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## Solution to Linear Algebra Hoffman & Kunze Chapter 3.5

#### Exercise 3.5.1

In $\mathbb R^3$ let $\alpha_1=(1,0,1)$, $\alpha_2=(0,1,-2)$, $\alpha_3=(-1,-1,0)$.

(a) If $f$ is a linear functional on $\mathbb R^3$ such that$$f(\alpha_1)=1,\quad f(\alpha_2)=-1,\quad f(\alpha_3)=3,$$and if $\alpha=(a,b,c)$, find $f(\alpha)$.
(b) Describe explicitly a linear functional $f$ on $\mathbb R^3$ such that$$f(\alpha_1)=f(\alpha_2)=0\quad\text{but \quad f(\alpha_3)\not=0}.$$(c) Let $f$ be any linear functional such that $$f(\alpha_1)=f(\alpha_2)=0\quad\text{and \quad f(\alpha_3)\not=0}.$$If $\alpha=(2,3,-1)$, show that $f(\alpha)\not=0$.

Solution:

(a) We need to write $(a,b,c)$ in terms of $\alpha_1,\alpha_2,\alpha_3$. We can do this by row reducing the following augmented matrix whose colums are the $\alpha_i$'s.
$$\left[\begin{array}{ccc|c}1&0&-1&a\\0&1&-1&b\\1&-2&0&c\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|c}1&0&-1&a\\0&1&-1&b\\0&-2&-1&c-a\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|c}1&0&-1&a\\0&1&-1&b\\0&0&-1&c-a+2b\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|c}1&0&-1&a\\0&1&-1&b\\0&0&1&a-2b-c\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|c}1&0&0&2a-2b-c\\0&1&0&a-b-c\\0&0&1&a-2b-c\end{array}\right]$$Thus if $(a,b,c)=x_1\alpha_1+x_2\alpha_2+x_3\alpha_3$ then $x_1=2a-2b-c$, $x_2=a-b-c$ and $x_3=a-2b-c$. Now \begin{align*}f(a,b,c)&=f(x_1\alpha_1+x_2\alpha_2+x_3\alpha_3) \\&= x_1f(\alpha_1)+x_2f(\alpha_2)+x_3f(\alpha_3)\\&=(2a-2b-c)\cdot 1 + (a-b-c) \cdot (-1) + (a-2b-c) \cdot 3\\&=(2a-2b-c)-(a-b-c)+(3a-6b-3c)\\&=4a-7b-3c.\end{align*} In summary
$$f(\alpha)=4a-7b-3c.$$(b) Let $f(x,y,z)=x-2y-z$. The $f(1,0,1)=0$, $f(0,1,-2)=0$, and $f(-1,-1,0)=1$.

(c) Using part (a) we know that $\alpha=(2,3,-1)=-\alpha_1-3\alpha_3$ (plug in $a=2$, $b=3$, $c=-1$ for the formulas for $x_1,x_2,x_3$). Thus $f(\alpha)=-f(\alpha_1)-3f(\alpha_3)=0- 3f(\alpha_3)$ and since $f(\alpha_3)\not=0$, $-3f(\alpha_3)\not=0$ and thus $f(\alpha)\not=0$.

#### Exercise 3.5.2

Let $\mathcal B=\{\alpha_1,\alpha_2,\alpha_3\}$ be the basis for $\mathbb C^3$ defined by
$$\alpha_1=(1,0,-1),\quad\alpha_2=(1,1,1),\quad\alpha_3=(2,2,0).$$Find the dual basis of $\mathcal B$.

Solution: The dual basis $\{f_1,f_2,f_3\}$ are given by $f_i(x_1,x_2,x_3)=\sum_{j=1}^3 A_{ij}x_j$ where $(A_{1,1},A_{1,2},A_{1,3})$ is the solution to the system
$$\left[\begin{array}{ccc|c}1&0&-1&1\\1&1&1&0\\2&2&0&0\end{array}\right],$$$(A_{2,1},A_{2,2},A_{2,3})$ is the solution to the system
$$\left[\begin{array}{ccc|c}1&0&-1&0\\1&1&1&1\\2&2&0&0\end{array}\right],$$and $(A_{3,1},A_{3,2},A_{3,3})$ is the solution to the system
$$\left[\begin{array}{ccc|c}1&0&-1&0\\1&1&1&0\\2&2&0&1\end{array}\right],$$We row reduce the generic matrix
$$\left[\begin{array}{ccc|c}1&0&-1&a\\1&1&1&b\\2&2&0&c\end{array}\right] \rightarrow\left[\begin{array}{ccc|c}1&0&0&a+b-\frac12c\\0&1&0&c-b-a\\0&0&1&b-\frac12c\end{array}\right].$$$a=1$, $b=0$, $c=0$ $\Rightarrow$ $f_1(x_1,x_2,x_3)=x_1-x_2$.

$a=0$, $b=1$, $c=0$ $\Rightarrow$ $f_2(x_1,x_2,x_3)=x_1-x_2+x_3$.

$a=0$, $b=0$, $c=1$ $\Rightarrow$ $f_3(x_1,x_2,x_3)=-\frac12x_1+x_2-\frac12x_3$.

Then $\{f_1,f_2,f_3\}$ is the dual basis to $\{\alpha_1,\alpha_2,\alpha_3\}$.

#### Exercise 3.5.3

If $A$ and $B$ are $n\times n$ matrices over the field $F$, show that trace$(AB)$ $=$ trace$(BA)$. Now show that similar matrices have the same trace.

Solution: We have $(AB)_{ij}=\sum_{k=1}^n A_{ik}B_{kj}$ and $(BA)_{ij}=\sum_{k=1}^n B_{ik}A_{kj}$. Thus
\begin{align*}\text{trace}(AB)&=\sum_{i=1}^n(AB)_{ii}=\sum_{i=1}^n\sum_{k=1}^n A_{ik}B_{ki}\\&=\sum_{i=1}^n\sum_{k=1}^n B_{ki}A_{ik}
=\sum_{k=1}^n\sum_{i=1}^n B_{ki}A_{ik}
\\&=\sum_{k=1}^n(BA)_{kk}=\text{trace}(BA).\end{align*}Suppose $A$ and $B$ are similar. Then $\exists$ an invertible $n\times n$ matrix $P$ such that $A=PBP^{-1}$. Thus \begin{align*}\text{trace}(A)&=\text{trace}(PBP^{-1})=\text{trace}((P)(BP^{-1}))\\&=\text{trace}((BP^{-1})(P))=\text{trace}(B).\end{align*}

#### Exercise 3.5.4

Let $V$ be the vector space of all polynomial functions $p$ from $R$ into $R$ which have degree $2$ or less:
$$p(x)=c_0+c_1x+c_2x^2.$$Define three linear functionals on $V$ by
$$f_1(p)=\int_0^1p(x)dx,\quad f_2(x)=\int_0^2p(x)dx,\quad f_3(x)=\int_0^3p(x)dx.$$Show that $\{f_1,f_2,f_3\}$ is a basis for $V^{*}$ by exhibiting the basis for $V$ of which it is the dual.

Solution: We have $$\int_0^ac_0+c_1x+c_2x^2dx$$$$=c_0x+\frac12c_1x^2+\frac13c_2x^3 \mid_0^a$$$$=c_0a+\frac12c_1a^2+\frac13c_2a^3.$$Thus
$$\int_0^1p(x)dx=c_1+\frac12c_1+\frac13c_2$$$$\int_0^2p(x)dx=2c_1+2c_1+\frac83c_2$$$$\int_0^3p(x)dx=3c_1+\frac92c_1+9c_2$$Thus we need to solve the following system three times
$$\left\{\begin{array}{l} c_1+\frac12c_1+\frac13c_2=u\\ 2c_1+2c_1+\frac83c_2=v\\ 3c_1+\frac92c_1+9c_2=w \end{array}\right.$$Once when $(u,v,w)=(1,0,0)$, once when $(u,v,w)=(0,1,0)$ and once when $(u,v,w)=(0,0,1)$.

We therefore row reduce the following matrix
$$\left[\begin{array}{ccc|ccc} 1 & 1/2 & 1/3 & 1 & 0 & 0\\ 2 & 2 & 8/3 & 0 & 1 & 0\\ 3 & 9/2 & 9 & 0 & 0 & 1 \end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc} 1 & 1/2 & 1/3 & 1 & 0 & 0\\ 0 & 1 & 2 & -2 & 1 & 0\\ 0 & 3 & 8 & -3 & 0 & 1 \end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc} 1 & 0 & -2/3 & 2 & -1/2 & 0\\ 0 & 1 & 2 & -2 & 1 & 0\\ 0 & 0 & 2 & 3 & -3 & 1 \end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc} 1 & 0 & -2/3 & 2 & -1/2 & 0\\ 0 & 1 & 2 & -2 & 1 & 0\\ 0 & 0 & 1 & 3/2 & -3/2 & 1/2 \end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 3 & -3/2 & 1/3\\ 0 & 1 & 0 & -5 & 4 & -1\\ 0 & 0 & 1 & 3/2 & -3/2 & 1/2 \end{array}\right].$$Thus
$$\alpha_1=3-5x+\frac32x^2$$$$\alpha_2=-\frac32+4x-\frac32x^2$$$$\alpha_3=\frac13-x+\frac12x^2.$$

#### Exercise 3.5.5

If $A$ and $B$ are $n\times n$ complex matrices, show that $AB-BA=I$ is impossible.

Solution: Recall for $n\times n$ matrices $M$, $\text{trace}(M)=\sum_{i=1}^nM_{ii}$. The trace is clearly additive $$\text{trace}(M_1+M_2)=\text{trace}(M_1)+\text{trace}(M_2).$$ We know from Exercise $3$ that $\text{trace}(AB)=\text{trace}(BA)$. Thus \begin{align*}\text{trace}(AB-BA)&=\text{trace}(AB)-\text{trace}(BA)\\&=\text{trace}(AB)-\text{trace}(AB)=0.\end{align*} But $\text{trace}(I)=n$ and $n\not=0$ in $\mathbb C$.

#### Exercise 3.5.6

Let $m$ and $n$ be positive integers and $F$ a field. Let $f_1,\dots,f_m$ be linear functionals on $F^n$. For $\alpha$ in $F^n$ define
$$T(\alpha)=(f_1(\alpha),\dots,f_m(\alpha)).$$Show that $T$ is a linear transformation from $F^n$ into $F^m$. Then show that every linear transformation from $F^n$ into $F^m$ is of the above form, for some $f_1,\dots,f_m$.

Solution: Clearly $T$ is a well defined function from $F^n$ into $F^m$. We must just show it is linear. Let $\alpha,\beta\in F^n$, $c\in\mathbb C$. Then
$$T(c\alpha+\beta)=(f_1(c\alpha+\beta),\dots,f_m(c\alpha+\beta))$$$$=(cf_1(\alpha)+f_1(\beta),\dots,cf_n(\alpha)+f_n(\beta))$$$$=c(f_1(\alpha),\dots,f_n(\alpha))+(f_1(\beta),\dots,f_n(\beta))$$$$=cT(\alpha)+T(\beta).$$Thus $T$ is a linear transformation.

Let $S$ be any linear transformation from $F^n$ to $F^m$. Let $M$ be the matrix of $S$ with respect to the standard bases of $F^n$ and $F^m$. Then $M$ is an $m\times n$ matrix and $S$ is given by $X\mapsto MX$ where we identify $F^n$ as $F^{n\times1}$ and $F^m$ with $F^{m\times1}$. Now for each $i=1,\dots,m$ let $$f_i(x_1,\dots,x_n)=\sum_{j=1}^nM_{ij}x_j.$$ Then $X\mapsto MX$ is the same as $$X\mapsto(f_1(X),\dots,f_m(x))$$ (keeping in mind our identification of $F^m$ with $F^{m\times1}$). Thus $S$ has been written in the desired form.

#### Exercise 3.5.7

Let $\alpha_1=(1,0,-1,2)$ and $\alpha_2=(2,3,1,1)$, and let $W$ be the subspace of $\mathbb R^4$ spanned by $\alpha_1$ and $\alpha_2$. Which linear functionals $f$:
$$f(x_1,x_2,x_3,x_4)=c_1x_1+c_2x_2+c_3x_3+c_4x_4$$are in the annihilator of $W$?

Solution: The two vectors $\alpha_1$ and $\alpha_2$ are linearly independent since neither is a multiple of the other. Thus $W$ has dimension $2$ and $\{\alpha_1,\alpha_2\}$ is a basis for $W$. Therefore a functional $f$ is in the annihilator of $W$ if and only if $f(\alpha_1)=f(\alpha_2)=0$. We find such $f$ by solving the system
$$\left\{\begin{array}{l}f(\alpha_1)=0\\ f(\alpha_2)=0\end{array}\right.$$or equivalently
$$\left\{\begin{array}{l} c_1-c_3+2c_4=0\\ 2c_1+3c_2+c_3+c_4=0 \end{array}\right.$$We do this by row reducing the matrix
$$\left[\begin{array}{cccc} 1 & 0 & -1 & 2\\ 2 & 3 & 1 & 1 \end{array}\right]$$$$\rightarrow\left[\begin{array}{cccc} 1 & 0 & -1 & 2\\ 0 & 1 & 1 & -1 \end{array}\right]$$Therefore
$$c_1=c_3-2c_4$$$$c_2=-c_3+c_4.$$
The general element of $W^0$ is therefore
$$f(x_1,x_2,x_3,x_4)=(c_3-2c_4)x_1+(c_3+c_4)x_2+c_3x_3+c_4x_4,$$for arbitrary elements $c_3$ and $c_4$. Thus $W^0$ has dimension $2$ as expected.

#### Exercise 3.5.8

Let $W$ be the subspace of $\mathbb R^5$ which is spanned by the vectors
$$\alpha_1=\epsilon_1+2\epsilon_2+\epsilon_3,\quad \alpha_2=\epsilon_2+3\epsilon_3+3\epsilon_4+\epsilon_5$$$$\alpha_3=\epsilon_1+4\epsilon_2+6\epsilon_3+4\epsilon_4+\epsilon_5.$$Find a basis for $W^0$.

Solution: The vectors $\alpha_1$, $\alpha_2$, $\alpha_3$ are linearly independent as can be seen by row reducing the matrix
$$\left[\begin{array}{ccccc} 1 & 2 & 1 & 0 & 0\\ 0 & 1 & 3 & 3 & 1\\ 1 & 4 & 6 & 4 & 1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccccc} 1 & 2 & 1 & 0 & 0\\ 0 & 1 & 3 & 3 & 1\\ 0 & 2 & 5 & 4 & 1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccccc} 1 & 0 & -5 & -6 & -2\\ 0 & 1 & 3 & 3 & 1\\ 0 & 0 & -1 & -2 & -1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccccc} 1 & 0 & -5 & -6 & -2\\ 0 & 1 & 3 & 3 & 1\\ 0 & 0 & 1 & 2 & 1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccccc} 1 & 0 & 0 & 4 & 3\\ 0 & 1 & 0 & -3 & -2\\ 0 & 0 & 1 & 2 & 1\end{array}\right].$$Thus $W$ has dimension $3$ and $\{\alpha_1,\alpha_2,\alpha_3\}$ is a basis for $W$. We know every functional is given by $$f(x_1,x_2,x_3,x_4,x_5)=c_1x_2+c_2x_2+c_3x_3+c_4x_4+c_5x_5$$ for some $c_1,\dots,c_5$. From the row reduced matrix we see that the general solution for an element of $W^0$ is
$$f(x_1,x_2,x_3,x_4,x_5)=(-4c_4-3c_5)x_1+(3c_4+2c_5)x_2-(2c_4+c_5)x_3+c_4x_4+c_5x_5.$$

#### Exercise 3.5.9

Let $V$ be the vector space of all $2\times2$ matrices over the field of real numbers, and let
$$B=\left[\begin{array}{cc} 2&-2\\-1&1\end{array}\right].$$Let $W$ be the subspace of $V$ consisting of all $A$ such that $AB=0$. Let $f$ be a linear functional on $V$ which is in the annihilator of $W$. Suppose that $f(I)=0$ and $f(C)=3$, where $I$ is the $2\times2$ identity matrix and
$$C=\left[\begin{array}{cc}0&0\\0&1\end{array}\right].$$Find $f(B)$.

Solution: The general linear functional on $V$ is of the form $f(A)=aA_{11}+bA_{12}+cA_{21}+dA_{22}$ for some $a,b,c,d\in\mathbb R$. If $A\in W$ then
$$\left[\begin{array}{cc} x & y\\z & w\end{array}\right]\left[\begin{array}{cc} 2 & -2\\-1 & 1\end{array}\right]=\left[\begin{array}{cc} 0 & 0\\0 & 0\end{array}\right]$$implies $y=2x$ and $w=2y$. So $W$ consists of all matrices of the form
$$\left[\begin{array}{cc} x & 2x\\y & 2y\end{array}\right]$$Now $f\in W^0$ $\Rightarrow$ $f\left(\left[\begin{array}{cc} x & 2x\\y & 2y\end{array}\right]\right)=0$ $\forall$ $x,y\in\mathbb R$ $\Rightarrow$ $ax+2bx+cy+2dy=0$ $\forall$ $x,y\in\mathbb R$ $\Rightarrow$ $(a+2b)x+(c+2d)y=0$ $\forall$ $x,y\in\mathbb R$ $\Rightarrow$ $b = -\frac12a$ and $d=-\frac12c$. So the general $f\in W^0$ is of the form $$f\left(A\right)=aA_{11}-\frac12aA_{12}+cA_{21}-\frac12cA_{22}.$$Now $f(C)=3$ $\Rightarrow$ $d=3$ $\Rightarrow$ $-\frac12c=3$ $\Rightarrow$ $c=-6$. And $f(I)=0$ $\Rightarrow$ $a-\frac12c=0$ $\Rightarrow$ $c=2a$ $\Rightarrow$ $a=-3$. Thus $$f(A)=-3A_{11}+\frac32A_{12}-6A_{21}+3A_{22}.$$ Thus $$f(B)=-3\cdot2+\frac32\cdot(-2)-6\cdot(-1)+3\cdot1=0.$$

#### Exercise 3.5.10

Let $F$ be a subfield of the complex numbers. We define $n$ linear functionals on $F^n$ $(n\geq2)$ by
$$f_k(x_1,\dots,x_n)=\sum_{j=1}^n(k-j)x_j,\quad 1\leq k\leq n.$$What is the dimension of the subspace annihilated by $f_1,\dots,f_n$?

Solution: $N_{f_k}$ is the subspace annihilated by $f_k$. By the comments on page 101, $N_{f_k}$ has dimension $n-1$. Now the standard basis vector $\epsilon_2$ is in $N_{f_2}$ but is not in $N_{f_1}$. Thus $N_{f_1}$ and $N_{f_2}$ are distinct hyperspaces. Thus their intersection has dimension $n-2$. Now $\epsilon_3$ is in $N_{f_3}$ but is not in $N_{f_1}\cup N_{f_2}$. Thus $N_{f_1}\cap N_{f_2}\cap N_{f_3}$ is the intersection of three distinct hyperspaces and so has dimension $n-3$. Continuing in this way, $\epsilon_i\not\in\cup_{j=1}^{i-1} N_{f_i}$. Thus $\cup_{j=1}^{i} N_{f_i}$ is the intersection of $i$ distinct hyperspaces and so has dimension $n-i$. Thus when $i=n$ we have $\cup_{j=1}^{n} N_{f_i}$ has dimension $0$.

#### Exercise 3.5.11

Let $W_1$ and $W_2$ be subspace of a finite-dimensional vector space $V$.

(a) Prove that $(W_1+W_2)^0=W_1^0\cap W_2^0$.
(b) Prove that $(W_1\cap W_2)^0=W_1^0+ W_2^0$.

Solution:

(a) $f\in(W_1+W_2)^0$ $\Rightarrow$ $f(v)=0$ $\forall$ $v\in W_1+W_2$ $\Rightarrow$ $f(w_1+w_2)=0$ $\forall$ $w_1\in W_1$, $w_2\in W_2$ $\Rightarrow$ $f(w_1)=0$ $\forall$ $w_1\in W_1$ (take $w_2=0$) and $f(w_2)=0$ $\forall$ $w_2\in W_2$ (take $w_1=0$). Thus $f\in W_1^0$ and $f\in W_2^0$. Thus $f\in W_1^0\cap W_2^0$. Thus $(W_1+W_2)^0\subseteq W_1^0\cap W_2^0$.

Conversely, let $f\in W_1^0\cap W_2^0$. Let $v\in W_1+W_2$. Then $v=w_1+w_2$ where $w_i\in W_i$. Thus $$f(v)=f(w_1+w_2)=f(w_1)+f(w_2)=0+0$$ (since $f\in W_1^0$ and $f\in W_2^0$). Thus $f(v)=0$ $\forall$ $v\in W_1+W_2$. Thus $f\in(W_1+W_2)^0$. Thus $W_1^0\cap W_2^0\subseteq (W_1+W_2)^0$.

Since $(W_1+W_2)^0\subseteq W_1^0\cap W_2^0$ and $W_1^0\cap W_2^0\subseteq (W_1+W_2)^0$ it follows that $$W_1^0\cap W_2^0= (W_1+W_2)^0.$$(b) $f\in W_1^0+W_2^0$ $\Rightarrow$ $f=f_1+f_2$, for some $f_i\in W_i^0$. Now let $v\in W_1\cap W_2$. Then $$f(v)=(f_1+f_2)(v)=f_1(v)+f_2(v)=0+0.$$ Thus $f\in(W_1\cap W_2)^0$. Thus $W_1^0+W_2^0 \subseteq (W_1\cap W_2)^0$.

Now let $f\in(W_1\cap W_2)^0$. In the proof of Theorem 6 on page 46 it was shown that we can choose a basis for $W_1+W_2$
$$\{\alpha_1,\dots,\alpha_k,\quad \beta_1,\dots,\beta_m,\quad \gamma_1,\dots,\gamma_n\}$$where $\{\alpha_1,\dots,\alpha_k\}$ is a basis for $W_1\cap W_2$, $\{\alpha_1,\dots,\alpha_k,\quad \beta_1,\dots,\beta_m\}$ is a basis for $W_1$ and $\{\alpha_1,\dots,\alpha_k,\,\gamma_1,\dots,\gamma_n\}$ is a basis for $W_2$. We expand this to a basis for all of $V$
$$\{\alpha_1,\dots,\alpha_k,\quad \beta_1,\dots,\beta_m,\quad \gamma_1,\dots,\gamma_n,\quad \lambda_1,\dots,\lambda_{\ell}\}.$$Now the general element $v\in V$ can be written as
\begin{equation}
v=\sum_{i=1}^k x_i\alpha_i+\sum_{i=1}^m y_i\beta_i+\sum_{i=1}^n z_i\gamma_i + \sum_{i=1}^{\ell}w_i\lambda_i
\label{wefe2}
\end{equation}and $f$ is given by
$$f(v)=\sum_{i=1}^k a_ix_i+\sum_{i=1}^m b_iy_i+\sum_{i=1}^n c_iz_i+ \sum_{i=1}^{\ell}d_iw_i$$for some constants $a_i$, $b_i$, $c_i$, $d_i$. Since $f(v)=0$ for all $v\in W_1\cap W_2$, it follows that $a_1=\cdots=a_k=0$. So
$$f(v)=\sum b_iy_i+\sum c_iz_i+ \sum d_iw_i.$$ Define
$$f_1(v)=\sum c_iz_i+ \sum d_iw_i$$and
$$f_2(v)=\sum b_iy_i.$$Then $f=f_1+f_2$. Now if $v\in W_1$ then
$$v=\sum_{i=1}^k x_i\alpha_i+\sum_{i=1}^m y_i\beta_i,$$so that the coefficients $z_i$ and $w_i$ in (\ref{wefe2}) are all zero. Thus $f_1(v)=0$. Thus $f_1\in W_1^0$. Similarly if $v\in W_2$ then the coefficients $y_i$ and $w_i$ in (\ref{wefe2}) are all zero and thus $f_2(v)=0$. So $f_2\in W_2$. Thus $f=f_1+f_2$ where $f_1\in W_1^0$ and $f_2\in W_2^0$. Thus $f\in W_1^0+W_2^0$. Thus $(W_1\cap W_2)^0\subseteq W_1^0+W_2^0$.

Thus $(W_1\cap W_2)^0\subseteq W_1^0+W_2^0$.

#### Exercise 3.5.12

Let $V$ be a finite-dimensional vector space over the field $F$ and let $W$ be a subspace of $V$. If $f$ is a linear functional on $W$, prove that there is a linear functional $g$ on $V$ suvch that $g(\alpha)=f(\alpha)$ for each $\alpha$ in the subspace $W$.

Solution: Let $\mathcal B$ be a basis for $W$ and let $\mathcal B'$ be a basis for $V$ such that $\mathcal B\subseteq \mathcal B'$. A linear function on a vector space is uniquely determined by its values on a basis, and conversely any function on the basis can be extended to a linear function on the space. Thus we define $g$ on $\mathcal B$ by $g(\beta)=f(\beta)$ $\forall$ $\beta\in\mathcal B$. Then define $g(\beta)=0$ for all $\beta\in\mathcal B'\setminus\mathcal B$. Since we have defined $g$ on $\mathcal B'$ it defines a linear functional on $V$ and since it agrees with $f$ on a basis for $W$ it agrees with $f$ on all of $W$.

#### Exercise 3.5.13

Let $F$ be a subfield of the field of complex numbers and let $V$ be any vector space over $F$. Suppose that $f$ and $g$ are linear functionals on $V$ such that the function $h$ defined by $h(\alpha)=f(\alpha)g(\alpha)$ is also a linear functional on $V$. Prove that either $f=0$ or $g=0$.

Solution: Suppose neither $f$ nor $g$ is the zero function. We will derive a contradiction. Let $v\in V$. Then $h(2v)=f(2v)g(2v)=4f(v)g(v)$. But also $h(2v)=2h(v)=2f(v)g(v)$. Therefore $f(v)g(v)=2f(v)g(v)$ $\forall$ $v\in V$. Thus $f(v)g(v)=0$ $\forall$ $v\in V$. Let $\mathcal B$ be a basis for $V$. Let $\mathcal B_1=\{\beta\in\mathcal B\mid f(\beta)=0\}$ and $\mathcal B_2=\{\beta\in\mathcal B\mid g(\beta)=0\}$. Since $f(\beta)g(\beta)=0$ $\forall$ $\beta\in\mathcal B$, we have $\mathcal B=\mathcal B_1\cup\mathcal B_2$. Suppose $\mathcal B_1\subseteq \mathcal B_2$. Then $\mathcal B_2=\mathcal B$ and consequently $g$ is the zero function. Thus $\mathcal B_1\not\subseteq \mathcal B_2$. And similarly $\mathcal B_2\not\subseteq \mathcal B_1$. Thus we can choose $\beta_1\in\mathcal B_1\setminus\mathcal B_2$ and $\beta_2\in\mathcal B_2\setminus\mathcal B_1$. So we have $f(\beta_2)\not=0$ and $g(\beta_1)\not=0$. Then $$f(\beta_1+\beta_2)g(\beta_1+\beta_2)=f(\beta_1)g(\beta_1)+f(\beta_2)g(\beta_1)+f(\beta_1)g(\beta_2)+f(\beta_2)g(\beta_2).$$Since $f(\beta_1)=g(\beta_2)=0$, this equals $f(\beta_2)g(\beta_1)$ which is non-zero since each term is non-zero. And this contradicts the fact that $f(v)g(v)=0$ $\forall$ $v\in V$.

#### Exercise 3.5.14

Let $F$ be a field of characteristic zero and let $V$ be a finite-dimensional vector space over $F$. If $\alpha_1,\dots,\alpha_m$ are finitely many vectors in $V$, each different from the zero vector, prove that there is a linear functional $f$ on $V$ such that
$$f(\alpha_i)\not=0,\quad i=1,\dots,m.$$

Solution: Re-index if necessary so that $\{\alpha_1,\dots,\alpha_k\}$ is a basis for the subspace generated by $\{\alpha_1,\dots,\alpha_m\}$. So each $\alpha_{k+1},\dots,\alpha_{m}$ can be written in terms of $\alpha_1,\dots,\alpha_k$. Extend $\{\alpha_1,\dots,\alpha_k\}$ to a basis for $V$
$$\{\alpha_1,\dots,\alpha_k,\beta_1,\dots,\beta_n\}.$$ For each $i=k+1,\dots,m$ write $\alpha_i=\sum_{j=1}^kA_{ij}\alpha_j$. Since $\alpha_{k+1},\dots,\alpha_{m}$ are all non-zero, for each $i=k+1,\dots,m$ $\exists$ $j_i\leq k$ such that $A_{ij_i}\not=0$. Now define $f$ by mapping $\alpha_1,\dots,\alpha_k$ to $k$ arbitrary non-zero values and map $\beta_i$ to zero $\forall$ $i$. Then $f(\alpha_{k+1})=\sum_{j=1}^kA_{k+1,j}f(\alpha_j)$. If $f(\alpha_{k+1})=0$ then leaving $f(\alpha_i)$ fixed for all $i\leq k$ and adjusting $f(\alpha_{j_{k+1}})$, it equals zero for exactly one possible value of $f(\alpha_{j_{k+1}})$ (since $A_{k+1,j_{k+1}}\not=0$). Thus we can redefine $f(\alpha_{j_{k+1}})$ so that $f(\alpha_{k+1})\not=0$ while maintaining $f(\alpha_{j_{k+1}})\not=0$.

Now if $f(\alpha_{k+2})=0$, then leaving $f(\alpha_i)$ fixed for $i\not=j_{k+2}$, it equals zero for exactly one possible value of $f(\alpha_{j_{k+2}})$ (since $A_{k+2,j_{k+2}}\not=0$) So we can adjust $f(\alpha_{j_{k+2}})$ so that $f(\alpha_{k+2})\not=0$ and $f(\alpha_{k+1})\not=0$ and $f(\alpha_{k+2})\not=0$ simultaneously.

Continuing in this way we can adjust $f(\alpha_{j_{k+3}}),\dots,f(\alpha_{j_m})$ as necessary until all $f(\alpha_{k+1}),\dots,f(\alpha_{m})$ are non-zero and also all of $f(\alpha_1),\dots,f(\alpha_k)$ are non-zero.

#### Exercise 3.5.15

According to Exercise 3, similar matrices have the same trace. Thus we can define the trace of a linear operator on a finite-dimensional space to be the trace of any matrix which represents the operator in an ordered basis. This is well-defined since all such representing matrices for one operator are similar.

Now let $V$ be the space of all $2\times2$ matrices over the field $F$ and let $P$ be a fixed $2\times2$ matrix. Let $T$ be the linear operator on $V$ defined by $T(A)=PA$. Prove that $\text{trace}(T)=2\text{trace}(P)$.

Solution: Write
$$P=\left[\begin{array}{cc}P_{11}&P_{12}\\P_{21}&P_{22}\end{array}\right].$$Let
$$e_{11}=\left[\begin{array}{cc}1&0\\0&0\end{array}\right],\quad e_{12}=\left[\begin{array}{cc}0&1\\0&0\end{array}\right]$$$$e_{21}=\left[\begin{array}{cc}0&0\\1&0\end{array}\right],\quad e_{22}=\left[\begin{array}{cc}0&0\\0&1\end{array}\right]$$Then $\mathcal B=\{e_{11},e_{12},e_{21},e_{22}\}$ is an ordered basis for $V$. We find the matrix of the linear transformation with respect to this basis.
$$T(e_{11})=\left[\begin{array}{cc}P_{11}&0\\P_{21}&0\end{array}\right]=P_{11}e_{11}+P_{21}e_{21}$$$$T(e_{12})=\left[\begin{array}{cc}0&P_{11}\\0&P_{21}\end{array}\right]=P_{11}e_{12}+P_{21}e_{22}$$$$T(e_{21})=\left[\begin{array}{cc}P_{21}&0\\P_{22}&0\end{array}\right]=P_{12}e_{11}+P_{22}e_{21}$$$$T(e_{22})=\left[\begin{array}{cc}0&P_{12}\\0&P_{22}\end{array}\right]=P_{12}e_{12}+P_{22}e_{22}.$$Thus the matrix of $T$ with respect to $\mathcal B$ is
$$\left[\begin{array}{cccc} P_{11} & 0 & P_{12} & 0\\ 0 & P_{11} & 0 & P_{12}\\ P_{21} & 0 & P_{22} & 0\\ 0 & P_{21} & 0 & P_{22} \end{array}\right].$$The trace of this matrix is $2P_{11} + 2P_{22}=2\text{trace}(P)$.

#### Exercise 3.5.16

Show that the trace functional on $n\times n$ matrices is unique in the following sense. If $W$ is the space of $n\times n$ matrices over the field $F$ and if $f$ is a linear functional on $W$ such that $f(AB)=f(BA)$ for each $A$ and $B$ in $W$, then $f$ is a scalar multiple of the trace function. If, in addition, $f(I)=n$ then $f$ is the trace function.

Solution: Let $A$ and $B$ be $n\times n$ matrices. The $\ell,m$ entry in $AB$ is
\begin{equation}
(AB)_{\ell m}=\sum_{k=1}^nA_{\ell k}B_{km}
\label{fj9320}
\end{equation}and the $\ell,m$ entry in $BA$ is
\begin{equation}
(BA)_{\ell m}=\sum_{k=1}^nB_{\ell k}A_{km}.
\label{fjw9320}
\end{equation}Fix $i,j\in\{1,\dots,n\}$ such that $i>j$. Let $A$ be the matrix where $A_{ij}=1$ and all other entries are zero. Let $B$ be the matrix where $B_{ii}=1$ and all other entries are zero. Consider the general element of $AB$
$$(AB)_{\ell m}=\sum_{k=1}^nA_{\ell k}B_{km}.$$The only non-zero $A$ in the sum on the right is $A_{ij}$. But $B_{jm}=0$ since $j>i$ and only $B_{ii}\not=0$. Thus $AB$ is the zero matrix.

Now we compute $BA$. From (\ref{fjw9320}) the only non-zero term is when $\ell=i$, $m=j$ and $k=i$.

Thus the matrix $AB$ has zeros in every position except for the $i,j$ position where it equals one.

Now the general functional on $n\times n$ matrices is of the form
$$f(M)=\sum_{\ell=1}^n\sum_{m=1}^n c_{\ell m}M_{\ell m}$$for some constants $c_{\ell m}$. Now $f(AB)=f(0)=0$ and $f(BA)=c_{ij}$. So if $f(AB)=f(BA)$ then it follows that $c_{ij}=0$.

Thus we have shown that $c_{ij}=0$ for all $i>j$. Similarly $c_{ij}=0$ for all $i<j$. Thus the only possible non-zero coefficients are $c_{11},\dots,c_{nn}$.$$f(M)=\sum_{i=1}^n c_{ii}M_{ii}.$$We will be done if we show $c_{11}=c_{mm}$ for all $m=2,\dots,n$. Fix $2\leq i\leq n$. Let $A$ be the matrix such that $A_{11}=A_{i1}=1$ and $A_{\ell m}=0$ in all other positions. Let $B=A^{\text{T}}$. Then $AB$ is zero in every position except $A_{11}=A_{1i}=A_{i1}=A_{ii}=1$. And $BA$ is zero in every position except $(BA)_{11}=2$. Thus $f(AB)=c_{11}+c_{ii}$ and $f(BA)=2c_{11}$. Thus if $f(AB)=f(BA)$ then $c_{11}+c_{ii}=2c_{11}$ which implies $c_{11}=c_{ii}$. Thus there's a constant $c$ such that $c_{ii}=c$ for all $i$.

Thus $f$ is given by
$$f(M)=\sum_{k=1}^n cM_{ii}.$$If $f(I)=n$ then $c=1$ and we have the trace function.

#### Exercise 3.5.17

Let $W$ be the space of $n\times n$ matrices over the field $F$, and let $W_0$ be the subspace spanned by the matrices $C$ of the form $C=AB-BA$. Prove that $W_0$ is exactly the subspace of matrices which have trace zero. (Hint: What is the dimension of the space of matrices of trace zero? Use the matrix 'units,' i.e., matrices with exactly one non-zero entry, to construct enough linearly independent matrices of the form $AB-BA$.)

Solution: Let $W'=\{w\in W\mid \text{trace}(w)=0\}$. We want to show $W'=W_0$. We know from Exercise 3 that $\text{trace}(AB-BA)=0$ for all matrices $A,B$. Since matrices of the form $AB-BA$ span $W_0$, it follows that $\text{trace}(M)=0$ for all $M\in W_0$. Thus $W_0\subseteq W'$.

Since the trace function is a linear functional, the dimension of $W'$ is $\text{dim}(W)-1=n^2-1$. Thus if we show the dimension of $W_0$ is also $n^2-1$ then we will be done. We do this by exhibiting $n^2-1$ linearly independent elements of $W_0$. Denote by $E_{ij}$ the matrix with a one in the $i,j$ position and zeros in all other positions. Let $H_{ij}=E_{ii}-E_{jj}$. Let $$\mathcal B=\{E_{ij}\mid i\not=j\}\cup\{H_{1,i}\mid 2\leq i\leq n\}.$$ We will show that $\mathcal B\subseteq W_0$ and that $\mathcal B$ is a linearly independent set. First, it clear that they are linearly independent because $E_{ij}$ is the only vector in $\mathcal B$ with a non-zero value in the $i,j$ position and $H_{1,i}$ is the only vector in $\mathcal B$ with a non-zero value in the $i,i$ position. Now $2E_{ij}=H_{ij}E_{ij}-E_{ij}H_{ij}$ and $H_{ij}=E_{ij}E_{ji}-E_{ji}E_{ij}$. Thus $E_{ij}\in W_0$ and $H_{ij}\in W_0$. Now $$|\mathcal B|=|\{E_{ij}\mid i\not=j\}|+|\{H_{1,i}\mid 2\leq i\leq n\}|=(n^2-n)+(n-1)=n^2-1$$ Thus we are done.

### This Post Has 4 Comments

1. 3.5.10 is incorrect. Here is a counter example; Let f1, f2, f3 be linear functionals such that
f1(x, y, z) = y - z
f2(x, y, z) = x + z
f3(x, y, z) = x + y
If ei is the i-th standard basis of R^3, then clearly fi(ei) = 0. By the proof given, the subspace annihilated by the given functionals is the zero subspace. However, if v = (-x, x, x), then f1(v) = f2(v) = f3(v) = 0 which is a contradiction.

1. The dimension appears to be (n - 2), and this is how I proved it:
Let A be the matrix representing the system of equations f1(v) = 0, ..., fn(v) = 0. I don't know how to properly format matrices in the comments so I will skip the actual reduction, but it is easy to see the rank of A is 2 and therefore the null space (and therefore, the subspace annihilated by our given functionals) must have dimension n - 2.

1. The idea is clear. These linear functionals can be expressed by two linear functionals, namely $$x_1+x_2+\cdots+x_n$$ and $$x_1+2x_2+\cdots+nx_n.$$Clearly, they are linearly independent. Hence the dimension of the subspace annihilated by $f_1,\dots,f_n$ is $n-2$.

2. Yes, you are right. Some solutions from http://greggrant.org/ have low quality. I will consider double check and rewrite most of them.