Challenge Problems for Chapter 3

Suppose \(\{\vec {v}_{1}, \dots , \vec {v}_{m}\}\) is a linearly independent set in \(\RR ^n\), and that \(\vec {w}\) is not in \(\mbox {span}\left (\vec {v}_{1}, \dots , \vec {v}_{m}\right )\).
1.
Is \(\vec {w}\) in \(\mbox {span}\left (\vec {v}_{1}+\vec {w}, \dots , \vec {v}_{m}+\vec {w}\right )\) YES, NO
Suppose \(\vec {w}\) is in \(\mbox {span}\left (\vec {v}_{1}+\vec {w}, \dots , \vec {v}_{m}+\vec {w}\right )\). Then we can write
\begin{align*} \vec {w} &= a_1 (\vec {v}_{1}+\vec {w}) + \dots + a_m (\vec {v}_{m}+\vec {w}) \\ \vec {w} &= a_1\vec {v}_{1} + \dots + a_m\vec {v}_{m}+ (a_1 + \dots + a_m)\vec {w}. \end{align*}

Now consider two cases separately: either \(a_1 + \dots + a_m = 0\) or \(a_1 + \dots + a_m \ne 0\). In either case, arrive at a contradiction and conclude that \(\vec {w}\) is not in \(\mbox {span}\left (\vec {v}_{1}+\vec {w}, \dots , \vec {v}_{m}+\vec {w}\right )\).

2.
Is \(\{\vec {v}_{1}+\vec {w}, \dots , \vec {v}_{m}+\vec {w}\}\) linearly independent? YES, NO
If you assume linear dependence, you should be able to show \(\vec {w}\) is in the span of the original set, which is a contradiction.
Suppose \(\vec {n}_1\), \(\vec {n}_2\), and \(\vec {n}_3\) are the rows of the \(3 \times 3\) matrix \(A\). Then we can interpret the solution to the system of equations \([A|\vec {0}]\) as the intersection of three planes containing the origin. Discuss what this intersection would look like geometrically if the reduced row echelon form of \([A|\vec {0}]\) is of the form:
1.
\begin{equation*} \left [ \begin{array}{ccc|c} 1 & 0 & * & 0 \\ 0 & 1 & * & 0 \\ 0 & 0 & 0 & 0 \end{array} \right ] \end{equation*}
2.
\begin{equation*} \left [ \begin{array}{ccc|c} 1 & * & * & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right ] \end{equation*}
3.
Are there any other possibilities?
Show that for any matrix \(A\), \(\text {rref}(A)\) has a row of zeros if and only if one of the rows of \(A\) can be expressed as a linear combination of the others.
Recall that a plane in \(\RR ^3\) has an equation of the form \(ax+by+cz=d\), where \(a\), \(b\), \(c\) are the components of a normal vector to the plane. Suppose three planes intersect in a line, as shown below.

[Picture]

We can find the line of intersection by solving a system of equations. Suppose the augmented matrix \([A | \vec {d}]\) corresponds to this system.

1.
Explain why \(\text {rref}[A | \vec {d}]\) will have a row of zeros.
2.
Consider the normal vectors to the three planes. What geometric property of these particular three normal vectors explains the row of zeros in the reduced row-echelon form?
All three normal vectors lie in the same plane, so one of the normal vectors (rows) must be a linear combination of the other two.

Bibliography

The first problem came from the end of Chapter 1 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, pp. 33–34.