1553 Exam 2 2.5-3.4

Linear Independence

If the only solution is x1=0 x1=0 and x2=0 x2=0, the vectors are linearly independent.

But if you can find a solution where, for example, x1=2x1=2 and x2=−1x2=−1, and the equation still equals the zero vector, the vectors are linearly dependent.

Easy way to remember

Independent = no vector is redundant.

Dependent = at least one vector can be made from a combination of the other vectors.

Linear Dependence

In other words, {v1, v2, . . . , vp} is linearly dependent if there exist numbers x1, x2, . . . , xp, not all equal to zero, such that x1v1 + x2v2 + · · · + xpvp = 0

So the vectors are linearly dependent, and an equation of linear dependence

is (taking z = 1), which results in the zero vector

The columns of a matrix are linearly independent if the equation has the trivial solution.

False

If is a set of linearly dependent vectors, then every vector in can be written as a linear combination of the other vectors in .

False

The columns of a matrix with dimensions mxn , where m

True

If a set of vectors contains fewer vectors than there are entries in the vectors, then the set must be linearly independent.

False

Two vectors are linearly dependent if and only if they are collinear.

True

Let A be a matrix with linearly independent columns.

The equation AX=b has a solution for all b precisely when it is a square matrix.

Let be a matrix with more rows than columns.

The columns of A could be either linearly dependent or linearly independent.

When there is a unique solution

Linearly Independent

This is true if and only if the matrix A has a pivot in each column.

Linearly Independent

Solving the matrix equation Ax = 0 will either

verify that the columns v1, v2, . . . , vp of A are linearly independent, or will produce a linear dependence relation.

A set of vectors {v1, v2, . . . , vp} is linearly dependent

if and only if one of the vectors is in the span of the other ones

A set of vectors {v1, v2, . . . , vp} is linearly dependent if and only if

you can remove one of the vectors without shrinking the span

A set of (nonzero) vectors is linearly independent if and only if, every time you add another vector to the set

the span gets bigger.

A subspace of R n is a subset V of R n satisfying:

. The zero vector is in V. "not empty" 2. If u and v are in V, then u + v is also in V. "closed under addition" 3. If u is in V and c is in R, then cu is in V. "closed under × scalars

Every subspace is a span, and

every span is a subspace

The line through any nonzero vector in V is also in V.

This means that if a vector vv is inside a subspace V, then every scalar multiple of that vector must also be inside V.

"If u,vu,v are in V, then cu+dv is in V."

This is saying that if two vectors are in a subspace, you can:

Multiply each vector by any number.

Add the resulting vectors together.

The answer must still be inside the subspace.

For example, if u,v∈Vu,v∈V, then

Closure under scalar multiplication:

Multiplying a vector in V by a scalar keeps it in V.

Closure under addition:

Adding vectors in VV keeps the result in VV.

Span{u,v}{u,v} is contained in V

Since every linear combination of u and v stays inside V, their entire span must be inside V.

If a subspace contains your vectors

it must also contain everything you can build from those vectors using scalar multiplication and addition.