Cancellation of scalars

mark

Let $V$ be a vector space, let $\vec{u},\vec{v}\in V$, let $\alpha\in\mathbb R$ be non-zero, and suppose that
$$
\alpha \vec{u} = \alpha \vec{v}.
$$
Show that $\vec{u}=\vec{v}$.

---

Comment

Try to use a linear string of justified equalities.

Student27

I'm assuming $\alpha$ is non-zero. If $\alpha=0$, I don't know if there is a meaningful solution.
Let $\alpha$ be non-zero. By property Z, $$\alpha\vec{u}=\alpha\vec{v} + 0$$
Thus, $$\alpha\vec{u}-\alpha\vec{v}=0$$
By distribution over scalar addition, this is equivalent to:
$$\alpha(\vec{u}-\vec{v})=0$$
Since alpha is non-zero, $\vec{u}-\vec{v}$ must be equal to 0, thus $\vec{u}=\vec{v}$.

mark

It seems that I'm getting a lot of questions about how to structure this linearly. All of the trial submissions for the HW I'm getting take a "balance the equations" approach. For this one (when written out carefully), that looks like so:

$$
\begin{align}
\alpha \vec{u} &= \alpha \vec{v} \\
\frac{1}{\alpha}\left(\alpha \vec{u}\right) &= \frac{1}{\alpha} \left(\alpha \vec{v}\right) \\
\left(\frac{1}{\alpha}\alpha\right) \vec{u} &= \left(\frac{1}{\alpha}\alpha\right) \vec{v} \\
1 \vec{u} &= 1 \vec{v} \\
\vec{u} &= \vec{v}.
\end{align}
$$

As it turns out, we can construct the linear string of equalitites that we want from here. Start at the lower left $\vec{u}$, follow the left side up (ignoring the first line), then turn right and follow the right side down to the $\vec{v}$, like so:

$$
\begin{align}
\frac{1}{\alpha}\left(\alpha \vec{u}\right) &\rightarrow \frac{1}{\alpha} \left(\alpha \vec{v}\right) \\
\uparrow &\cdots \downarrow \\
\left(\frac{1}{\alpha}\alpha\right) \vec{u} &= \left(\frac{1}{\alpha}\alpha\right) \vec{v} \\
\uparrow &\cdots \downarrow \\
1 \vec{u} &= 1 \vec{v} \\
\uparrow &\cdots \downarrow \\
\vec{u} &= \vec{v}.
\end{align}
$$

That should yield a string of equalities that looks like:

$$
\vec{u} = 1 \vec{u} = \left(\frac{1}{\alpha}\alpha\right) \vec{v} = \frac{1}{\alpha}\left(\alpha \vec{u}\right) = \frac{1}{\alpha} \left(\alpha \vec{v}\right) = \left(\frac{1}{\alpha}\alpha\right) \vec{v} = 1 \vec{v} = \vec{v}.
$$

We could write this out in a way that we justify each equality as follows:

$$
\begin{align}
\vec{u} &= 1 \vec{u} && \text{Property O on page 198} \\
&= \left(\frac{1}{\alpha}\alpha\right) \vec{v} && \text{Multiplicative inverses of scalars} \\
&= \frac{1}{\alpha}\left(\alpha \vec{u}\right) && \text{Associativity} \\
&= \frac{1}{\alpha}\left(\alpha \vec{v}\right) && \text{Hypothesis} \\
&= \frac{1}{\alpha}\left(\alpha \vec{v}\right) && \text{Associativity} \\
&= 1 \vec{v} && \text{Multiplicative inverses of scalars} \\
&= \vec{v} && \text{Property O on page 198}.
\end{align}
$$

To type that up in LaTeX we can use the align* environment:

\begin{align*}
  \vec{u} &= 1 \vec{u} && \text{Property O on page 198} \\
    &= \left(\frac{1}{\alpha}\alpha\right) \vec{v} && \text{Multiplicative inverses of scalars} \\
    &= \frac{1}{\alpha}\left(\alpha \vec{u}\right) && \text{Associativity} \\
    &= \frac{1}{\alpha}\left(\alpha \vec{v}\right) && \text{Hypothesis} \\
    &= \frac{1}{\alpha}\left(\alpha \vec{v}\right) && \text{Associativity} \\
    &= 1 \vec{v} && \text{Multiplicative inverses of scalars}  \\
    &= \vec{v}  && \text{Property O on page 198}.
\end{align*}

Note that the alignment symbol is indicated with an ampersand (&) and the space is indicated with a double ampersand (&&). That should compile just fine in straight up LaTeX. When posting to the forum, don't forget to wrap it in double dollar signs ($$ ... $$) and that line breaks need four slashes (\\\\). Also, you should use align, rather than align*.