$, $c_2$, $c_3$ such that $$ c_1 \vv_1 + c_2 \vv_2 + c_3 \vv_3 = \vec 0 \qquad \Box $$

A very similar argument works for the general case.

Definition: The number of vectors in a basis for a subspace $S$ is called the dimension of $S$, denoted $\dim S$.

Example: $\dim \R^n = \query{n}$

Example: If $S$ is a line through the origin in $\R^2$ or $\R^3$, then $\dim S = \query{1}$

Example: If $S$ is a plane through the origin in $\R^3$, then $\dim S = \query{2}$

Example: If $S = \span(\colll 3 0 2, \colll {-2} 1 1, \colll 1 1 3)$, then $\dim S = \query{2}$

Example: Let $A$ be the matrix from last class whose reduced row echelon form is $$ R = \bmat{rrrrr} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & 2 & 0 & 3 \\ 0 & 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 & 0 \emat $$ Then: $\ \dim \row(A) = \query{3}$ $\ \dim \col(A) = \query{3}$ $\ \dim \null(A) = \query{2}$

Note that $\dim \row(A) = \rank(A)$, since we defined the rank of $A$ to be the number of nonzero rows in $R$. The above theorem shows that this number doesn't depend on how you row reduce $A$.

We call the dimension of the null space the nullity of $A$ and write $\nullity(A) = \dim \null(A)$. This is what we called the "number of free variables" in Chapter 2.

From the way we find the basis for $\row(A)$, $\col(A)$ and $\null(A)$, can you deduce any relationships between their dimensions?

Theorems 3.24 and 3.26: Let $A$ be an $m \times n$ matrix. Then $$ \dim \row(A) = \dim \col(A) = \rank(A) $$ and $$ \rank(A) + \nullity(A) = n . $$

Very important!

Questions:

True/false: if $A$ is $2 \times 5$, then the nullity of $A$ is 3. False. We know that $\rank(A) \leq 2$ and $\rank(A) + \nullity(A) = 5$, so $\nullity(A) \geq 3$ (and $\leq 5$).

True/false: if $A$ is $5 \times 2$, then $\nullity(A) \geq 3$. False. $\rank(A) + \nullity(A) = 2$, so $\nullity(A) = 0$, $1$ or $2$.

Example: Find the nullity of $$ M = \bmat{rrrrrrr} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 8 & 9 & 10 & 11 & 12 & 13 & 14 \emat $$ and of $M^T$. Any guesses? The rows of $M$ are linearly independent, so the rank is $2$, so the nullity is $7 - 2 = 5$. The rank of $M^T$ is also $2$, so the nullity of $M^T$ is $2 - 2 = 0$.

For larger matrices, you would compute the rank by row reduction.

Fundamental Theorem of Invertible Matrices, Version 2

Theorem 3.27: 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 $I_n$.
f. $\rank(A) = n$
g. $\nullity(A) = 0$
h. The columns of $A$ are linearly independent.
i. The columns of $A$ span $\R^n$.
j. The columns of $A$ are a basis for $\R^n$.

In fact, since $\rank(A) = \rank(A^T)$, all of the statements are also equivalent to the statements with $A$ replaced by $A^T$. In particular, we can add the following:

k. The rows of $A$ are linearly independent.
l. The rows of $A$ span $\R^n$.
m. The rows of $A$ are a basis for $\R^n$.

Example 3.52: Show that the vectors $\colll 1 2 3$, $\colll {-1} 0 1$ and $\colll 4 9 7$ form a basis for $\R^3$.

Solution: Show that matrix $A$ with these vectors as the columns has rank 3. On board.

Not covering Theorem 3.28.

Questions

True/false: Every subspace of $\R^3$ has dimension 0, 1, 2 or 3.

True/false: If a matrix $A$ has row echelon form $$ R=\bmat{rrr} 0 & 2 & 3 \\ 0 & 0 & 4 \\ \emat $$ then a basis for $\col(A)$ is given by $\coll 2 0$ and $\coll 3 4$.

Question: What's a basis for $\row(A)$?