Hoffman & Kunze

Exercise 3.1.1 (a) $T$ is not a linear transformation because $T(0,0)=(1,0)$ and according to the comments after Example 5 on page 68 we know that it must always be that…

Exercise 2.6.1 Let $\alpha_1$, $\alpha_2$, $\dots$, $\alpha_n$ be the colunms of $A$. Then $\alpha_i\in F^{s\times1}$ $\forall$ $i$. Thus $\{\alpha_1,\dots,\alpha_n\}$ are $n$ vectors in $F^{s\times1}$. But $F^{s\times1}$ has dimension $s<n$ thus…

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…

Exercise 2.3.1 Suppose $v_1$ and $v_2$ are linearly dependent. If one of them, say $v_1$, is the zero vector then it is a scalar multiple of the other one $v_1=0\cdot…

Exercise 2.2.1 (a) This is not a subspace because for $(1,\dots,1)$ the additive inverse is $(-1,\dots,-1)$ which does not satisfy the condition. (b) Suppose $(a_1,a_2,a_3,\dots,a_n)$ and $(b_1,b_2,b_3,\dots,b_n)$ satisfy the condition…

Exercise 2.1.1 Example 1 starts with any field and defines the objects, the addition rule and the scalar multiplication rule. We must show the set of $n$-tuples satisfies the eight…

Exercise 1.6.1 As in Exercise 4, Section 1.5, we row reduce and keep track of the elementary matrices involved. It takes nine steps to put $A$ in row-reduced form resulting…

Exercise 1.5.1 We have $$AB=\left[\begin{array}{c} 4\\4\end{array}\right],$$so $$ABC= \left[\begin{array}{c} 4\\4\end{array}\right]\cdot [1\ \ -1] =\left[\begin{array}{cc} 4 & -4\\ 4 & -4 \end{array}\right].$$and $$CBA=[1\ \ -1] \cdot \left[\begin{array}{c} 4\\4\end{array}\right] = [0].$$ Exercise 1.5.2…

Exercise 1.4.1 The coefficient matrix is $$\left[\begin{array}{ccc} \frac13 & 2 & -6\\ -4& 0& 5\\ -3&6&-13\\ -\frac73&2&-\frac83 \end{array}\right] $$This reduces as follows: $$\rightarrow\left[\begin{array}{ccc} 1 & 6 & -18\\ -4& 0&…

Exercise 1.3.1 The matrix of coefficients is $$\left[\begin{array}{cc}1-i&-i\\2&1-i\end{array}\right].$$Row reducing $$\rightarrow \left[\begin{array}{cc}2&1-i\\1-i&-i\end{array}\right]\rightarrow\left[\begin{array}{cc}2&1-i\\0&0\end{array}\right] $$Thus $2x_1+(1-i)x_2=0$. Thus for any $x_2\in\mathbb C$, $(\frac12(i-1)x_2,x_2)$ is a solution and these are all solutions. Exercise 1.3.2 We…