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Problem #37 from pg. 61 and problem #1 from pg. 68 are graded..
#37,pg.61: True or false; if false construct a specific counterexample, if true justify. If $\vec{v_1},\ldots,\vec{v_4}$ are in $\mathbb{R}^{4\times 1}$ and $\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}$ form a linearly dependent set, then $\{\vec{v}_1,\vec{v}_2,\vec{v}_3,\vec{v}_4\}$ also forms a linearly dependent set.
Solution: True. Linear dependence can be characterized as "one of the vectors can be written in terms of the others" (see Theorem 7, pg. 58). Hence if $\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}$ is linearly dependent, one of those can be written in terms of the others. Assume ("without loss of generality") that $\vec{v}_2$ can be written in terms of $\vec{v}_1$ and $\vec{v}_3$. In this case we know there exist weights $\alpha$ and $\beta$ so that
$$\vec{v}_3 = \alpha \vec{v}_1 + \beta \vec{v}_2.$$
Now consider the set $\{\vec{v}_1,\vec{v}_2,\vec{v}_3,\vec{v}_4\}$. Consider the following linear combination:
$$\alpha \vec{v}_1 + \beta \vec{v}_3 + 0 \vec{v}_4 = \alpha \vec{v}_1 + \beta\vec{v}_3= \vec{v}_2,$$
and we see that we can still "write one of the vectors in terms of the others". Hence the set $\{\vec{v}_1,\vec{v}_2,\vec{v}_3,\vec{v}_4\}$ is always a linearly dependent set if $\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}$ is, no matter what the vector $\vec{v}_4$ is .
#1, pg.68: Let $A = \left[ \begin{array}{ll}
2 & 0 \\
0 & 2
\end{array} \right]$ and define $T \colon \mathbb{R}^{2 \times 1} \rightarrow \mathbb{R}^{2 \times 1}$ by $T(\vec{x}) = A\vec{x}$. Find the images under $T$ of $\vec{u} = \left[ \begin{array}{ll}
1 \\
-3
\end{array} \right]$ and $\vec{v} = \left[ \begin{array}{ll}
a \\
b
\end{array} \right]$.
Solution: We compute directly:
$$\begin{array}{ll}
T(\vec{u}) &= \left[ \begin{array}{ll}
2 & 0 \\
0 & 2
\end{array} \right] \left[ \begin{array}{ll}
1 \\
-3
\end{array} \right] \\
&= 1 \left[ \begin{array}{ll}
2 \\
0
\end{array} \right] - 3 \left[ \begin{array}{ll}
0 \\
2
\end{array} \right] \\
&= \left[ \begin{array}{ll}
2 \\
-6,
\end{array} \right]
\end{array}$$
and
$$\begin{array}{ll}
T(\vec{v}) &= \left[ \begin{array}{ll}
2 & 0 \\
0 & 2
\end{array} \right] \left[ \begin{array}{ll}
a \\
b
\end{array} \right] \\
&= a \left[ \begin{array}{ll}
2 \\
0
\end{array} \right] + b \left[ \begin{array}{ll}
0 \\
2
\end{array} \right] \\
&= \left[ \begin{array}{ll}
2a \\
2b
\end{array} \right].
\end{array}$$
