Adam Panagos / Engineer / Lecturer

###### Applied Linear Algebra: Vectors

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###### Vectors is the next topic we investigate. We first look at how to compute the matrix product Ax = y. Then, we discuss a variety of topics such as the span of a set of vectors, linear combinations of vectors, and how to determine if vectors are linearly independent or dependent.

The Matrix Product Ax

3/1/15

Running Time: 6:20

This video provides several different examples of computing the matrix-vector product Ax where A is a matrix and x is a vector. This product is only defined when the number of columns in A equals the number of rows in x. When this is the case, the product Ax is defined as a linear combination of the columns of A where each column is weighted by the entries of the vector x.

Spanning Vectors #1

3/8/15

Running Time: 4:31

In this problem we work with the vectors v1 and v2 and determine if the set {v1, v2} spans R2.If the set of vectors {v1,v2} spans R2, then ANY vector from R2 can be written as a linear combination of these vectors. To see if this is true, an arbitrary vector from R2 is selected an and an augmented matrix is constructed and solved.

Spanning Vectors #2

3/8/15

Running Time: 4:27

In this problem we work with the vectors v1 and v2 and determine if the set {v1, v2} spans R2.If the set of vectors {v1,v2} spans R2, then ANY vector from R2 can be written as a linear combination of these vectors. To see if this is true, an arbitrary vector from R2 is selected an and an augmented matrix is constructed and solved. If there is no solution to this system of equations, then the set of vectors {v1,v2} does not span R2.Obviously, this approach can be extended to spaces with higher dimension.

Linear Combination of Vectors #1

3/8/15

Running Time: 3:37

Given the vectors v1, v2, and v3, we see if the vector b can be written as a linear combination of the vectors.This can be easily determined by constructing an augmented matrix, performing row operations, and finding the coefficients such that a1*v1 + a2*v2 + a3*v3 = b. If values for a1, a2, and a3 can be found, then b is a linear combination of {v1,v2,v3} and we say that b is in the Span{v1,v2,v3}. If the augmented matrix has no solution, then b is NOT a linear combination of the vectors.For this example, b CAN be written as a linear combination.

Linear Combination of Vectors #2

3/8/15

Running Time: 3:53

Given the vectors v1, v2, and v3, we see if the vector b can be written as a linear combination of the vectors.This can be easily determined by constructing an augmented matrix, performing row operations, and finding the coefficients such that a1*v1 + a2*v2 + a3*v3 = b. If values for a1, a2, and a3 can be found, then b is a linear combination of {v1,v2,v3} and we say that b is in the Span{v1,v2,v3}. If the augmented matrix has no solution, then b is NOT a linear combination of the vectors.For this example, b CANNOT be written as a linear combination.

Linearly Independent Vectors #1

3/19/15

Running Time: 4:59

Given a set of vectors we want to determine if they are linearly independent or not (i.e. linear dependent). A set of vectors is linearly independent when the linear combination of the vectors is equal to an all-zero vector only in the trivial case when all combining coefficients are zero.Thus, to determine if a set of vectors is linearly independent, we just have to construct and solve a homogeneous system of equations. If the only solution is the trivial solution, the vectors are linearly independent. If there are more solutions than just the trivial solution, the vectors are linearly dependent.

Linearly Independent Vectors #2

3/19/15

Running Time: 5:37

This example also works with a set of vectors, however, one of these vectors contains the variable “alpha” as one of its elements. We find the value of alpha that makes the set of vectors linearly dependent. A set of vectors is linearly dependent when their linear combination is equal to an all-zero vector without all combining coefficients being zero.So, to determine when these vectors are dependent, we just have to construct and solve a homogeneous system of equations. We row reduce the corresponding augmented matrix and find the value of alpha that results in a free variable. With a free variable in the solution, we have an infinite family of solutions, not just the trivial all-zero vector, resulting in a set of vectors that is linearly dependent.