Math 1600A Lecture 20, Section 2, 25 Oct 2013

Announcements:

Read Section 3.6 for Monday. Work through recommended homework questions.

Tutorials: No tutorials next week!

We're more than halfway done the lectures! This is lecture 20 out of 37.

Office hour: Monday, 1:30-2:30, MC103B.
Help Centers: Monday-Friday 2:30-6:30 in MC 106.

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

1. Find the reduced row echelon form $R$ of $A$.
2. The nonzero rows of $R$ form a basis for $\row(A) = \row(R)$.
3. The columns of $A$ that correspond to the columns of $R$ with leading 1's form a basis for $\col(A)$.
4. Use back substitution to solve $R \vx = \vec 0$; the vectors that arise are a basis for $\null(A) = \null(R)$.

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 $$ \kern-5ex \begin{aligned} &c_1 \vv_1 + c_2 \vv_2 + c_3 \vv_3 \\ = &c_1 ( a_{11} \vu_1 + a_{21} \vu_2 ) + c_2 ( a_{12} \vu_1 + a_{22} \vu_2 ) + c_3 ( a_{13} \vu_1 + a_{23} \vu_2 ) \\ = &(c_1 a_{11} + c_2 a_{12} + c_3 a_{13}) \vu_1 + (c_1 a_{21} + c_2 a_{22} + c_3 a_{23}) \vu_2 \end{aligned} $$ 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: $\quad\dim \row(A) = \query{3}$ $\quad\dim \col(A) = \query{3}$ $\quad\dim \null(A) = \query{2}$

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 $$ \kern-8ex \dim \row(A) = \dim \col(A) = \rank(A) \qtext{and} \rank(A) + \nullity(A) = n . $$

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 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 $\rank(A) = \rank(A^T)$, 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 whiteboard.

Not covering Theorem 3.28.

Coordinates

Theorem 3.29: For every vector $v$ in $S$, there is exactly one way to write $v$ as a linear combination of the vectors in $\cB$: $$ \vv = c_1 \vv_1 + \cdots + c_k \vv_k $$

Proof: Try to work it out yourself! It's a good exercise.

We call the coefficients $c_1, c_2, \ldots, c_k$ the coordinates of $\vv$ with respect to $\cB$, and write $$ [\vv]_{\cB} = \collll {c_1} {c_2} {\vdots} {c_k} $$

Example: Let $S$ be the plane through the origin in $\R^3$ spanned by $\vv_1 = \colll 1 2 3$ and $\vv_2 = \colll 4 5 6$, so $\cB = \{ \vv_1, \vv_2 \}$ is a basis for $S$. Let $\vv = \colll 6 9 {12}$. Then $$ \vv = 2 \vv_1 + 1 \vv_2 \qqtext{so} [\vv]_{\cB} = \coll 2 1 $$ Note that while $\vv$ is a vector in $\R^3$, it only has two coordinates with respect to $\cB$.

Example: Let $\cB = \{ \ve_1, \ve_2, \ve_3 \}$ be the standard basis for $\R^3$, and consider $\vv = \colll 6 9 {12}$. Then $$ \vv = 6 \ve_1 + 9 \ve_2 + 12\ve_3 \qqtext{so} [\vv]_{\cB} = \colll 6 9 {12} $$ We've implicitly been using the standard basis everywhere, but often in applications it is better to use a basis suited to the problem.

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