If you find any mistakes, please make a comment! Thank you.

Solution to Linear Algebra Hoffman & Kunze Chapter 2.4


Exercise 2.4.1

Using Theorem 7, page 52, if we calculate the inverse of
$$P=\left[\begin{array}{cccc}1&0&1&0\\1&0&0&0\\0&1&0&0\\0&1&4&2\end{array}\right].$$then the columns of $P^{-1}$ will give the coefficients to write the standard basis vectors in terms of the $\alpha_i$'s. We do this by row-reducing the augmented matrix
$$\left[\begin{array}{cccc|cccc}1&0&1&0&1&0&0&0\\1&0&0&0&0&1&0&0\\0&1&0&0&0&0&1&0\\0&1&4&2&0&0&0&1\end{array}\right].$$ The left side must reduce to the identity whlie the right side transforms to the inverse of $P$. Row reduction gives
$$\left[\begin{array}{cccc|cccc}1&0&1&0&1&0&0&0\\1&0&0&0&0&1&0&0\\0&1&0&0&0&0&1&0\\0&1&4&2&0&0&0&1\end{array}\right].$$$$\rightarrow\left[\begin{array}{cccc|cccc}1&0&1&0&1&0&0&0\\0&0&-1&0&-1&1&0&0\\0&1&0&0&0&0&1&0\\0&1&4&2&0&0&0&1\end{array}\right].$$$$\rightarrow\left[\begin{array}{cccc|cccc}1&0&1&0&1&0&0&0\\0&1&0&0&0&0&1&0\\0&0&-1&0&-1&1&0&0\\0&1&4&2&0&0&0&1\end{array}\right].$$$$\rightarrow\left[\begin{array}{cccc|cccc}1&0&1&0&1&0&0&0\\0&1&0&0&0&0&1&0\\0&0&-1&0&-1&1&0&0\\0&0&4&2&0&0&-1&1\end{array}\right].$$$$\rightarrow\left[\begin{array}{cccc|cccc}1&0&1&0&1&0&0&0\\0&1&0&0&0&0&1&0\\0&0&1&0&1&-1&0&0\\0&0&4&2&0&0&-1&1\end{array}\right].$$$$\rightarrow\left[\begin{array}{cccc|cccc}1&0&0&0&0&1&0&0\\0&1&0&0&0&0&1&0\\0&0&1&0&1&-1&0&0\\0&0&0&2&-4&4&-1&1\end{array}\right].$$$$\rightarrow\left[\begin{array}{cccc|cccc}1&0&0&0&0&1&0&0\\0&1&0&0&0&0&1&0\\0&0&1&0&1&-1&0&0\\0&0&0&1&-2&2&-1/2&1/2\end{array}\right].$$Thus $\{\alpha_1,\dots,\alpha_4\}$ is a basis. Call this basis $\beta$.Thus $(1,0,0,0)= \alpha_3-2\alpha_4$, $(0,1,0,0)=\alpha_1-\alpha_3+2\alpha_4$, $(0,0,1,0)=\alpha_2-\frac12\alpha_4$ and $(0,0,0,1)=\frac12\alpha_4$.

Thus $[(1,0,0,0)]_\beta=(0,0,1,-2)$, $[(0,1,0,0)]_\beta=(1,0,-1,2)$, $[(0,0,1,0)]_\beta=(0,1,0,-1/2)$ and $[(0,0,0,1)]_\beta=(0,0,0,1/2)$.


Exercise 2.4.2

Using Theorem 7, page 52, the answer is $P^{-1}\left[\begin{array}{c}1\\0\\1\end{array}\right]$ where
$$P=\left[\begin{array}{cccc}2i&2&0\\1&-1&1+i\\0&0&1-i\end{array}\right].$$We find $P^{-1}$ by row-reducing the augmented matrix
$$\left[\begin{array}{ccc|ccc}2i&2&0&1&0&0\\1&-1&1+i&0&1&0\\0&0&1-i&0&0&1\end{array}\right].$$The right side will transform into the $P^{-1}$. Row reducing:
$$\left[\begin{array}{ccc|ccc}2i&2&0&1&0&0\\1&-1&1+i&0&1&0\\0&0&1-i&0&0&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-1&1+i&0&1&0\\2i&2&0&1&0&0\\0&0&1-i&0&0&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-1&1+i&0&1&0\\0&2+2i&2-2i&1&-2i&0\\0&0&1-i&0&0&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-1&1+i&0&1&0\\0&1&-i&\frac{1-i}{4}&\frac{-1-i}{2}&0\\0&0&1-i&0&0&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&1&\frac{1-i}{4}&\frac{1-i}{2}&0\\0\rule{0mm}{4mm}&1&-i&\frac{1-i}{4}&\frac{-1-i}{2}&0\\0\rule{0mm}{4mm}&0&1-i&0&0&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&1&\frac{1-i}{4}&\frac{1-i}{2}&0\\0\rule{0mm}{4mm}&1&-i&\frac{1-i}{4}&\frac{-1-i}{2}&0\\0\rule{0mm}{4mm}&0&1&0&0&\frac{1+i}{2}\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&0&\frac{1-i}{4}&\frac{1-i}{2}&\frac{-1-i}{2}\\0\rule{0mm}{4mm}&1&0&\frac{1-i}{4}&\frac{-1-i}{2}&\frac{-1+i}{2}\\0\rule{0mm}{4mm}&0&1&0&0&\frac{1+i}{2}\end{array}\right]$$Therefore$$P^{-1}\left[\begin{array}{c}1\\0\\1\end{array}\right]=\left[\begin{array}{ccc}\frac{1-i}{4}&\frac{1-i}{2}&\frac{-1-i}{2}\\\rule{0mm}{4mm}\frac{1-i}{4}&\frac{-1-i}{2}&\frac{-1+i}{2}\\0&0&\rule{0mm}{4mm}\frac{1+i}{2}\end{array}\right]\left[\begin{array}{c}1\\0\rule{0mm}{4mm}\\1\rule{0mm}{4mm}\end{array}\right]=\left[\begin{array}{c}\frac{-1-3i}{4}\\ \frac{-1+i}{4}\rule{0mm}{4mm}\\\rule{0mm}{4mm} \frac{1+i}{2}\end{array}\right]$$Thus $(1,0,1)=\frac{-1-3i}{4}(2i,1,0)+\frac{-1+i}{4}(2,-1,0)+\frac{1+i}{2}(0,1+i,1-i)$.


Exercise 2.4.3

Using Theorem 7, page 52, the answer is $P^{-1}\left[\begin{array}{c}a\\b\\c\end{array}\right]$ where
$$P=\left[\begin{array}{cccc}1&1&1\\0&1&0\\-1&1&0\end{array}\right].$$We find $P^{-1}$ by row-reducing the augmented matrix
$$\left[\begin{array}{ccc|ccc}1&1&1&1&0&0\\0&1&0&0&1&0\\-1&1&0&0&0&1\end{array}\right].$$The right side will transform into the $P^{-1}$. Row reducing:
$$\left[\begin{array}{ccc|ccc}1&1&1&1&0&0\\0&1&0&0&1&0\\-1&1&0&0&0&1\end{array}\right].$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&1&1&1&0&0\\0&1&0&0&1&0\\0&2&1&1&0&1\end{array}\right].$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&1&1&-1&0\\0&1&0&0&1&0\\0&0&1&1&-2&1\end{array}\right].$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&0&0&1&-1\\0&1&0&0&1&0\\0&0&1&1&-2&1\end{array}\right].$$Therefore,
$$P^{-1}\left[\begin{array}{c}a\\b\\c\end{array}\right]=\left[\begin{array}{ccc}0&1&-1\\0&1&0\\1&-2&1\end{array}\right]\left[\begin{array}{c}a\\b\\c\end{array}\right]=\left[\begin{array}{c}b-c\\b\\a-2b+c\end{array}\right]$$Thus the answer is$$[(a,b,c)]_{\mathcal B}=(b-c,\ b,\ a-2b+c).$$


Exercise 2.4.4

(a) To show $\alpha_1$ and $\alpha_2$ form a basis of the space they generate we must show they are linearly independent. In other words that $a\alpha_1+b\alpha_2=0$ $\Rightarrow$ $a=b=0$. Equivalently we need to show neither is a multiple of the other. If $\alpha_2=c\alpha_1$ then from the second coordinate it follows that $c=0$ which would imply $\alpha_2=(0,0,0)$, which it does not. So $\{\alpha_1,\alpha_2\}$ is a basis for the space they genearate.

(b) Since the first coordinate of both $\beta_1$ and $\beta_2$ is one, it's clear that neither is a multiple of the other. So they generate a two dimensional subspace of $\mathbb C^3$. If we show $\beta_1$ and $\beta_2$ can be written as linear combinations of $\alpha_1$ and $\alpha_2$ then since the spaces generated by them both have dimension two, by Corollary 1, page 46, they must be equal. To show $\beta_1$ and $\beta_2$ can be written as linear combinations of $\alpha_1$ and $\alpha_2$ we row-reduce the augmented matrix
$$\left[\begin{array}{cc|cc}1&1+i&1&1\\0&1&1&i\\i&-1&0&1+i\end{array}\right].$$Row reduction follows:
$$\left[\begin{array}{cc|cc}1&1+i&1&1\\0&1&1&i\\i&-1&0&1+i\end{array}\right]
\rightarrow\left[\begin{array}{cc|cc}1&1+i&1&1\\0&1&1&i\\0&-i&-i&1\end{array}\right]
\rightarrow\left[\begin{array}{cc|cc}1&0&-i&2-i\\0&1&1&i\\0&0&0&0\end{array}\right]$$Thus $\beta_1=-i\alpha_1+\alpha_2$ and $\beta_2=(2-i)\alpha_1+i\alpha_2$.

(c) We have to write the $\beta_i$'s in terms of the $\alpha_i$'s, basically the opposite of what we did in part b. In this case we row-reduce the augmented matrix
$$\left[\begin{array}{cc|cc}1&1&1&1+i\\1&i&0&1\\0&1+i&i&-1\end{array}\right]
\rightarrow\left[\begin{array}{cc|cc}1&1&1&1+i\\0&-1+i&-1&-i\\0&1+i&i&-1\end{array}\right]
\rightarrow\left[\begin{array}{cc|cc}1&1&1&1+i\\0&-1+i&-1&-i\\0&0&0&0\end{array}\right]$$$$\rightarrow\left[\begin{array}{cc|cc}1&1&1&1+i\\0&1&\frac{1+i}{2}&\frac{-1+i}{2}\\0&0&0&0\rule{0mm}{4mm}\end{array}\right]
\rightarrow\left[\begin{array}{cc|cc}1&0&\frac{1-i}{2}&\frac{3+i}{2}\\0\rule{0mm}{4mm}&1&\frac{1+i}{2}&\frac{-1+i}{2}\\0\rule{0mm}{4mm}&0&0&0\end{array}\right]$$Thus $\alpha_1=\frac{1-i}{2}\beta_1+\frac{1+i}{2}\beta_2$ and $\alpha_2=\frac{3+i}{2}\beta_1+\frac{-1+i}{2}\beta_2$. So finally, if $\mathcal B$ is the basis $\{\beta_1,\beta_2\}$ then
$$[\alpha_1]_{\mathcal B}=\left(\frac{1-i}{2},\frac{1+i}{2}\right)$$$$[\alpha_2]_{\mathcal B}=\left(\frac{3+i}{2},\frac{-1+i}{2}\right).$$


Exercise 2.4.5

It suffices by Corollary 1, page 46, to show $\alpha$ and $\beta$ are linearly indepdenent, because then they generate a subspace of $\mathbb R^2$ of dimension two, which therefore must be all of $\mathbb R^2$. The second condition on $x_1,x_2,y_1,y_2$ implies that neither $\alpha$ nor $\beta$ are the zero vector. To show two vectors are linearly independent we only need show neither is a non-zero scalar multiple of the other. Suppose WLOG that $\beta=c\alpha$ for some $c\in\mathbb R$, and since neither vector is the zero vector, $c\not=0$. Then $y_1=cx_1$ and $y_2=cx_2$. Thus the conditions on $x_1,x_2,y_1,y_2$ implies
$$0=x_1y_1+x_2y_2=cx_1^2+cx_2^2=c(x_1^2+x_2^2)=c\cdot 1=c.$$Thus $c=0$, a contradiction.

It remains to find the coordinates of the arbitrary vector $(a,b)$ in the ordered basis $\{\alpha,\beta\}$. To find the coordinates of $(a,b)$ we can row-reduce the augmented matrix
$$\left[\begin{array}{cc|c}x_1&y_1&a\\x_2&y_2&b\end{array}\right].$$It cannot be that both $x_1=x_2=0$ so assume WLOG that $x_1\not=0$. Also it cannot be that both $y_1=y_2=0$. Assume first that $y_1\not=0$. Since order matters we cannot assume $y_1\not=0$ WLOG, so we must consider both cases. Then note that $x_1y_1+x_2y_2=0$ implies
\begin{equation}
\frac{x_2y_2}{x_1y_1}=-1
\label{2.5.5}
\end{equation} Thus if $x_1y_2-x_2y_1=0$ then $\frac{x_2}{x_1}=\frac{y_2}{y_1}$ from which (\ref{2.5.5}) implies $\left(\frac{x_2}{x_1}\right)^2=-1$, a contradiction. Thus we can conclude that $x_1y_2-x_2y_1\not=0$. We use this in the following row reduction to be sure we are not dividing by zero.
$$\left[\begin{array}{cc|c}x_1&y_1&a\\x_2&y_2&b\end{array}\right]
\rightarrow\left[\begin{array}{cc|c}1&y_1/x_1&a/x_1\\x_2&y_2&b\end{array}\right]
\rightarrow\left[\begin{array}{cc|c}1&y_1/x_1&a/x_1\\0&y_2-\frac{x_2y_1}{x_1}&b-\frac{x_2a}{x_1}\end{array}\right]
=\left[\begin{array}{cc|c}1&y_1/x_1&a/x_1\\0&\frac{x_1y_2-x_2y_1}{x_1}&\frac{bx_1-ax_2}{x_1}\end{array}\right]$$$$\rightarrow\left[\begin{array}{cc|c}1&y_1/x_1&a/x_1\\0&1&\frac{bx_1-ax_2}{x_1y_2-x_2y_1}\end{array}\right]
\rightarrow\left[\begin{array}{cc|c}1&0&\frac{ay_2-by_1}{x_1y_2-x_2y_1}\\0&1&\frac{bx_1-ax_2}{x_1y_2-x_2y_1}\rule{0mm}{5mm}\end{array}\right]$$Now if we substitute $y_1=-x_2y_2/x_1$ into the numerator and denominator of $\frac{ay_2-by_1}{x_1y_2-x_2y_1}$ and use $x_1^2+x_2^2=1$ it simplifies to $ax_1+bx_2$. Similarly $\frac{ay_2-by_1}{x_1y_2-x_2y_1}$ simplifies to $ay_1+by_2$. So we get
$$\left[\begin{array}{cc|c}1&0&ax_1+bx_2\\0&1&ay_1+by_2\end{array}\right].$$Now assume $y_2\not=0$ (and we continue to assume $x_1\not=0$ since we assumed that WLOG). In this case
\begin{equation}
\frac{y_1}{y_2}=-\frac{x_2}{x_1}
\label{2323r2}
\end{equation}So if $x_1y_2-x_2y_1=0$ then $\frac{x_2y_1}{x_1y_2}=1$. But then (\ref{2323r2}) implies $\left(\frac{x_2}{x_1}\right)^2=-1$ a contradition. So also in this case we can assume $x_1y_2-x_2y_1\not=0$ and so we can do the same row-reduction as before. Thus in all cases
$$(ax_1+bx_2)\alpha+(ay_1+by_2)\beta=(a,b)$$or equivalently
$$(ax_1+bx_2)(x_1,x_2)+(ay_1+by_2)(y_1,y_2)=(a,b).$$


Exercise 2.4.6

Suppose $a+be^{ix}+ce^{-ix}=0$ as functions of $x\in\mathbb R$. In other words $a+be^{ix}+ce^{-ix}=0$ for all $x\in\mathbb R$. Let $y=e^{ix}$. Then $y\not=0$ and $a+by+\frac{c}{y}=0$ which implies $ay+by^2+c=0$. This is at most a quadratic polynomial in $y$ thus can be zero for at most two values of $y$. But $e^{ix}$ takes infinitely many different values as $x$ varies in $\mathbb R$, so $ay+by^2+c$ cannot be zero for all $y=e^{ix}$, so this is a contradiction.

We know that $e^{ix}=\cos(x)+i\sin(x)$. Thus $e^{-ix}=\cos(x)-i\sin(x)$. Adding these gives $2\cos(x)=e^{ix}+e^{-ix}$. Thus $\cos(x)=\frac12e^{ix}+\frac12e^{-ix}$. Subtracting instead of adding the equations gives $e^{ix}-e^{-ix}=2i\sin(x)$. Thus $\sin(x)=\frac{1}{2i}e^{ix}-\frac{1}{2i}e^{-ix}$ or equivalently $\sin(x)=-\frac{i}{2}e^{ix}+\frac{i}{2}e^{-ix}$. Thus the requested matrix is
$$P=\left[\begin{array}{ccc}1 & 0 & 0\\ 0 & 1/2 & -i/2\\ 0 & 1/2 & i/2\end{array}\right].$$


Exercise 2.4.7

We know $V$ has dimension three (it follows from Example 16, page 43, that $\{1,x,x^2\}$ is a basis). Thus by Corollary 2 (b), page 45, it suffices to show $\{g_1,g_2,g_3\}$ span $V$. We need to solve for $u,v,w$ the equation
$$c_2x^2+c_1x+c_0=u+v(x+t)+w(x+t)^2.$$Rearranging
$$c_2x^2+c_1x+c_0=wx^2+(v+2wt)x+(u+vt+wt^2).$$It follows that
$$w=c_2,\quad v=c_1-2c_2t,$$$$u=c_0-c_1t+c_2t^2.$$Thus $\{g_1,g_2,g_3\}$ do span $V$ and the coordinates of $f(x)=c_2x^2+c_1x+c_0$ are $$(c_2,\ \ c_1-2c_2t,\ \ c_0-c_1t+c_2t^2).$$

From http://greggrant.org

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This Post Has 2 Comments

  1. For 2.4.6 is P supposed to be the transpose?

  2. For 2.4.2 I started with a 1 in the middle of the bottom row of P and got the final answer (-1/2 - i/2, i/2, 3/4 + i/4)

Leave a Reply

Close Menu