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Linear Algebra: Help & Tutorials6 chapters | 44 lessons

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Lesson Transcript

Instructor:
*Damien Howard*

Damien has a master's degree in physics and has taught physics lab to college students.

In this lesson we'll start by reviewing matrix reduced row echelon form, which is integral to finding a basis of a vector space. Then we'll work through a problem together to see exactly how finding a basis is accomplished.

In math, we often work with sets, or collections, of expressions. For example, you could have an ordered set of numbers (a sequence) and have a problem that tells you to find the next number in the sequence. Another common example is working with a set of equations to solve the variables in them.

In linear algebra, you might find yourself working with a set of vectors. When the operations of scalar multiplication and vector addition hold for a set of vectors, we call it a **vector space**. When working with a vector space, one thing you might want to do is identify the vectors that form a basis for it. A vector space's **basis** is a subset of vectors within the space that are linearly independent and span the space.

A basis is **linearly independent** because the vectors in it cannot be defined as a linear combination of any of the other vectors in the basis. By **spanning** the vector space, we mean that the vectors in that space can be defined as a linear combination of the vectors in the basis.

Before we go on with vectors, we need to do a quick review of matrix operations. The most essential step to finding the basis of a vector space actually involves a matrix. More specifically, you'll need to be able to put a matrix in reduced row echelon form, which adheres to the following four conditions:

- All non-zero rows are above any rows containing only zeros.
- The first non-zero entry in a row is to the right of the first non-zero entry of the row above it.
- All elements in the same column as the first non-zero element in a row are zeroes.
- The first non-zero entry of each row is a one.

In order to get a matrix in reduced row echelon form, there are three different operations that can be performed on the rows of a matrix.

- You can swap any two rows within a matrix.
- A row in the matrix can be multiplied by a constant.
- A row can have a multiple of another row added to it.

The process of using these three operations to get a matrix into reduced row echelon form is called row reduction.

Now we have everything we need to find a basis of a vector space. Let's work through an example problem. Note that vector spaces can have multiple bases, and here we will find one possible basis for the following vector space.

In order to find a basis for this vector space, we start by putting these vectors into the columns of a matrix.

Once this is done, we use the row reduction operations we just went over to put the matrix into reduced row echelon form. This will be the most mathematically intensive step in finding the basis of a vector space.

At this point, you're almost done. We just need to do a little comparison of our reduced row echelon matrix to the original matrix we started with.

First, identify each column that has a leading one for a row in it. Then compare these columns in the reduced row echelon matrix to their corresponding columns in the original matrix. Finally, the vectors that make up the columns of that original matrix will be a basis for your vector space.

A **vector space** is a set, or collection, of vectors under which scalar multiplication and vector addition hold true. When working with a vector space, it can be useful to find a basis for it - a **basis** being a subset of vectors within that space that are **linearly independent** (cannot be viewed as a linear combination of one another) and **spanning** (the vectors in that vector space can be viewed as a linear combination of the vectors in the basis).

To find the basis of a vector space, start by taking the vectors in it and turning them into columns of a matrix. We then use row reduction to get this matrix in reduced row echelon form, for which the following four conditions must be met:

- All non-zero rows are above the rows in the matrix containing only zeros.
- The leading non-zero entry of every row is to the right of the leading non-zero entry of the row above it.
- The other elements in the same column as any leading non-zero entry for a row must all be zeroes.
- The leading non-zero entry of a row must be a one.

Then match the columns of the reduced row echelon matrix that contain non-zero leading row elements with their corresponding columns in the original matrix. The set of vectors that make up these columns in the original matrix is a basis for the vector space.

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Linear Algebra: Help & Tutorials6 chapters | 44 lessons

- Scalars and Vectors: Definition and Difference 3:23
- Performing Operations on Vectors in the Plane 5:28
- The Dot Product of Vectors: Definition & Application 6:21
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