Compute \begin{equation*} 4\colvector{2\\-3\\4\\1\\0}+ (-2)\colvector{1\\2\\-5\\2\\4}+ \colvector{-1\\3\\0\\1\\2} \end{equation*}
Solve the given vector equation for $x$, or explain why no solution exists: \begin{equation*} 3\colvector{1\\2\\-1}+ 4\colvector{2\\0\\x}= \colvector{11\\6\\17} \end{equation*}
Solve the given vector equation for $\alpha$, or explain why no solution exists: \begin{equation*} \alpha\colvector{1\\2\\-1}+ 4\colvector{3\\4\\2} = \colvector{-1\\0\\4} \end{equation*}
Solve the given vector equation for $\alpha$, or explain why no solution exists: \begin{equation*} \alpha\colvector{3\\2\\-2}+ \colvector{6\\1\\2} = \colvector{0\\-3\\6} \end{equation*}
Find $\alpha$ and $\beta$ that solve the vector equation. \begin{equation*} \alpha\colvector{1\\0}+ \beta\colvector{0\\1} = \colvector{3\\2} \end{equation*}
Find $\alpha$ and $\beta$ that solve the vector equation. \begin{equation*} \alpha\colvector{2\\1}+ \beta\colvector{1\\3 }= \colvector{5\\0} \end{equation*}
Provide reasons (mostly vector space properties) as justification for each of the seven steps of the following proof.
Theorem
For any vectors $\vect{u},\,\vect{v},\,\vect{w}\in\complex{m}$,
if $\vect{u} + \vect{v} = \vect{u} + \vect{w}$, then $\vect{v} = \vect{w}$.
Proof: Let $\vect{u},\,\vect{v},\,\vect{w}\in\complex{m}$,
and suppose $\vect{u} + \vect{v} = \vect{u} + \vect{w}$.
\begin{align*}
\vect{v}
&= \zerovector + \vect{v}\\
&= (-\vect{u} + \vect{u}) + \vect{v}\\
&= -\vect{u} + (\vect{u} + \vect{v})\\
&= -\vect{u} + (\vect{u} + \vect{w})\\
&= (-\vect{u} + \vect{u}) + \vect{w}\\
&= \zerovector + \vect{w}\\
&= \vect{w}
\end{align*}
Provide reasons (mostly vector space properties) as justification for each of the six steps of the following proof.
Theorem
For any vector $\vect{u}\in\complex{m}$, $0\vect{u}=\zerovector.$
Proof: Let $\vect{u}\in\complex{m}$.
\begin{align*}
\zerovector
&= 0\vect{u} +(-0\vect{u})\\
&= (0+0)\vect{u} + (-0\vect{u})\\
&= (0\vect{u}+0\vect{u}) + (-0\vect{u})\\
&= 0\vect{u} + (0\vect{u} + (-0\vect{u}))\\
&= 0\vect{u} + \zerovector\\
&= 0\vect{u}
\end{align*}
Provide reasons (mostly vector space properties) as justification for each of the six steps of the following proof.
Theorem
For any scalar $c$, $c\,\zerovector = \zerovector$.
Proof: Let $c$ be an arbitrary scalar.
\begin{align*}
\zerovector
&= c\zerovector + (-c\zerovector)\\
&= c(\zerovector + \zerovector) + (-c\zerovector)\\
&= (c\zerovector + c\zerovector) + (-c\zerovector)\\
&= c\zerovector + (c\zerovector + (-c\zerovector))\\
&= c\zerovector + \zerovector\\
&= c\zerovector
\end{align*}
Prove Property CC of Theorem VSPCV. Write your proof in the style of the proof of Property DSAC given in this section.
Prove Property SMAC of Theorem VSPCV. Write your proof in the style of the proof of Property DSAC given in this section.
Prove Property DVAC of Theorem VSPCV. Write your proof in the style of the proof of Property DSAC given in this section.
Exercises T30, T31 and T32 are about making a careful definition of “vector subtraction”.
Suppose $\vect{u}$ and $\vect{v}$ are two vectors in $\complex{m}$. Define a new operation, called “subtraction,” as the new vector denoted $\vect{u}-\vect{v}$ and defined by \begin{align*} \vectorentry{\vect{u}-\vect{v}}{i}=\vectorentry{\vect{u}}{i}-\vectorentry{\vect{v}}{i}&&1\leq i\leq m \end{align*} Prove that we can express the subtraction of two vectors in terms of our two basic operations. More precisely, prove that $\vect{u}-\vect{v}=\vect{u}+(-1)\vect{v}$. So in a sense, subtraction is not something new and different, but is just a convenience. Mimic the style of similar proofs in this section.
Prove, by giving counterexamples, that vector subtraction is not commutative and not associative.
Prove that vector subtraction obeys a distributive property. Specifically, prove that $\alpha\left(\vect{u}-\vect{v}\right)=\alpha\vect{u}-\alpha\vect{v}$.
Can you give two different proofs? Distinguish your two proofs by using the alternate descriptions of vector subtraction provided by Exercise VO.T30.