This section introduces perhaps the most important ideas in the entire book. This is where we start studying dimension, which we've already mentioned, what rank means/represents, smallest possible linear independent sets, subspaces, like a plane through the origin in \mathbb{R}^3.
A subspace of \mathbb{R}^n is any collection S of vectors in \mathbb{R}^n such that:
1. The zero vector 0 is in S.
2. If \vec{u} and \vec{v} are in S, then \vec{u}+\vec{v} is in S.
(We say S is closed under addition.)
3. If \vec{u} is in S and c is a scalar, then c\vec{u} is in S.
(We say S is closed under scalar multiplication.)
Note: We could have combined properties (2) and (3) and required, equivalently, that S be closed under linear combinations:
If \vec{u}_1,\vec{u}_2,...,\vec{u}_k are in S and c_1,c_2,...,c_k are scalars,
then c_1 \vec{u}_1+c_2 \vec{u}_2+\cdots+c_k \vec{u}_k is in S.
Example: What are the subspaces of \mathbb{R}^3?
1. S = \{0\}
2. All lines through the origin.
3. All planes through the origin.
4. \mathbb{R}^3 itself.
Exercse: Prove that a plane in \mathbb{R}^3 through the origin is in fact a subspace of \mathbb{R}^3.
Hint: Let \vec{v}_1,\vec{v}_2 be the direction vectors for the plane. What is span(\vec{v}_1,\vec{v}_2)?
Example: Is a plane that doesn't go through the origin a subspace?
Let \vec{v}_1,...,\vec{v}_k be vectors in \mathbb{R}^n. Then span(\vec{v}_1,...,\vec{v}_k) is a subspace of \mathbb{R}^n.
Example: Show that the set of all vectors \left[ \begin{array}{c} x \\ y \\z \end{array} \right] that satisfy the conditions x = 2y and z = -3y forms a subspace of \mathbb{R}^3.
What do vectors in this set look like? Substituting the two coniditons:
y \left[ \begin{array}{c} 2 \\ 1 \\ -3 \end{array} \right]
Since y is arbitrary, the given set of vectors is span\left( \left[ \begin{array}{c} 2 \\ 1 \\ -3 \end{array} \right]\right) and is thus a subspace of \mathbb{R}^3.
Example: Is the set of all vectors \left[ \begin{array}{c} x \\ y \\z \end{array} \right] such that x = 2y+1 and z = -3y a subspace of \mathbb{R}^3?
Example: Is the set of all vectors \left[ \begin{array}{c} x \\ y \end{array} \right], where y = x^2 a subspace of \mathbb{R}^3?
In order for a set S to be a subspace of some \mathbb{R}^n, we must prove that properties (1) through (3) hold in general. However, for S to fail to be a subspace of \mathbb{R}^3, it is enough to show that one of these three properties fails to hold. The easiest course is usually to find a single, specific counterexample to illustrate the failure of the property. Once you have done so, there is no need to consider the other properties.
Let A be an m \times n matrix.
1. The row space of A is the subspace row(A) of \mathbb{R}^n spanned by the rows of A.
2. The column space of A is the subspace col(A) of \mathbb{R}^m spanned by the columns of A.
Example: Consider the matrix A = \left[ \begin{array}{cc} 2 & 1 \\ 0 & -1 \\ 1 & 3 \end{array}\right].
(a) Determine whether \vec{b} = \left[ \begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right] is in the column space of A.
(b) Determine whether \vec{w} = \left[ \begin{array}{cc} 4 &5 \end{array} \right] is in the row space of A.
(c) Describe row(A) and col(A).
Again, all three of these processes come down to row reduction. The only trick is remembering how to set them up.
(a) Here we just augment A with \vec{b}, and row reduce the result: [A|\vec{b}].
Note that during the process, if \vec{b} is in col(A), then we will also get the coefficients so that we can write \vec{b} as a linear combination of the columns of A.
(b) There are two ways to see if \vec{w} is in row(A). We could first "translate" the rows of A and the vector \vec{w} into column vectors and then solve as above. This would mean row reducing [A^T|\vec{w}^T]. We would also get the coefficients to write \vec{w} as a linear combination of the rows of A.
The other way is to augment A with \vec{w}, but in a new way. Instead of placing \vec{w} on the right, we augment on the bottom: \left[ \begin{array}{c} A \\ \vec{w}\end{array} \right], and then row reduce this new matrix. Since we are row reducing, all the "new" rows will always be linear combinations of the "original" rows. If we can row reduce to the point where the bottom row is all zeros, we know that \vec{w} is a linear combination of the other rows. Unless you keep track of your row operations, this method does not tell you the coefficients of the linear combination.
(c) Finding the row and column spaces of a matrix, is exactly the same as finding the span of a set of vectors. Here the vectors are just the rows or columns of A.
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Let B be any matrix that is row equivalent to a matrix A. Then row(A) = row(B).
Let A be an m \times n matrix and let N be the set of solutions to the homogeneous linear system A \vec{x} = 0.
Then N is a subspace of \mathbb{R}^n.
Let A be an m \times n matrix. The null space of A is the subspace of \mathbb{R}^n consisting of solutions of the homogeneous linear system A\vec{x} = 0. It is denoted by null(A).
Let A be a matrix whose entries are real numbers. For any system of linear equations A\vec{x} = \vec{b}, exactly one of the following is true:
a. There is no solution.
b. There is a unique solution.
c. There are infinitely many solutions.
Remark: We now have three different subspaces attached to a matrix, the row space, the column space, and the null space. If A is an m \times n matrix, the row space is a subspace of \mathbb{R}^n, the column space is a subspace of \mathbb{R}^m, and the null space is a subspace of \mathbb{R}^n.
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