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Subspaces

A subspace is a vector space that is contained within another vector space. So every subspace is a vector space in its own right, but it is also defined relative to some other (larger) vector space.

Subspaces

A subspace is a vector space that is contained within another vector space. So every subspace is a vector space in its own right, but it is also defined relative to some other (larger) vector space. We will discover shortly that we are already familiar with a wide variety of subspaces from previous sections.

Summarize the three conditions that allow us to quickly test if a set is a subspace.
W&=\setparts{\colvector{a\\b\\c}}{3a-2b+c=5}
Name five general constructions of sets of column vectors (subsets of $\complex{m}$) that we now know as subspaces.

Subspaces

Here is the principal definition for this section.

Let us look at an example of a vector space inside another vector space.

Condensed: the full section is in Beezer, A First Course in Linear Algebra.

Testing Subspaces

In we proceeded through all ten of the vector space properties before believing that a subset was a subspace. But six of the properties were easy to prove, and we can lean on some of the properties of the vector space (the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. A shortcut if there ever was one.

Theorem: Testing Subsets for Subspaces

Suppose that $V$ is a vector space and $W$ is a subset of $V$, $W\subseteq V$. Endow $W$ with the same operations as $V$. Then $W$ is a subspace if and only if three conditions are met

$W$ is nonempty, $W\neq\emptyset$.
If $\vect{x}\in W$ and $\vect{y}\in W$, then $\vect{x}+\vect{y}\in W$.
If $\alpha\in\complex{\null}$ and $\vect{x}\in W$, then $\alpha\vect{x}\in W$.

Proof

We have the hypothesis that $W$ is a subspace, so by we know that $W$ contains a zero vector. This is enough to show that $W\neq\emptyset$. Also, since $W$ is a vector space it satisfies the additive and scalar multiplication closure properties (, ), and so exactly meets the second and third conditions. If that was easy, the other direction might require a bit more work.

We have three properties for our hypothesis, and from this we should conclude that $W$ has the ten defining properties of a vector space. The second and third conditions of our hypothesis are exactly and . Our hypothesis that $V$ is a vector space implies that , , , , and all hold. They continue to be true for vectors from $W$ since passing to a subset, and keeping the operation the same, leaves their statements unchanged. Eight down, two to go.

Suppose $\vect{x}\in W$. Then by the third part of our hypothesis (scalar closure), we know that $(-1)\vect{x}\in W$. By $(-1)\vect{x}=\vect{-x}$, so together these statements show us that $\vect{-x}\in W$. $\vect{-x}$ is the additive inverse of $\vect{x}$ in $V$, but will continue in this role when viewed as an element of the subset $W$. So every element of $W$ has an additive inverse that is an element of $W$ and is completed. Just one property left.

While we have implicitly discussed the zero vector in the previous paragraph, we need to be certain that the zero vector (of $V$) really lives in $W$. Since $W$ is nonempty, we can choose some vector $\vect{z}\in W$. Then by the argument in the previous paragraph, we know $\vect{-z}\in W$. Now by for $V$ and then by the second part of our hypothesis (additive closure) we see that \zerovector=\vect{z}+(\vect{-z})\in W

So $W$ contains the zero vector from $V$. Since this vector performs the required duties of a zero vector in $V$, it will continue in that role as an element of $W$. This gives us, , the final property of the ten required. ( contributed to this proof.)

Condensed: the full section is in Beezer, A First Course in Linear Algebra.

The Span of a Set

The span of a set of column vectors got a heavy workout in and . The definition of the span depended only on being able to formulate linear combinations. In any of our more general vector spaces we always have a definition of vector addition and of scalar multiplication. So we can build linear combinations and manufacture spans. This subsection contains two definitions that are just mild variants of definitions we have seen earlier for column vectors. If you have not already, compare them with and .

Example: A linear combination of matrices

In the vector space $M_{23}$ of $2\times 3$ matrices, we have the vectors \vect{x}&= \begin{bmatrix} 1&3&-2\\ 2&0&7 \end{bmatrix} & \vect{y}&= \begin{bmatrix} 3&-1&2\\ 5&5&1 \end{bmatrix} & \vect{z}&= \begin{bmatrix} 4&2&-4\\ 1&1&1 \end{bmatrix} and we can form linear combinations such as 2\vect{x}+4\vect{y}+(-1)\vect{z}&= 2 \begin{bmatrix} 1&3&-2\\ 2&0&7 \end{bmatrix} +4 \begin{bmatrix} 3&-1&2\\ 5&5&1 \end{bmatrix} +(-1) \begin{bmatrix} 4&2&-4\\ 1&1&1 \end{bmatrix}\\ &= \begin{bmatrix} 2&6&-4\\ 4&0&14 \end{bmatrix} + \begin{bmatrix} 12&-4&8\\ 20&20&4 \end{bmatrix} + \begin{bmatrix} -4&-2&4\\ -1&-1&-1 \end{bmatrix}\\ &= \begin{bmatrix} 10&0&8\\ 23&19&17 \end{bmatrix} or, 4\vect{x}-2\vect{y}+3\vect{z}&= 4 \begin{bmatrix} 1&3&-2\\ 2&0&7 \end{bmatrix} -2 \begin{bmatrix} 3&-1&2\\ 5&5&1 \end{bmatrix} +3 \begin{bmatrix} 4&2&-4\\ 1&1&1 \end{bmatrix}\\ &= \begin{bmatrix} 4&12&-8\\ 8&0&28 \end{bmatrix} + \begin{bmatrix} -6&2&-4\\ -10&-10&-2 \end{bmatrix} + \begin{bmatrix} 12&6&-12\\ 3&3&3 \end{bmatrix}\\ &= \begin{bmatrix} 10&20&-24\\ 1&-7&29 \end{bmatrix}

When we realize that we can form linear combinations in any vector space, then it is natural to revisit our definition of the span of a set, since it is the set of all possible linear combinations of a set of vectors.

Let us again examine membership in a span.

Notice how and contained questions about membership in a span, but these questions quickly became questions about solutions to a system of linear equations. This will be a common theme going forward.

Condensed: the full section is in Beezer, A First Course in Linear Algebra.

Subspace Constructions

Several of the subsets of vectors spaces that we worked with in are also subspaces they are closed under vector addition and scalar multiplication in $\complex{m}$.

That was easy! Notice that we could have used this same approach to prove that the null space is a subspace, since provided a description of the null space of a matrix as the span of a set of vectors. However, I much prefer the current proof of . Speaking of easy, here is a very easy theorem that exposes another of our constructions as creating subspaces.

One more.

So the span of a set of vectors, and the null space, column space, row space and left null space of a matrix are all subspaces, and hence are all vector spaces, meaning they have all the properties detailed in and in the basic theorems presented in . We have worked with these objects as just sets in and , but now we understand that they have much more structure. In particular, being closed under vector addition and scalar multiplication means a subspace is also closed under linear combinations.

Condensed: the full section is in Beezer, A First Course in Linear Algebra.

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Kysymyksiä

What does a determinant mean geometrically?

It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.

What is an eigenvector?

A direction the matrix does not turn: it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.

Why is matrix multiplication not commutative?

Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.

Osaa tästä sivusta on muokattu Beezer, A First Course in Linear Algebra (GFDL 1.2). Tässä on tiivistetty ja selitetty uudelleen, virheet ovat meidän.

Lisää Linear Algebra

Determinants Matrix inverse Eigenvalues and eigenvectors Matrix multiplication Row reduction Vector spaces, span and linear independence Orthogonality, projections and least squares Diagonalisation and matrix powers Linear transformations and change of basis