Lecture 5. Transposes, permutations, spaces $R^n$

讲座摘要（lecture summary）

In this lecture we introduce vector spaces and their subspaces.

Permutations

Multiplication by a permutation matrix $P$ swaps the rows of a matrix; when applying the method of elimination we use permutation matrices to move zeros out of pivot positions. Our factorization $A = LU$ then becomes $PA = LU$, where $P$ is a permutation matrix which reorders any number of rows of $A$. Recall that $P^{−1} = P^T$, i.e. that $P^{T}P = I$.

Transposes

When we take the transpose of a matrix, its rows become columns and its columns become rows. If we denote the entry in row $i$ column j of matrix $A$ by $A_{ij}$, then we can describe $A^T$ by: $(A^T)_{ij} = A_{ji}$. For example:

$$\begin{bmatrix*}[c] 1 & 3 \\ 2 & 3 \\ 4 & 1 \end{bmatrix*}^T = \begin{bmatrix*}[c] 1 & 2 & 4 \\ 3 & 3 & 1 \end{bmatrix*}$$

$A$ matrix $A$ is symmetric if $A^T = A$. Given any matrix $R$ (not necessarily square) the product $R^TR$ is always symmetric, because $(R^TR)^T = R^T(R^T)^T = R^TR$. (Note that $(R^T)^T = R$.)

Vector spaces

We can add vectors and multiply them by numbers, which means we can discuss linear combinations of vectors. These combinations follow the rules of a vector space.

One such vector space is $\R ^ 2$, the set of all vectors with exactly two real number components. We depict the vector $\begin{bmatrix} a \\ b\end{bmatrix}$ by drawing an arrow from the origin to the point $(a, b)$ which is a units to the right of the origin and $b$ units above it, and we call $\R^2$ the “$x − y$ plane”.

Another example of a space is $\R^n$, the set of (column) vectors with n real number components.

Closure

The collection of vectors with exactly two positive real valued components is not a vector space. The sum of any two vectors in that collection is again in the collection, but multiplying any vector by, say, $−5$, gives a vector that’s not in the collection. We say that this collection of positive vectors is closed under addition but not under multiplication.

If a collection of vectors is closed under linear combinations (i.e. under addition and multiplication by any real numbers), and if multiplication and addition behave in a reasonable way, then we call that collection a vector space.

Subspaces

A vector space that is contained inside of another vector space is called a subspace of that space. For example, take any non-zero vector $v$ in $\R^2$. Then the set of all vectors $c\mathbf{v}$, where c is a real number, forms a subspace of $\R2$. This collection of vectors describes a line through $\begin{bmatrix} 0 \\ 0\end{bmatrix}$ in $R^2$ and is closed under addition.

A line in $\R^2$ that does not pass through the origin is not a subspace of $\R^2$. Multiplying any vector on that line by $0$ gives the zero vector, which does not lie on the line. Every subspace must contain the zero vector because vector spaces are closed under multiplication.

The subspaces of $\R^2$ are:

1. all of $\R^2$,
2. any line through $\begin{bmatrix} 0 \\ 0\end{bmatrix}$ and
3. the zero vector alone (Z).

The subspaces of $\R^3$ are:

1. all of $\R^3$,
2. any plane through the origin,
3. any line through the origin, and
4. the zero vector alone (Z).

Column space

Given a matrix $A$ with columns in $R^3$, these columns and all their linear combinations form a subspace of $R^3$. This is the column space $C(A)$. If $A = \begin{bmatrix*}[c] 1 & 3 \\ 2 & 3 \\ 4 & 1 \end{bmatrix*}$, the column space of A is the plane through the origin in $\R^3$ containing $\begin{bmatrix*}[c] 1 \\ 2 \\ 4 \end{bmatrix*}$ and $\begin{bmatrix*}[c] 3 \\ 3 \\ 1 \end{bmatrix*}$.

Our next task will be to understand the equation $A\mathbf{x} = b$ in terms of subspaces and the column space of $A$.

Problems and Solutions（习题及答案）

Problem 5.1: (2.7 #13. Introduction to Linear Algebra: Strang)

• a) Find a $3$ by $3$ permutation matrix with $P^3 = I$ (but not $P = I$).
• b) Find a $4$ by $4$ permutation $\hat{P}$ with $\hat{P}^4 \ne I$.

Solution

• a) Let $P$ move the rows in a cycle: the first to the second, the second to the third, and the third to the first. So

$$P = \begin{bmatrix*}[c] 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix*}, P^2 = \begin{bmatrix*}[c] 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{bmatrix*}, \text{and} \enspace P^3 = I.$$

• b) Let $\hat{P}$ be the block diagonal matrix with 1 and P on the diagonal; $\hat{P} = \begin{bmatrix} 1 & 0 \\ 0 & P \end{bmatrix}$ .Since $P^3 = I$, also $\hat{P}^3 = I$. So $\hat{P}^4 = \hat{P} \ne I$.

Problem 5.2: Suppose $A$ is a four by four matrix. How many entries of $A$ can be chosen independently if:

• a) $A$ is symmetric?
• b) $A$ is skew-symmetric? $(A^T = −A)$

Solution

• a) The most general form of a four by four symmetric matrix is:

$$A = \begin{bmatrix*}[c] a & e & f & g\\ e & b & h & i\\ f & h & c & j\\ g & i & j & d \end{bmatrix*}.$$

Therefore $\mathbf{10}$ entries can be chosen independently.

• b) The most general form of a four by four skew-symmetric matrix is:

$$A = \begin{bmatrix*}[r] 0 & -a & -b & -c\\ a & 0 & -d & -e\\ b & d & 0 & -f\\ c & e & f & 0 \end{bmatrix*}.$$

Therefore $\mathbf{6}$ entries can be chosen independently.

Problem 5.3: (3.1 #18.) True or false (check addition or give a counterexample):

• a) The symmetric matrices in $M$ (with $A^T = A$) form a subspace.
• b) The skew-symmetric matrices in $M$ (with $A^T = −A$) form a subspace.
• c) The unsymmetric matrices in $M$ (with $A^T \ne A$) form a subspace.

Solution

• a) True: $A^T = A$ and $B^T = B$ lead to:

$$(A + B)^T = A^T + B^T = A + B, \text{and} \enspace (cA)^T = cA.$$

• b) True: $A^T = −A$ and $B^T = −B$ lead to:

$$(A + B)^T = A^T + B^T = −A − B = −(A + B), \text{and} \enspace (cA)^T = −cA.$$

• c) False: $\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.