For Problems 15-18, determine whether the given set S of...

For Problems $15-18$, determine whether the given set $S$ of vectors is a basis for $M_{m \times n}(\mathbb{R})$. $$ \begin{array}{l} m=n=2: S=\left\{\left[\begin{array}{rr} -2 & -8 \\ 1 & 4 \end{array}\right],\left[\begin{array}{rr} 0 & 1 \\ -1 & 1 \end{array}\right],\right. \\ \left.\left[\begin{array}{rr} -5 & 0 \\ 5 & -4 \end{array}\right],\left[\begin{array}{ll} 3 & -2 \\ 4 & -1 \end{array}\right]\right\} \end{array} $$

For Problems $15-18$, determine whether the given set $S$ of vectors is a basis for $M_{m \times n}(\mathbb{R})$.
$$
\begin{array}{l}
m=n=2: S=\left\{\left[\begin{array}{rr}
-2 & -8 \\
1 & 4
\end{array}\right],\left[\begin{array}{rr}
0 & 1 \\
-1 & 1
\end{array}\right],\right. \\
\left.\left[\begin{array}{rr}
-5 & 0 \\
5 & -4
\end{array}\right],\left[\begin{array}{ll}
3 & -2 \\
4 & -1
\end{array}\right]\right\}
\end{array}
$$

Key Concepts

Vector Space

A vector space is a collection of objects, called vectors, which can be added together and scaled by numbers (scalars) while satisfying a list of axioms (such as associativity, commutativity, etc.). In the context of matrices, the set of all m×n matrices over ? forms a vector space with standard matrix addition and scalar multiplication.

Basis

A basis of a vector space is a set of vectors that is both linearly independent and spans the entire space. This means every element in the vector space can be uniquely expressed as a linear combination of the basis vectors, and the basis set is minimal in the sense that removing any vector would cause the loss of the spanning property.

Linear Independence

A set of vectors is said to be linearly independent if the only solution to a linear combination of these vectors equaling the zero vector is when all the coefficients in the combination are zero. This property ensures that none of the vectors in the set can be written as a linear combination of the others, which is essential for forming a basis.

Spanning Set

A spanning set of a vector space is a collection of vectors such that any vector in the space can be expressed as a linear combination of the vectors in the set. If a set spans the vector space and is also linearly independent, it forms a basis.

Dimension

The dimension of a vector space is defined as the number of vectors in a basis for that space. For example, the space of all m×n matrices over ? has dimension m·n because each entry of the matrix can be independently assigned, corresponding to m·n degrees of freedom.

Transcript

00:01 Okay, this time they gave us four matrices and they want us to know if they span the vector space m2r.

00:08 Now this is a better question to solve.

00:10 What do we know from our theory? well, if we have four vectors, let's say v1, b2, b3, v4, and we're guessing that they span a four -dimensional vector space, then you should be able to write the zero vector uniquely.

00:29 So what we can do is solve this as a linear system if there are no coefficients like a, b, c, d, such that when you multiply these, add them together, you get zero.

00:47 That aren't trivial, just zero.

00:49 Then they're definitely independent.

00:51 So that's what we're going to do.

00:53 Translate them into vectors, negative 2, negative 8, 1.

01:03 01 -1 -1 -1 -1 -9 -5 -0 -5 -negative 4 3 -negative 2 -4 -1 of negative 1 so let's see do these does this correspond to my first matrices yes what i did was i took the first two row entries wrote them as the first two vector entries and the second rows entries as the second two entries of the vector okay, and then we want to solve for the zero vector in terms of these.

01:41 Hopefully we should just get the trivial solution because that would mean they're independent and a vectoring.

01:47 So i'm going to choose three is my pivot row since there's a one right there.

01:54 And what i'm going to do is replace the first row by adding two times our pivot row.

02:09 Then i'm going to add 8 times our pivot row to the second row.

02:18 We get to leave our pivot row alone.

02:22 And i'm going to subtract 4 times the pivot row to row 4.

02:36 Okay, let's move on to the pivot.

02:44 So we have a 0 -0 here.

02:50 And i'm going to choose row 1 is our pivot row this time by dividing the whole thing by negative half...

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