# Linear Algebra/Basis/Solutions
## Solutions

    This exercise is recommended for all readers.

### Problem 1

Decide if each is a basis for $\mathbb {R} ^{3}$.
* $\langle {\begin{pmatrix}1\\2\\3\end{pmatrix}},{\begin{pmatrix}3\\2\\1\end{pmatrix}},{\begin{pmatrix}0\\0\\1\end{pmatrix}}\rangle$
* $\langle {\begin{pmatrix}1\\2\\3\end{pmatrix}},{\begin{pmatrix}3\\2\\1\end{pmatrix}}\rangle$
* $\langle {\begin{pmatrix}0\\2\\-1\end{pmatrix}},{\begin{pmatrix}1\\1\\1\end{pmatrix}},{\begin{pmatrix}2\\5\\0\end{pmatrix}}\rangle$
* $\langle {\begin{pmatrix}0\\2\\-1\end{pmatrix}},{\begin{pmatrix}1\\1\\1\end{pmatrix}},{\begin{pmatrix}1\\3\\0\end{pmatrix}}\rangle$

### Answer

By Theorem 1.12, each is a basis if and only if each vector in the space can be given in a unique way as a linear combination of the given vectors.
* Yes this is a basis. The relation gives which has the unique solution $c_{3}=x-2y+z$, $c_{2}=x/2-y/4$, and $c_{1}=-x/2+3y/4$.
    $c_{1}{\begin{pmatrix}1\\2\\3\end{pmatrix}}+c_{2}{\begin{pmatrix}3\\2\\1\end{pmatrix}}+c_{3}{\begin{pmatrix}0\\0\\1\end{pmatrix}}={\begin{pmatrix}x\\y\\z\end{pmatrix}}$
    $\left({\begin{array}{*{3}{c}|c}1&3&0&x\\2&2&0&y\\3&1&1&z\end{array}}\right){\xrightarrow[{-3\rho _{1}+\rho _{3}}]{-2\rho _{1}+\rho _{2}}}\;{\xrightarrow[]{2\rho _{2}+\rho _{3}}}\left({\begin{array}{*{3}{c}|c}1&3&0&x\\0&-4&0&-2x+y\\0&0&1&x-2y+z\end{array}}\right)$
* This is not a basis. Setting it up as in the prior item gives a linear system whose solution is possible if and only if the three-tall vector's components $x$, $y$, and $z$ satisfy $x-2y+z=0$. For instance, we can find the coefficients $c_{1}$ and $c_{2}$ that work when $x=1$, $y=1$, and $z=1$. However, there are no $c$'s that work for $x=1$, $y=1$, and $z=2$. Thus this is not a basis; it does not span the space.
    $c_{1}{\begin{pmatrix}1\\2\\3\end{pmatrix}}+c_{2}{\begin{pmatrix}3\\2\\1\end{pmatrix}}={\begin{pmatrix}x\\y\\z\end{pmatrix}}$
    $\left({\begin{array}{*{2}{c}|c}1&3&x\\2&2&y\\3&1&z\end{array}}\right){\xrightarrow[{-3\rho _{1}+\rho _{3}}]{-2\rho _{1}+\rho _{2}}}\;{\xrightarrow[]{2\rho _{2}+\rho _{3}}}\left({\begin{array}{*{2}{c}|c}1&3&x\\0&-4&-2x+y\\0&0&x-2y+z\end{array}}\right)$
* Yes, this is a basis. Setting up the relationship leads to this reduction which has a unique solution for each triple of components $x$, $y$, and $z$.
    $\left({\begin{array}{*{3}{c}|c}0&1&2&x\\2&1&5&y\\-1&1&0&z\end{array}}\right){\xrightarrow[]{rho_{1}\leftrightarrow \rho _{3}}}\;{\xrightarrow[]{\rho _{1}+\rho _{2}}}\;{\xrightarrow[]{(1/3)\rho _{2}+\rho _{3}}}\left({\begin{array}{*{3}{c}|c}-1&1&0&z\\0&3&5&y+2z\\0&0&1/3&x-y/3-2z/3\end{array}}\right)$
* No, this is not a basis. The reduction which does not have a solution for each triple $x$, $y$, and $z$. Instead, the span of the given set includes only those three-tall vectors where $x=y/3+2z/3$.
    $\left({\begin{array}{*{3}{c}|c}0&1&1&x\\2&1&3&y\\-1&1&0&z\end{array}}\right){\xrightarrow[]{rho_{1}\leftrightarrow \rho _{3}}}\;{\xrightarrow[]{2\rho _{1}+\rho _{2}}}{\xrightarrow[]{-1/3)\rho _{2}+\rho _{3}}}\left({\begin{array}{*{3}{c}|c}-1&1&0&z\\0&3&3&y+2z\\0&0&0&x-y/3-2z/3\end{array}}\right)$
    This exercise is recommended for all readers.

### Problem 2

Represent the vector with respect to the basis.
* ${\begin{pmatrix}1\\2\end{pmatrix}}$, $B=\langle {\begin{pmatrix}1\\1\end{pmatrix}},{\begin{pmatrix}-1\\1\end{pmatrix}}\rangle \subseteq \mathbb {R} ^{2}$
* $x^{2}+x^{3}$, $D=\langle 1,1+x,1+x+x^{2},1+x+x^{2}+x^{3}\rangle \subseteq {\mathcal {P}}_{3}$
* ${\begin{pmatrix}0\\-1\\0\\1\end{pmatrix}}$, ${\mathcal {E}}_{4}\subseteq \mathbb {R} ^{4}$

### Answer

* We solve with and conclude that $c_{2}=1/2$ and so $c_{1}=3/2$. Thus, the representation is this.
    $c_{1}{\begin{pmatrix}1\\1\end{pmatrix}}+c_{2}{\begin{pmatrix}-1\\1\end{pmatrix}}={\begin{pmatrix}1\\2\end{pmatrix}}$
    $\left({\begin{array}{*{2}{c}|c}1&-1&1\\1&1&2\end{array}}\right){\xrightarrow[]{\rho _{1}+\rho _{2}}}\left({\begin{array}{*{2}{c}|c}1&-1&1\\0&2&1\end{array}}\right)$
    ${\rm {Rep}}_{B}({\begin{pmatrix}1\\2\end{pmatrix}})={\begin{pmatrix}3/2\\1/2\end{pmatrix}}_{B}$
* The relationship $c_{1}\cdot (1)+c_{2}\cdot (1+x)+c_{3}\cdot (1+x+x^{2})+c_{4}\cdot (1+x+x^{2}+x^{3})=x^{2}+x^{3}$ is easily solved by eye to give that $c_{4}=1$, $c_{3}=0$, $c_{2}=-1$, and $c_{1}=0$.
    ${\rm {Rep}}_{D}(x^{2}+x^{3})={\begin{pmatrix}0\\-1\\0\\1\end{pmatrix}}_{D}$
* ${\rm {Rep}}_{{\mathcal {E}}_{4}}({\begin{pmatrix}0\\-1\\0\\1\end{pmatrix}})={\begin{pmatrix}0\\-1\\0\\1\end{pmatrix}}_{{\mathcal {E}}_{4}}$

### Problem 3

Find a basis for ${\mathcal {P}}_{2}$, the space of all quadratic polynomials. Must any such basis contain a polynomial of each degree:~degree zero, degree one, and degree two?

### Answer

One basis is $\langle 1,x,x^{2}\rangle$. There are bases for ${\mathcal {P}}_{2}$ that do not contain any polynomials of degree one or degree zero. One is $\langle 1+x+x^{2},x+x^{2},x^{2}\rangle$. (Every basis has at least one polynomial of degree two, though.)

### Problem 4

Find a basis for the solution set of this system.
    ${\begin{array}{*{4}{rc}r}x_{1}&-&4x_{2}&+&3x_{3}&-&x_{4}&=&0\\2x_{1}&-&8x_{2}&+&6x_{3}&-&2x_{4}&=&0\end{array}}$

### Answer

The reduction
    $\left({\begin{array}{*{4}{c}|c}1&-4&3&-1&0\\2&-8&6&-2&0\end{array}}\right){\xrightarrow[]{2\rho _{1}+\rho _{2}}}\left({\begin{array}{*{4}{c}|c}1&-4&3&-1&0\\0&0&0&0&0\end{array}}\right)$
gives that the only condition is that $x_{1}=4x_{2}-3x_{3}+x_{4}$. The solution set is
    $\{{\begin{pmatrix}4x_{2}-3x_{3}+x_{4}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix}}\,{\big |}\,x_{2},x_{3},x_{4}\in \mathbb {R} \}=\{x_{2}{\begin{pmatrix}4\\1\\0\\0\end{pmatrix}}+x_{3}{\begin{pmatrix}-3\\0\\1\\0\end{pmatrix}}+x_{4}{\begin{pmatrix}1\\0\\0\\1\end{pmatrix}}\,{\big |}\,x_{2},x_{3},x_{4}\in \mathbb {R} \}$
and so the obvious candidate for the basis is this.
    $\langle {\begin{pmatrix}4\\1\\0\\0\end{pmatrix}},{\begin{pmatrix}-3\\0\\1\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\0\\1\end{pmatrix}}\rangle$
We've shown that this spans the space, and showing it is also linearly independent is routine.
    This exercise is recommended for all readers.

### Problem 5

Find a basis for ${\mathcal {M}}_{2\!\times \!2}$, the space of $2\!\times \!2$ matrices.

### Answer

There are many bases. This is an easy one.
    $\langle {\begin{pmatrix}1&0\\0&0\end{pmatrix}},{\begin{pmatrix}0&1\\0&0\end{pmatrix}},{\begin{pmatrix}0&0\\1&0\end{pmatrix}},{\begin{pmatrix}0&0\\0&1\end{pmatrix}}\rangle$
    This exercise is recommended for all readers.

### Problem 6

Find a basis for each.
* The subspace $\{a_{2}x^{2}+a_{1}x+a_{0}\,{\big |}\,a_{2}-2a_{1}=a_{0}\}$ of ${\mathcal {P}}_{2}$
* The space of three-wide row vectors whose first and second components add to zero
* This subspace of the $2\!\times \!2$ matrices
    $\{{\begin{pmatrix}a&b\\0&c\end{pmatrix}}\,{\big |}\,c-2b=0\}$

### Answer

For each item, many answers are possible.
* One way to proceed is to parametrize by expressing the $a_{2}$ as a combination of the other two $a_{2}=2a_{1}+a_{0}$. Then $a_{2}x^{2}+a_{1}x+a_{0}$ is $(2a_{1}+a_{0})x^{2}+a_{1}x+a_{0}$ and suggests $\langle 2x^{2}+x,x^{2}+1\rangle$. This only shows that it spans, but checking that it is linearly independent is routine.
    $\{(2a_{1}+a_{0})x^{2}+a_{1}x+a_{0}\,{\big |}\,a_{1},a_{0}\in \mathbb {R} \}=\{a_{1}\cdot (2x^{2}+x)+a_{0}\cdot (x^{2}+1)\,{\big |}\,a_{1},a_{0}\in \mathbb {R} \}$
* Parametrize $\{{\begin{pmatrix}a&b&c\end{pmatrix}}\,{\big |}\,a+b=0\}$ to get $\{{\begin{pmatrix}-b&b&c\end{pmatrix}}\,{\big |}\,b,c\in \mathbb {R} \}$, which suggests using the sequence $\langle {\begin{pmatrix}-1&1&0\end{pmatrix}},{\begin{pmatrix}0&0&1\end{pmatrix}}\rangle$. We've shown that it spans, and checking that it is linearly independent is easy.
* Rewriting suggests this for the basis.
    $\{{\begin{pmatrix}a&b\\0&2b\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {R} \}=\{a\cdot {\begin{pmatrix}1&0\\0&0\end{pmatrix}}+b\cdot {\begin{pmatrix}0&1\\0&2\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {R} \}$
    $\langle {\begin{pmatrix}1&0\\0&0\end{pmatrix}},{\begin{pmatrix}0&1\\0&2\end{pmatrix}}\rangle$

### Problem 7

Check Example 1.6.

### Answer

We will show that the second is a basis; the first is similar. We will show this straight from the definition of a basis, because this example appears before Theorem 1.12.
To see that it is linearly independent, we set up $c_{1}\cdot (\cos \theta -\sin \theta )+c_{2}\cdot (2\cos \theta +3\sin \theta )=0\cos \theta +0\sin \theta$. Taking $\theta =0$ and $\theta =\pi /2$ gives this system
    ${\begin{array}{*{2}{rc}r}c_{1}\cdot 1&+&c_{2}\cdot 2&=&0\\c_{1}\cdot (-1)&+&c_{2}\cdot 3&=&0\end{array}}\;{\xrightarrow[]{\rho _{1}+\rho _{2}}}\;{\begin{array}{*{2}{rc}r}c_{1}&+&2c_{2}&=&0\\&+&5c_{2}&=&0\end{array}}$
which shows that $c_{1}=0$ and $c_{2}=0$.
The calculation for span is also easy; for any $x,y\in \mathbb {R}$, we have that $c_{1}\cdot (\cos \theta -\sin \theta )+c_{2}\cdot (2\cos \theta +3\sin \theta )=x\cos \theta +y\sin \theta$ gives that $c_{2}=x/5+y/5$ and that $c_{1}=3x/5-2y/5$, and so the span is the entire space.
    This exercise is recommended for all readers.

### Problem 8

Find the span of each set and then find a basis for that span.
* $\{1+x,1+2x\}$ in ${\mathcal {P}}_{2}$
* $\{2-2x,3+4x^{2}\}$ in ${\mathcal {P}}_{2}$

### Answer

* Asking which $a_{0}+a_{1}x+a_{2}x^{2}$ can be expressed as $c_{1}\cdot (1+x)+c_{2}\cdot (1+2x)$ gives rise to three linear equations, describing the coefficients of $x^{2}$, $x$, and the constants. Gauss' method with back-substitution shows, provided that $a_{2}=0$, that $c_{2}=-a_{0}+a_{1}$ and $c_{1}=2a_{0}-a_{1}$. Thus, with $a_{2}=0$, we can compute appropriate $c_{1}$ and $c_{2}$ for any $a_{0}$ and $a_{1}$. So the span is the entire set of linear polynomials $\{a_{0}+a_{1}x\,{\big |}\,a_{0},a_{1}\in \mathbb {R} \}$. Parametrizing that set $\{a_{0}\cdot 1+a_{1}\cdot x\,{\big |}\,a_{0},a_{1}\in \mathbb {R} \}$ suggests a basis $\langle 1,x\rangle$ (we've shown that it spans; checking linear independence is easy).
    ${\begin{array}{*{2}{rc}r}c_{1}&+&c_{2}&=&a_{0}\\c_{1}&+&2c_{2}&=&a_{1}\\&&0&=&a_{2}\end{array}}$
* With we get this system. Thus, the only quadratic polynomials $a_{0}+a_{1}x+a_{2}x^{2}$ with associated $c$'s are the ones such that $0=(-4/3)a_{0}-(4/3)a_{1}+a_{2}$. Hence the span is $\{(-a_{1}+(3/4)a_{2})+a_{1}x+a_{2}x^{2}\,{\big |}\,a_{1},a_{2}\in \mathbb {R} \}$. Parametrizing gives $\{a_{1}\cdot (-1+x)+a_{2}\cdot ((3/4)+x^{2})\,{\big |}\,a_{1},a_{2}\in \mathbb {R} \}$, which suggests $\langle -1+x,(3/4)+x^{2}\rangle$ (checking that it is linearly independent is routine).
    $a_{0}+a_{1}x+a_{2}x^{2}=c_{1}\cdot (2-2x)+c_{2}\cdot (3+4x^{2})=(2c_{1}+3c_{2})+(-2c_{1})x+(4c_{2})x^{2}$
    ${\begin{array}{*{2}{rc}r}2c_{1}&+&3c_{2}&=&a_{0}\\-2c_{1}&&&=&a_{1}\\&&4c_{2}&=&a_{2}\end{array}}\;{\xrightarrow[]{\rho _{1}+\rho _{2}}}\;{\xrightarrow[]{-4/3)\rho _{2}+\rho _{3}}}\;{\begin{array}{*{2}{rc}r}2c_{1}&+&3c_{2}&=&a_{0}\\&&3c_{2}&=&a_{0}+a_{1}\\&&0&=&(-4/3)a_{0}-(4/3)a_{1}+a_{2}\end{array}}$
    This exercise is recommended for all readers.

### Problem 9

Find a basis for each of these subspaces of the space ${\mathcal {P}}_{3}$ of cubic polynomials.
* The subspace of cubic polynomials $p(x)$ such that $p(7)=0$
* The subspace of polynomials $p(x)$ such that $p(7)=0$ and $p(5)=0$
* The subspace of polynomials $p(x)$ such that $p(7)=0$, $p(5)=0$, and~$p(3)=0$
* The space of polynomials $p(x)$ such that $p(7)=0$, $p(5)=0$, $p(3)=0$, and~$p(1)=0$

### Answer

* The subspace is $\{a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\,{\big |}\,a_{0}+7a_{1}+49a_{2}+343a_{3}=0\}$. Rewriting $a_{0}=-7a_{1}-49a_{2}-343a_{3}$ gives $\{(-7a_{1}-49a_{2}-343a_{3})+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\,{\big |}\,a_{1},a_{2},a_{3}\in \mathbb {R} \}$, which, on breaking out the parameters, suggests $\langle -7+x,-49+x^{2},-343+x^{3}\rangle$ for the basis (it is easily verified).
* The given subspace is the collection of cubics $p(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}$ such that $a_{0}+7a_{1}+49a_{2}+343a_{3}=0$ and $a_{0}+5a_{1}+25a_{2}+125a_{3}=0$. Gauss' method gives that $a_{1}=-12a_{2}-109a_{3}$ and that $a_{0}=35a_{2}+420a_{3}$. Rewriting $(35a_{2}+420a_{3})+(-12a_{2}-109a_{3})x+a_{2}x^{2}+a_{3}x^{3}$ as $a_{2}\cdot (35-12x+x^{2})+a_{3}\cdot (420-109x+x^{3})$ suggests this for a basis $\langle 35-12x+x^{2},420-109x+x^{3}\rangle$. The above shows that it spans the space. Checking it is linearly independent is routine. (Comment. A worthwhile check is to verify that both polynomials in the basis have both seven and five as roots.)
    ${\begin{array}{*{4}{rc}r}a_{0}&+&7a_{1}&+&49a_{2}&+&343a_{3}&=&0\\a_{0}&+&5a_{1}&+&25a_{2}&+&125a_{3}&=&0\end{array}}\;{\xrightarrow[]{\rho _{1}+\rho _{2}}}\;{\begin{array}{*{4}{rc}r}a_{0}&+&7a_{1}&+&49a_{2}&+&343a_{3}&=&0\\&&-2a_{1}&-&24a_{2}&-&218a_{3}&=&0\end{array}}$
* Here there are three conditions on the cubics, that $a_{0}+7a_{1}+49a_{2}+343a_{3}=0$, that $a_{0}+5a_{1}+25a_{2}+125a_{3}=0$,and that $a_{0}+3a_{1}+9a_{2}+27a_{3}=0$. Gauss' method yields the single free variable $a_{3}$, with $a_{2}=-15a_{3}$, $a_{1}=71a_{3}$, and $a_{0}=-105a_{3}$. The parametrization is this. Therefore, a good candidate for the basis is $\langle -105+71x-15x^{2}+x^{3}\rangle$. It spans the space by the work above. It is clearly linearly independent because it is a one-element set (with that single element not the zero object of the space). Thus, any cubic through the three points $(7,0)$, $(5,0)$, and $(3,0)$ is a multiple of this one. (Comment. As in the prior question, a worthwhile check is to verify that plugging seven, five, and three into this polynomial yields zero each time.)
    ${\begin{array}{*{4}{rc}r}a_{0}&+&7a_{1}&+&49a_{2}&+&343a_{3}&=&0\\a_{0}&+&5a_{1}&+&25a_{2}&+&125a_{3}&=&0\\a_{0}&+&3a_{1}&+&9a_{2}&+&27a_{3}&=&0\end{array}}\;{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-\rho _{1}+\rho _{2}}}\;{\xrightarrow[]{2\rho _{2}+\rho _{3}}}\;{\begin{array}{*{4}{rc}r}a_{0}&+&7a_{1}&+&49a_{2}&+&343a_{3}&=&0\\&&-2a_{1}&-&24a_{2}&-&218a_{3}&=&0\\&&&&8a_{2}&+&120a_{3}&=&0\end{array}}$
    $\{(-105a_{3})+(71a_{3})x+(-15a_{3})x^{2}+(a_{3})x^{3}\,{\big |}\,a_{3}\in \mathbb {R} \}=\{a_{3}\cdot (-105+71x-15x^{2}+x^{3})\,{\big |}\,a_{3}\in \mathbb {R} \}$
* This is the trivial subspace of ${\mathcal {P}}_{3}$. Thus, the basis is empty $\langle \rangle$.
Remark. The polynomial in the third item could alternatively have been derived by multiplying out $(x-7)(x-5)(x-3)$.

### Problem 10

We've seen that it is possible for a basis to remain a basis when it is reordered. Must it always remain a basis?

### Answer

Yes. Linear independence and span are unchanged by reordering.

### Problem 11

Can a basis contain a zero vector?

### Answer

No linearly independent set contains a zero vector.
    This exercise is recommended for all readers.

### Problem 12

Let $\langle {\vec {\beta }}_{1},{\vec {\beta }}_{2},{\vec {\beta }}_{3}\rangle$ be a basis for a vector space.
* Show that $\langle c_{1}{\vec {\beta }}_{1},c_{2}{\vec {\beta }}_{2},c_{3}{\vec {\beta }}_{3}\rangle$ is a basis when $c_{1},c_{2},c_{3}\neq 0$. What happens when at least one $c_{i}$ is $0$?
* Prove that $\langle {\vec {\alpha }}_{1},{\vec {\alpha }}_{2},{\vec {\alpha }}_{3}\rangle$ is a basis where ${\vec {\alpha }}_{i}={\vec {\beta }}_{1}+{\vec {\beta }}_{i}$.

### Answer

* To show that it is linearly independent, note that $d_{1}(c_{1}{\vec {\beta }}_{1})+d_{2}(c_{2}{\vec {\beta }}_{2})+d_{3}(c_{3}{\vec {\beta }}_{3})={\vec {0}}$ gives that $(d_{1}c_{1}){\vec {\beta }}_{1}+(d_{2}c_{2}){\vec {\beta }}_{2}+(d_{3}c_{3}){\vec {\beta }}_{3}={\vec {0}}$, which in turn implies that each $d_{i}c_{i}$ is zero. But with $c_{i}\neq 0$ that means that each $d_{i}$ is zero. Showing that it spans the space is much the same; because $\langle {\vec {\beta }}_{1},{\vec {\beta }}_{2},{\vec {\beta }}_{3}\rangle$ is a basis, and so spans the space, we can for any ${\vec {v}}$ write ${\vec {v}}=d_{1}{\vec {\beta }}_{1}+d_{2}{\vec {\beta }}_{2}+d_{3}{\vec {\beta }}_{3}$, and then ${\vec {v}}=(d_{1}/c_{1})(c_{1}{\vec {\beta }}_{1})+(d_{2}/c_{2})(c_{2}{\vec {\beta }}_{2})+(d_{3}/c_{3})(c_{3}{\vec {\beta }}_{3})$. If any of the scalars are zero then the result is not a basis, because it is not linearly independent.
* Showing that $\langle 2{\vec {\beta }}_{1},{\vec {\beta }}_{1}+{\vec {\beta }}_{2},{\vec {\beta }}_{1}+{\vec {\beta }}_{3}\rangle$ is linearly independent is easy. To show that it spans the space, assume that ${\vec {v}}=d_{1}{\vec {\beta }}_{1}+d_{2}{\vec {\beta }}_{2}+d_{3}{\vec {\beta }}_{3}$. Then, we can represent the same ${\vec {v}}$ with respect to $\langle 2{\vec {\beta }}_{1},{\vec {\beta }}_{1}+{\vec {\beta }}_{2},{\vec {\beta }}_{1}+{\vec {\beta }}_{3}\rangle$ in this way ${\vec {v}}=(1/2)(d_{1}-d_{2}-d_{3})(2{\vec {\beta }}_{1})+d_{2}({\vec {\beta }}_{1}+{\vec {\beta }}_{2})+d_{3}({\vec {\beta }}_{1}+{\vec {\beta }}_{3})$.

### Problem 13

Find one vector ${\vec {v}}$ that will make each into a basis for the space.
* $\langle {\begin{pmatrix}1\\1\end{pmatrix}},{\vec {v}}\rangle$ in $\mathbb {R} ^{2}$
* $\langle {\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}},{\vec {v}}\rangle$ in $\mathbb {R} ^{3}$
* $\langle x,1+x^{2},{\vec {v}}\rangle$ in ${\mathcal {P}}_{2}$

### Answer

Each forms a linearly independent set if ${\vec {v}}$ is omitted. To preserve linear independence, we must expand the span of each. That is, we must determine the span of each (leaving ${\vec {v}}$ out), and then pick a ${\vec {v}}$ lying outside of that span. Then to finish, we must check that the result spans the entire given space. Those checks are routine.
* Any vector that is not a multiple of the given one, that is, any vector that is not on the line $y=x$ will do here. One is ${\vec {v}}={\vec {e}}_{1}$.
* By inspection, we notice that the vector ${\vec {e}}_{3}$ is not in the span of the set of the two given vectors. The check that the resulting set is a basis for $\mathbb {R} ^{3}$ is routine.
* For any member of the span $\{c_{1}\cdot (x)+c_{2}\cdot (1+x^{2})\,{\big |}\,c_{1},c_{2}\in \mathbb {R} \}$, the coefficient of $x^{2}$ equals the constant term. So we expand the span if we add a quadratic without this property, say, ${\vec {v}}=1-x^{2}$. The check that the result is a basis for ${\mathcal {P}}_{2}$ is easy.
    This exercise is recommended for all readers.

### Problem 14

Where $\langle {\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{n}\rangle$ is a basis, show that in this equation
    $c_{1}{\vec {\beta }}_{1}+\dots +c_{k}{\vec {\beta }}_{k}=c_{k+1}{\vec {\beta }}_{k+1}+\dots +c_{n}{\vec {\beta }}_{n}$
each of the $c_{i}$'s is zero. Generalize.

### Answer

To show that each scalar is zero, simply subtract $c_{1}{\vec {\beta }}_{1}+\dots +c_{k}{\vec {\beta }}_{k}-c_{k+1}{\vec {\beta }}_{k+1}-\dots -c_{n}{\vec {\beta }}_{n}={\vec {0}}$. The obvious generalization is that in any equation involving only the ${\vec {\beta }}$'s, and in which each ${\vec {\beta }}$ appears only once, each scalar is zero. For instance, an equation with a combination of the even-indexed basis vectors (i.e., ${\vec {\beta }}_{2}$, ${\vec {\beta }}_{4}$, etc.) on the right and the odd-indexed basis vectors on the left also gives the conclusion that all of the coefficients are zero.

### Problem 15

A basis contains some of the vectors from a vector space; can it contain them all?

### Answer

No; no linearly independent set contains the zero vector.

### Problem 16

Theorem 1.12 shows that, with respect to a basis, every linear combination is unique. If a subset is not a basis, can linear combinations be not unique? If so, must they be?

### Answer

Here is a subset of $\mathbb {R} ^{2}$ that is not a basis, and two different linear combinations of its elements that sum to the same vector.
    $\{{\begin{pmatrix}1\\2\end{pmatrix}},{\begin{pmatrix}2\\4\end{pmatrix}}\}\qquad 2\cdot {\begin{pmatrix}1\\2\end{pmatrix}}+0\cdot {\begin{pmatrix}2\\4\end{pmatrix}}=0\cdot {\begin{pmatrix}1\\2\end{pmatrix}}+1\cdot {\begin{pmatrix}2\\4\end{pmatrix}}$
Thus, when a subset is not a basis, it can be the case that its linear combinations are not unique.
But just because a subset is not a basis does not imply that its combinations must be not unique. For instance, this set
    $\{{\begin{pmatrix}1\\2\end{pmatrix}}\}$
does have the property that
    $c_{1}\cdot {\begin{pmatrix}1\\2\end{pmatrix}}=c_{2}\cdot {\begin{pmatrix}1\\2\end{pmatrix}}$
implies that $c_{1}=c_{2}$. The idea here is that this subset fails to be a basis because it fails to span the space; the proof of the theorem establishes that linear combinations are unique if and only if the subset is linearly independent.
    This exercise is recommended for all readers.

### Problem 17

A square matrix is symmetric if for all indices $i$ and $j$, entry $i,j$ equals entry $j,i$.
* Find a basis for the vector space of symmetric $2\!\times \!2$ matrices.
* Find a basis for the space of symmetric $3\!\times \!3$ matrices.
* Find a basis for the space of symmetric $n\!\times \!n$ matrices.

### Answer

* Describing the vector space as suggests this for a basis. Verification is easy.
    $\{{\begin{pmatrix}a&b\\b&c\end{pmatrix}}\,{\big |}\,a,b,c\in \mathbb {R} \}$
    $\langle {\begin{pmatrix}1&0\\0&0\end{pmatrix}},{\begin{pmatrix}0&0\\0&1\end{pmatrix}},{\begin{pmatrix}0&1\\1&0\end{pmatrix}}\rangle$
* This is one possible basis.
    $\langle {\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}},{\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}},{\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}},{\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}},{\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}},{\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}}\rangle$
* As in the prior two questions, we can form a basis from two kinds of matrices. First are the matrices with a single one on the diagonal and all other entries zero (there are $n$ of those matrices). Second are the matrices with two opposed off-diagonal entries are ones and all other entries are zeros. (That is, all entries in $M$ are zero except that $m_{i,j}$ and $m_{j,i}$ are one.)
    This exercise is recommended for all readers.

### Problem 18

We can show that every basis for $\mathbb {R} ^{3}$ contains the same number of vectors.
* Show that no linearly independent subset of $\mathbb {R} ^{3}$ contains more than three vectors.
* Show that no spanning subset of $\mathbb {R} ^{3}$ contains fewer than three vectors. (Hint. Recall how to calculate the span of a set and show that this method, when applied to two vectors, cannot yield all of $\mathbb {R} ^{3}$.)

### Answer

* Any four vectors from $\mathbb {R} ^{3}$ are linearly related because the vector equation gives rise to a linear system that is homogeneous (and so has a solution) and has four unknowns but only three equations, and therefore has nontrivial solutions. (Of course, this argument applies to any subset of $\mathbb {R} ^{3}$ with four or more vectors.)
    $c_{1}{\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+c_{2}{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}+c_{3}{\begin{pmatrix}x_{3}\\y_{3}\\z_{3}\end{pmatrix}}+c_{4}{\begin{pmatrix}x_{4}\\y_{4}\\z_{4}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}$
    ${\begin{array}{*{4}{rc}r}x_{1}c_{1}&+&x_{2}c_{2}&+&x_{3}c_{3}&+&x_{4}c_{4}&=&0\\y_{1}c_{1}&+&y_{2}c_{2}&+&y_{3}c_{3}&+&y_{4}c_{4}&=&0\\z_{1}c_{1}&+&z_{2}c_{2}&+&z_{3}c_{3}&+&z_{4}c_{4}&=&0\end{array}}$
* Given $x_{1}$, ..., $z_{2}$, to decide which vectors are in the span of $S$, set up and row reduce the resulting system. There are two variables $c_{1}$ and $c_{2}$ but three equations, so when Gauss' method finishes, on the bottomrow there will be some relationship of the form $0=m_{1}x+m_{2}y+m_{3}z$. Hence, vectors in the span of the two-element set $S$ must satisfy some restriction. Hence the span is not all of $\mathbb {R} ^{3}$.
    $S=\{{\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}},{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}\}$
    ${\begin{pmatrix}x\\y\\z\end{pmatrix}}$
    $c_{1}{\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+c_{2}{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}x\\y\\z\end{pmatrix}}$
    ${\begin{array}{*{2}{rc}r}x_{1}c_{1}&+&x_{2}c_{2}&=&x\\y_{1}c_{1}&+&y_{2}c_{2}&=&y\\z_{1}c_{1}&+&z_{2}c_{2}&=&z\end{array}}$

### Problem 19

One of the exercises in the Subspaces subsection shows that the set
    $\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\,{\big |}\,x+y+z=1\}$
is a vector space under these operations.
    ${\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}-1\\y_{1}+y_{2}\\z_{1}+z_{2}\end{pmatrix}}\qquad r{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}rx-r+1\\ry\\rz\end{pmatrix}}$
Find a basis.

### Answer

We have (using these peculiar operations with care)
    $\{{\begin{pmatrix}1-y-z\\y\\z\end{pmatrix}}\,{\big |}\,y,z\in \mathbb {R} \}=\{{\begin{pmatrix}-y+1\\y\\0\end{pmatrix}}+{\begin{pmatrix}-z+1\\0\\z\end{pmatrix}}\,{\big |}\,y,z\in \mathbb {R} \}=\{y\cdot {\begin{pmatrix}0\\1\\0\end{pmatrix}}+z\cdot {\begin{pmatrix}0\\0\\1\end{pmatrix}}\,{\big |}\,y,z\in \mathbb {R} \}$
and so a good candidate for a basis is this.
    $\langle {\begin{pmatrix}0\\1\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\end{pmatrix}}\rangle$
To check linear independence we set up
    $c_{1}{\begin{pmatrix}0\\1\\0\end{pmatrix}}+c_{2}{\begin{pmatrix}0\\0\\1\end{pmatrix}}={\begin{pmatrix}1\\0\\0\end{pmatrix}}$
(the vector on the right is the zero object in this space). That yields the linear system
    ${\begin{array}{*{3}{rc}r}(-c_{1}+1)&+&(-c_{2}+1)&-&1&=&1\\c_{1}&&&&&=&0\\&&c_{2}&&&=&0\end{array}}$
with only the solution $c_{1}=0$ and $c_{2}=0$. Checking the span is similar.