Spanning -- What does it Mean??

Before we can explain "spanning", we need to define a "linear combination" of vectors.

Def'n: if vector w can be expressed as k₁v₁ + k₂v₂ + .... + k_nv_n where ,
w is a linear combination of v₁, v₂, .... v_n .

, it is a linear combination of

-- so is EVERY VECTOR IN R ².

Now that we know that, the best way to understand what "Spanning" means is to call it
"Generating" instead of spanning. Spanning makes us think of a bridge reaching OVER a space -- whereas Generating evokes the image of a garden where things grow up or out FROM WITHIN a space.

Our text books define it this way:

Def'n: A set of vectors S = { v₁, v₂, .... v_n } spans a vector space V,
if every vector in V can be expressed as a linear combination of { v₁, v₂, .... v_n }.

The definition says that if every vector in V, can be GENERATED by a
combination of some or all vectors in set S, then S spans V.

Think of the vectors in S as SEEDS for the Garden (the vector space V). We can mix them together with soil, sun and water (our linear combinations) to Grow or Generate all the plants in the garden (all the vectors in the Space).

Or consider them the ingredients (elements) such as flour, eggs, milk, etc., that we combine to Generate Yummy Baked Stuff (cakes, breads and cookies). We use a unique combination of those ingredients to generate the specific cake, bread or cookies on the menu, so we can say that those ingredients SPAN the Vector Space of Yummy Baked Stuff (cakes, breads and cookies).

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Notation: lin(S) or lin { v₁, v₂, .... v_n } is the notation for the linear space spanned by S.

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Example: Determine if these vectors span R²

Solution: If every vector in R² can be expressed as k₁u + k₂v, for every a and b in R, then the set of vectors { u, v } spans R².

We rewrite as the system .

Since the determinant of the system matrix = – 2, the system has a unique solution for every a and b in R, so u and v Span R², the 2-dimensional vectors. We can generate any vector in the Cartesian plane by a linear combination of these 2 vectors.

However, we could do the same with these 3 vectors: .

The difference is that: vector w can be GENERATED by the other 2, so it is unnecessary.

We can show that 3u – 2v = w which means that 3u – 2v – w = 0. So, though this set of 3 vectors SPANS R², it is a linearly dependent (definition follows) set of vectors.

Linear Independence -- What does it Mean??

The "street-talk" explanation of linear independence says: the only way to get the zero vector from a combination of the given vectors is to TAKE NONE OF THEM.

That means the only solution to k₁v₁ + k₂v₂ + .... + k_nv_n = 0 is all the k's must = 0.

In the last example, we found 3u – 2v – w = 0 so, {u, v, w} is a linearly dependent set.

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Def'n: a set of vectors S = { v₁, v₂, .... v_n } is linearly independent if k₁v₁ + k₂v₂ + ... + k_nv_n = 0 has only the trivial solution. ie: all k's = 0.

If k₁v₁ + k₂v₂ + .... + k_nv_n = 0 has other solutions, S is a linearly dependent set.

Example: these vectors are linearly independent because
only 0u + 0v = 0 -- so they satisfy the definition.

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Example: these vectors
are linearly dependent because 3v₁ + v₂ – v₃ = 0.

Note: If any vector in the set is a linear combination of other vectors, the set is dependent.
Here we can see that 3v₁ + v₂ = v₃

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A familiar set of linearly independent vectors is , denoted { i, j, k }, from which every 3-dimensional vector can be uniquely generated.

This set therefore is linearly independent and it spans R³, so it forms a BASIS FOR R³.

Put 'Em Together & What've You Got?? A Basis!

Def'n: If S = { v₁, v₂, .... v_n } is a finite set of vectors in any vector space V,
and if S is linearly independent and S spans V, then S is a basis for V.

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The "Dimension" of vector space V is the number of vectors in a basis for V.

The vectors { i, j, k } form the STANDARD BASIS FOR R³.

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Basis Theorems

-- if there are n vectors in a basis for V, then any set in V with more than n vectors is linearly dependent.

-- any 2 bases for a finite dimensional vector space have the same number of vectors.

-- any set that includes the zero vector is linearly dependent.

If S = { v₁, v₂, .... v_n } is a set of n vectors in an n-dimensional vector space V then:

a) if S is linearly independent, it is a basis for V.

b) if S spans V, it is a basis for V.

c) Any linearly independent set in V, with r vectors (r < n)
can be expanded to form a basis for V.

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The first 2 statements of the last theorem tell us we need only show either linear independence or spanning but not both if a set has the right number of vectors in it. If the number of vectors in the set equals the dimension of the Vector Space, then the set need ONLY span or be linearly independent to be a basis.

The 3rd statement says that if we have a set with fewer vectors than the dimension, there are vectors in the space that we can add to the set to form a basis for the space. The set of vectors might span a subspace of V.

If the situation is square, with n vectors from an n-dimensional vector space, we find the determinant of the system matrix. If det(A) = 0, the set is dependent so the vectors DO NOT form a basis for the space. If det(A) is not 0, the vectors DO form a basis for the space.

In the previous example we saw that
are linearly dependent because 3v₁ + v₂ – v₃ = 0. Since we have 3 vectors from R³, which creates a SQUARE MATRIX, we could conclude the same thing from the zero determinant of the coefficient or system matrix.

; det(A) = 0; so the column vectors are linearly dependent

lessons lnalg4 and lnalg4.2
on finding a basis for the row, column and null spaces
cover more on this topic
Practice is found in Assignment #5 and Test #3 (from Index Table)

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