In the previous quiz, we explored the concept of sets of vectors *spanning* a vector space, meaning that every vector in the vector space can be written as a linear combination of the vectors in the span.

As we saw, there are multiple ways to span a vector space--for instance, both \(\left\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\right\}\) and \(\left\{\begin{pmatrix}4\\2\end{pmatrix}, \begin{pmatrix}2\\4\end{pmatrix}\right\}\) span the vector space \(\mathbb{R}^2\). Additionally, we saw that sometimes vectors in the span are “redundant,” meaning that the set would still span the vector space if it were removed.

In this quiz, we explore this idea further, looking at “minimal” spanning sets. As we’ll see, this leads to a number of related concepts that formalize some intuition we’ve previously seen.

The last few problems motivate a formal definition of “redundancy”: a set of vectors \(\{v_1, \ldots, v_n\}\) is called **linearly dependent** if there exist constants \(c_1, \ldots, c_n\), not all 0, such that
\[c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0\]
If no such constants exist, the vectors are called **linearly independent**.

For example, as we saw in the previous problem, the vectors \[\begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}1\\3\\4\end{pmatrix}, \begin{pmatrix}3\\8\\11\end{pmatrix}\] are linearly dependent. However, are the vectors \[\begin{pmatrix}0\\1\end{pmatrix}, \begin{pmatrix}1\\0\end{pmatrix}\] linearly independent? Why?

Suppose we have the equation
\[x + 2y = 3.\]
As we’ve seen before, this corresponds to the row vector \(\begin{pmatrix}1&2&3\end{pmatrix}\). Clearly, there are an infinite number of solutions to this simple equation. Suppose we did **one** of the following:

1)Added the row vector \(\begin{pmatrix}2&6&4\end{pmatrix}\) \((\)i.e. the equation \(2x + 6y = 4).\)

2)Added the row vector \(\begin{pmatrix}2&4&6\end{pmatrix}\) \((\)i.e. the equation \(2x + 4y = 6).\)

What would happen?

**A)** In both cases, the system would have a unique solution.

**B)** In case 1, the system would have a unique solution, and in case 2, the system would still have infinite solutions.

**C)** In case 1, the system would still have infinite solutions, and in case 2, the system would have a unique solution.

**D)** In both cases, the system would still have infinite solutions.

**not** have a unique solution?

Armed with these ideas, let’s revisit an old topic: **degrees of freedom**. Previously, we had to rely on the intuitive notion of “number of variables that could be anything,” e.g. the equation \(x + y + z = 6\) has 2 degrees of freedom \((\)since \(x\) and \(y\) “could be anything,” but \(z\) depends on those two\().\)

Now we can formalize this idea. The number of degrees of freedom is the number of variables, minus one for each linearly independent equation. A set of equations \( e_{1} , \dots , e_{m} \) is linearly independent if and only if the set cannot be expressed as \( a_{1} e_{1} + \dots + a_{m} e_{m} = 0 \) with constants \(a_{1} , \dots , a_{n} \) which are not all 0. This makes sense with our intuitive notion from before: with no equations, obviously each variable could be anything, and each equation “usually” reduces the number of degrees of freedom by 1.

Consider the system \[ \begin{align*} x + y + 2z &= 4 \\ 2x + y + z &= 4 \\ 5x + 3y + 4z &= 12. \end{align*} \] How many degrees of freedom does this system have?

In this chapter, we expanded on the idea of spanning sets, focusing on *minimal* spanning sets and how to find them. This leads naturally to the idea of linear independence, which in turn naturally leads to the idea of degrees of freedom. As we saw, adding equations (i.e. constraints) “usually” reduces the number of degrees of freedom by 1, but if the equation is “redundant”--or linearly dependent with the others--the number of degrees of freedom stays the same.

In the next chapter, we’ll take this idea further, exploring finding minimal spanning sets that span a given vector space.

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