Exercises 1A — Axler, Linear Algebra Done Right, 4th Edition
Exercise 1. Show that \(\alpha + \beta = \beta + \alpha\) for all \(\alpha,\beta \in \mathbb{C}\).
Proof.
\[ \text{Let } \alpha = a_R + a_I i,\quad \beta = b_R + b_I i, \quad \text{where } a_R,a_I,b_R,b_I \in \mathbb{R}. \]
\[ \begin{aligned} \alpha + \beta &= (a_R + a_I i) + (b_R + b_I i) && \text{[substitution]} \\ &= (a_R + b_R) + (a_I + b_I)i && \text{[def. of addition of complex numbers]} \\ &= (b_R + a_R) + (b_I + a_I)i && \text{[comm. prop. of addition in } \mathbb{R}\text{]} \\ &= (b_R + b_I i) + (a_R + a_I i) && \text{[def. of addition of complex numbers]} \\ &= \beta + \alpha && \text{[substitution]} \end{aligned} \]
\[ \therefore\ \alpha + \beta = \beta + \alpha \quad \text{for all } \alpha,\beta \in \mathbb{C}. \qquad \square \]
Exercise 2. Show that \((\alpha + \beta) + \lambda = \alpha + (\beta + \lambda)\) for all \(\alpha,\beta,\lambda \in \mathbb{C}\).
Proof.
\[ \text{Let } \alpha = a_R + a_I i,\quad \beta = b_R + b_I i,\quad \lambda = \ell_R + \ell_I i, \quad \text{where } a_R,a_I,b_R,b_I,\ell_R,\ell_I \in \mathbb{R}. \]
\[ \begin{aligned} (\alpha+\beta)+\lambda &= \bigl((a_R+a_I i)+(b_R+b_I i)\bigr)+(\ell_R+\ell_I i) && \text{[substitution]} \\ &= \bigl((a_R+b_R)+\ell_R\bigr)+\bigl((a_I+b_I)+\ell_I\bigr)i && \text{[def. of addition of complex numbers]} \\ &= \bigl(a_R+(b_R+\ell_R)\bigr)+\bigl(a_I+(b_I+\ell_I)\bigr)i && \text{[associative prop. of addition in } \mathbb{R}\text{]} \\ &= (a_R+a_I i)+\bigl((b_R+b_I i)+(\ell_R+\ell_I i)\bigr) && \text{[def. of addition of complex numbers]} \\ &= \alpha+(\beta+\lambda) && \text{[substitution]} \end{aligned} \]
\[ \therefore\ (\alpha+\beta)+\lambda = \alpha+(\beta+\lambda) \quad \text{for all } \alpha,\beta,\lambda \in \mathbb{C}. \qquad \square \]
Exercise 3. Show that \((\alpha\beta)\lambda = \alpha(\beta\lambda)\) for all \(\alpha,\beta,\lambda \in \mathbb{C}\).
Proof.
\[ \text{Let } \alpha = a_R + a_I i,\quad \beta = b_R + b_I i,\quad \lambda = \ell_R + \ell_I i, \quad \text{where } a_R,a_I,b_R,b_I,\ell_R,\ell_I \in \mathbb{R}. \]
\[ \begin{aligned} (\alpha\beta)\lambda &= \bigl((a_R+a_Ii)(b_R+b_Ii)\bigr)(\ell_R+\ell_Ii) && \text{[substitution]} \\ &= \bigl((a_Rb_R-a_Ib_I)+(a_Rb_I+a_Ib_R)i\bigr)(\ell_R+\ell_Ii) && \text{[def. of mult. of complex numbers]} \\ &= \bigl(a_Rb_R\ell_R - a_Ib_I\ell_R - a_Rb_I\ell_I - a_Ib_R\ell_I\bigr) \\ &\phantom{{}={}}\; + \bigl(a_Rb_R\ell_I - a_Ib_I\ell_I + a_Rb_I\ell_R + a_Ib_R\ell_R\bigr)i && \text{[def. of mult. of complex numbers]} \\ &= a_RX + a_RYi + a_I(-Y) + a_IXi && \text{[substitution, } X = b_R\ell_R-b_I\ell_I,\ Y = b_R\ell_I+b_I\ell_R\text{]} \\ &= (a_R+a_Ii)(X+Yi) \\ &= (a_R+a_Ii)\bigl((b_R\ell_R-b_I\ell_I)+(b_R\ell_I+b_I\ell_R)i\bigr) && \text{[substitution]} \\ &= (a_R+a_Ii)\bigl((b_R+b_Ii)(\ell_R+\ell_Ii)\bigr) && \text{[def. of mult. of complex numbers]} \\ &= \alpha(\beta\lambda) && \text{[substitution]} \end{aligned} \]
\[ \therefore\ (\alpha\beta)\lambda = \alpha(\beta\lambda) \quad \text{for all } \alpha,\beta,\lambda \in \mathbb{C}. \qquad \square \]
Exercise 4. Show that \(\lambda(\alpha+\beta) = \lambda\alpha + \lambda\beta\) for all \(\lambda,\alpha,\beta \in \mathbb{C}\).
Proof.
\[ \text{Let } \lambda = \ell_R + \ell_I i,\quad \alpha = a_R + a_I i,\quad \beta = b_R + b_I i, \quad \text{where } \ell_R,\ell_I,a_R,a_I,b_R,b_I \in \mathbb{R}. \]
\[ \begin{aligned} \lambda(\alpha+\beta) &= (\ell_R+\ell_I i)\bigl((a_R+a_I i)+(b_R+b_I i)\bigr) && \text{[substitution]} \\ &= (\ell_R+\ell_I i)\bigl((a_R+b_R)+(a_I+b_I)i\bigr) && \text{[def. of addition of complex numbers]} \\ &= \bigl(\ell_R(a_R+b_R) - \ell_I(a_I+b_I)\bigr) + \bigl(\ell_R(a_I+b_I) + \ell_I(a_R+b_R)\bigr)i && \text{[def. of mult. of complex numbers]} \\ &= \ell_Ra_R + \ell_Rb_R + \ell_Ra_Ii + \ell_Rb_Ii + \ell_Ia_Ri + \ell_Ib_Ri - \ell_Ia_I - \ell_Ib_I && \text{[distributive prop. in } \mathbb{R}\text{]} \\ &= \bigl(\ell_Ra_R + \ell_Ra_Ii + \ell_Ia_Ri - \ell_Ia_I\bigr) + \bigl(\ell_Rb_R + \ell_Rb_Ii + \ell_Ib_Ri - \ell_Ib_I\bigr) && \\ &= (\ell_R+\ell_Ii)(a_R+a_Ii) + (\ell_R+\ell_Ii)(b_R+b_Ii) && \text{[def. of mult. of complex numbers]} \\ &= \lambda\alpha + \lambda\beta && \text{[substitution]} \end{aligned} \]
\[ \therefore\ \lambda(\alpha+\beta) = \lambda\alpha + \lambda\beta \quad \text{for all } \lambda,\alpha,\beta \in \mathbb{C}. \qquad \square \]
Exercise 5. Show that for all \(\alpha \in \mathbb{C}\), there exists \(\beta \in \mathbb{C}\) such that \(\alpha + \beta = 0\), and that \(\beta\) is unique.
Proof.
Existence. Let \(\alpha = a_R + a_I i\) where \(a_R, a_I \in \mathbb{R}\). Let \(\beta = b_R + b_I i\) where \(b_R = -a_R\) and \(b_I = -a_I\).
\[ \begin{aligned} \alpha + \beta &= (a_R + a_I i) + (b_R + b_I i) && \text{[substitution]} \\ &= (a_R + b_R) + (a_I + b_I)i && \text{[def. of addition of complex numbers]} \\ &= (a_R - a_R) + (a_I - a_I)i && \text{[substitution]} \\ &= 0 + 0i = 0 \end{aligned} \]
So \(\beta = -a_R - a_I i\) is an additive inverse of \(\alpha\).
Uniqueness. Suppose \(\beta, \lambda \in \mathbb{C}\) both satisfy \(\alpha + \beta = 0\) and \(\alpha + \lambda = 0\). Then:
\[ \begin{aligned} \beta &= \beta + 0 && \text{[additive identity]} \\ &= \beta + (\alpha + \lambda) && \text{[since } \alpha + \lambda = 0\text{]} \\ &= (\beta + \alpha) + \lambda && \text{[associativity, Exercise 2]} \\ &= 0 + \lambda && \text{[since } \beta + \alpha = \alpha + \beta = 0\text{, by commutativity, Exercise 1]} \\ &= \lambda && \text{[additive identity]} \end{aligned} \]
\[ \therefore\ \beta = \lambda, \text{ so the additive inverse is unique.} \qquad \square \]
Exercise 6. Show that for all \(\alpha \in \mathbb{C}\) with \(\alpha \neq 0\), there exists \(\beta \in \mathbb{C}\) such that \(\alpha\beta = 1\), and that \(\beta\) is unique.
Proof.
Let \(\alpha = a_R + a_I i\) where \(a_R, a_I \in \mathbb{R}\). Since \(\alpha \neq 0\), at least one of \(a_R, a_I\) is nonzero. We consider three cases.
Existence.
Case 1: \(a_R \neq 0,\ a_I = 0\). Let \(\beta = \dfrac{1}{a_R} + 0 \cdot i\). Then:
\[ \alpha\beta = (a_R + 0 \cdot i)\!\left(\tfrac{1}{a_R} + 0 \cdot i\right) = a_R \cdot \tfrac{1}{a_R} = 1. \]
Case 2: \(a_R = 0,\ a_I \neq 0\). Let \(\beta = 0 + \dfrac{-1}{a_I} i\). Then:
\[ \alpha\beta = (0 + a_I i)\!\left(0 + \tfrac{-1}{a_I} i\right) = a_I \cdot \tfrac{-1}{a_I} \cdot i^2 = (-1)(-1) = 1. \]
Case 3: \(a_R \neq 0,\ a_I \neq 0\). Let \(\beta = b_R + b_I i\). By the definition of complex multiplication:
\[ \alpha\beta = (a_Rb_R - a_Ib_I) + (a_Rb_I + a_Ib_R)i. \]
For \(\alpha\beta = 1 = 1 + 0i\), we need:
\[ a_Rb_R - a_Ib_I = 1 \qquad \text{and} \qquad a_Rb_I + a_Ib_R = 0. \]
From the second equation: \(b_I = \dfrac{-a_Ib_R}{a_R}\). Substituting into the first:
\[ a_Rb_R - a_I\!\left(\tfrac{-a_Ib_R}{a_R}\right) = 1 \implies b_R\!\left(a_R + \tfrac{a_I^2}{a_R}\right) = 1 \implies b_R = \frac{a_R}{a_R^2 + a_I^2}. \]
Then \(b_I = \dfrac{-a_I}{a_R^2 + a_I^2}\), so \(\beta = \dfrac{a_R - a_Ii}{a_R^2 + a_I^2}\) satisfies \(\alpha\beta = 1\).
Uniqueness. Suppose \(\beta, \lambda \in \mathbb{C}\) both satisfy \(\alpha\beta = 1\) and \(\alpha\lambda = 1\). Then:
\[ \begin{aligned} \beta &= 1 \cdot \beta && \text{[multiplicative identity]} \\ &= (\alpha\lambda)\beta && \text{[since } \alpha\lambda = 1\text{]} \\ &= (\lambda\alpha)\beta && \text{[commutativity of mult. of complex numbers]} \\ &= \lambda(\alpha\beta) && \text{[associativity, Exercise 3]} \\ &= \lambda \cdot 1 && \text{[since } \alpha\beta = 1\text{]} \\ &= \lambda. && \text{[multiplicative identity]} \end{aligned} \]
\[ \therefore\ \beta = \lambda, \text{ so the multiplicative inverse is unique.} \qquad \square \]
Exercise 7. Show that \(\dfrac{-1+\sqrt{3}\,i}{2}\) is a cube root of \(1\).
Proof.
We show directly that \(\left(\dfrac{-1+\sqrt{3}\,i}{2}\right)^{\!3} = 1\).
\[ \begin{aligned} \left(\frac{-1+\sqrt{3}\,i}{2}\right)^{\!3} &= \left(\frac{-1+\sqrt{3}\,i}{2}\right)^{\!2}\!\left(\frac{-1+\sqrt{3}\,i}{2}\right) \\[6pt] &= \left(\frac{1 - 2\sqrt{3}\,i - 3}{4}\right)\!\left(\frac{-1+\sqrt{3}\,i}{2}\right) && \text{[def. of mult. of complex numbers]} \\[6pt] &= \frac{1}{8}\,(-2 - 2\sqrt{3}\,i)(-1+\sqrt{3}\,i) \\[6pt] &= \frac{1}{8}\,\bigl[2 - 2\sqrt{3}\,i + 2\sqrt{3}\,i + 6\bigr] && \text{[def. of mult. of complex numbers]} \\[6pt] &= \frac{1}{8}\cdot 8 = 1. \qquad \square \end{aligned} \]
Exercise 8. Find two distinct square roots of \(i\).
Solution.
Let \(\alpha = a_R + a_I i\) where \(a_R, a_I \in \mathbb{R}\) satisfy \(\alpha^2 = i\). Then:
\[ \alpha^2 = (a_R + a_I i)^2 = (a_R^2 - a_I^2) + 2a_Ra_I i = i = 0 + 1\cdot i. \]
Equating real and imaginary parts:
\[ a_R^2 - a_I^2 = 0 \qquad \text{and} \qquad 2a_Ra_I = 1. \]
From the first equation, \(a_R^2 = a_I^2\), so \(a_R = \pm\, a_I\).
- If \(a_R = -a_I\): substituting into the second equation gives \(-2a_I^2 = 1\), i.e. \(a_I^2 = -\tfrac{1}{2}\), which has no solution in \(\mathbb{R}\).
- If \(a_R = a_I\): substituting into the second equation gives \(2a_I^2 = 1\), so \(a_I^2 = \tfrac{1}{2}\), hence \(a_I = \pm\dfrac{1}{\sqrt{2}}\) and \(a_R = \pm\dfrac{1}{\sqrt{2}}\) (same sign).
This yields two solutions:
\[ \alpha = \frac{1+i}{\sqrt{2}}, \qquad \beta = -\frac{1+i}{\sqrt{2}}. \]
Verification. \(\alpha^2 = \dfrac{(1+i)^2}{2} = \dfrac{1+2i-1}{2} = \dfrac{2i}{2} = i\), and \(\beta^2 = \left(-\dfrac{1+i}{\sqrt{2}}\right)^{\!2} = \dfrac{(1+i)^2}{2} = i\). Since \(\alpha \neq \beta\), these are two distinct square roots of \(i\). \(\square\)
Exercise 9. Find \(x \in \mathbb{R}^4\) such that \((4,-3,1,7) + 2x = (5,9,-6,8)\).
Solution.
Let \(x = (x_1, x_2, x_3, x_4)\) where \(x_i \in \mathbb{R}\) for \(i \in \{1,2,3,4\}\). Then:
\[ (4,-3,1,7) + 2x = (4+2x_1,\,-3+2x_2,\,1+2x_3,\,7+2x_4) = (5,9,-6,8). \]
Equating components:
\[ \begin{aligned} 4 + 2x_1 &= 5 &&\Rightarrow& x_1 &= \tfrac{1}{2} \\ -3 + 2x_2 &= 9 &&\Rightarrow& x_2 &= 6 \\ 1 + 2x_3 &= -6 &&\Rightarrow& x_3 &= -\tfrac{7}{2} \\ 7 + 2x_4 &= 8 &&\Rightarrow& x_4 &= \tfrac{1}{2} \end{aligned} \]
\[ \therefore\ x = \left(\tfrac{1}{2},\, 6,\, -\tfrac{7}{2},\, \tfrac{1}{2}\right). \qquad \square \]
Exercise 10. Explain why there does not exist \(\lambda \in \mathbb{C}\) such that \[\lambda(2-3i,\,5+4i,\,-6+7i) = (12-5i,\,7+22i,\,-32-9i).\]
Proof.
Assume for contradiction that such a \(\lambda \in \mathbb{C}\) exists. Then \(\lambda\) must simultaneously satisfy:
\[ \lambda = \frac{12-5i}{2-3i}, \qquad \lambda = \frac{7+22i}{5+4i}, \qquad \lambda = \frac{-32-9i}{-6+7i}. \]
From the first two expressions, \((12-5i)(5+4i) = (7+22i)(2-3i)\). Computing each side:
\[ (12-5i)(5+4i) = 60+48i-25i+20 = 80+23i, \] \[ (7+22i)(2-3i) = 14-21i+44i+66 = 80+23i. \]
From the second and third expressions, \((7+22i)(-6+7i) = (-32-9i)(5+4i)\). Computing each side:
\[ (7+22i)(-6+7i) = -42+49i-132i-154 = -196-83i, \] \[ (-32-9i)(5+4i) = -160-128i-45i+36 = -124-173i. \]
But \(-196-83i \neq -124-173i\), a contradiction. Therefore no such \(\lambda \in \mathbb{C}\) exists. \(\square\)
Exercise 11. Show that \((x+y)+z = x+(y+z)\) for all \(x,y,z \in F^n\), where \(F\) is a field and \(n \in \mathbb{Z}^+\).
Proof.
Let \(x=(x_1,\ldots,x_n)\), \(y=(y_1,\ldots,y_n)\), \(z=(z_1,\ldots,z_n)\) where \(x_i,y_i,z_i \in F\).
\[ \begin{aligned} (x+y)+z &= \bigl((x_1,\ldots,x_n)+(y_1,\ldots,y_n)\bigr)+(z_1,\ldots,z_n) && \text{[substitution]} \\ &= (x_1+y_1,\ldots,x_n+y_n)+(z_1,\ldots,z_n) && \text{[def. of addition in } F^n\text{]} \\ &= \bigl((x_1+y_1)+z_1,\ldots,(x_n+y_n)+z_n\bigr) && \text{[def. of addition in } F^n\text{]} \\ &= \bigl(x_1+(y_1+z_1),\ldots,x_n+(y_n+z_n)\bigr) && \text{[associativity of addition in } F\text{]} \\ &= (x_1,\ldots,x_n)+(y_1+z_1,\ldots,y_n+z_n) && \text{[def. of addition in } F^n\text{]} \\ &= (x_1,\ldots,x_n)+\bigl((y_1,\ldots,y_n)+(z_1,\ldots,z_n)\bigr) && \text{[def. of addition in } F^n\text{]} \\ &= x+(y+z) && \text{[substitution]} \end{aligned} \]
\[ \therefore\ (x+y)+z = x+(y+z) \quad \text{for all } x,y,z \in F^n. \qquad \square \]
Exercise 12. Show that \((ab)x = a(bx)\) for all \(x \in F^n\) and \(a,b \in F\), where \(F\) is a field and \(n \in \mathbb{Z}^+\).
Proof.
Let \(x = (x_1,\ldots,x_n)\) where \(x_i \in F\) for \(i \in \{1,\ldots,n\}\).
\[ \begin{aligned} (ab)x &= (ab)(x_1,\ldots,x_n) && \text{[substitution]} \\ &= \bigl((ab)x_1,\ldots,(ab)x_n\bigr) && \text{[def. of scalar mult. in } F^n\text{]} \\ &= \bigl(a(bx_1),\ldots,a(bx_n)\bigr) && \text{[multiplicative associativity in } F\text{]} \\ &= a(bx_1,\ldots,bx_n) && \text{[def. of scalar mult. in } F^n\text{]} \\ &= a(bx) && \text{[substitution]} \end{aligned} \]
\[ \therefore\ (ab)x = a(bx) \quad \text{for all } x \in F^n,\ a,b \in F. \qquad \square \]
Exercise 13. Show that \(1x = x\) for all \(x \in F^n\), where \(F\) is a field and \(n \in \mathbb{Z}^+\).
Proof.
Let \(x = (x_1,\ldots,x_n)\) where \(x_i \in F\) for \(i \in \{1,\ldots,n\}\).
\[ \begin{aligned} 1x &= 1(x_1,\ldots,x_n) && \text{[substitution]} \\ &= (1x_1,\ldots,1x_n) && \text{[def. of scalar mult. in } F^n\text{]} \\ &= (x_1,\ldots,x_n) && \text{[multiplicative identity in } F\text{]} \\ &= x && \text{[substitution]} \end{aligned} \]
\[ \therefore\ 1x = x \quad \text{for all } x \in F^n. \qquad \square \]
Exercise 14. Show that \(\lambda(x+y) = \lambda x + \lambda y\) for all \(\lambda \in F\) and all \(x,y \in F^n\), where \(F\) is a field.
Proof.
Let \(x = (x_1,\ldots,x_n)\) and \(y = (y_1,\ldots,y_n)\) where \(x_i, y_i \in F\).
\[ \begin{aligned} \lambda(x+y) &= \lambda\bigl((x_1,\ldots,x_n)+(y_1,\ldots,y_n)\bigr) && \text{[substitution]} \\ &= \lambda(x_1+y_1,\ldots,x_n+y_n) && \text{[def. of addition in } F^n\text{]} \\ &= \bigl(\lambda(x_1+y_1),\ldots,\lambda(x_n+y_n)\bigr) && \text{[def. of scalar mult. in } F^n\text{]} \\ &= (\lambda x_1+\lambda y_1,\ldots,\lambda x_n+\lambda y_n) && \text{[distributive property of } F\text{]} \\ &= (\lambda x_1,\ldots,\lambda x_n)+(\lambda y_1,\ldots,\lambda y_n) && \text{[def. of addition in } F^n\text{]} \\ &= \lambda(x_1,\ldots,x_n)+\lambda(y_1,\ldots,y_n) && \text{[def. of scalar mult. in } F^n\text{]} \\ &= \lambda x + \lambda y && \text{[substitution]} \end{aligned} \]
\[ \therefore\ \lambda(x+y) = \lambda x + \lambda y \quad \text{for all } \lambda \in F,\ x,y \in F^n. \qquad \square \]
Exercise 15. Show that \((a+b)x = ax + bx\) for all \(a,b \in F\) and \(x \in F^n\), where \(F\) is a field.
Proof.
Let \(x = (x_1,\ldots,x_n)\) where \(x_i \in F\) for \(i \in \{1,\ldots,n\}\).
\[ \begin{aligned} (a+b)x &= (a+b)(x_1,\ldots,x_n) && \text{[substitution]} \\ &= \bigl((a+b)x_1,\ldots,(a+b)x_n\bigr) && \text{[def. of scalar mult. in } F^n\text{]} \\ &= (ax_1+bx_1,\ldots,ax_n+bx_n) && \text{[distributive property of } F\text{]} \\ &= (ax_1,\ldots,ax_n)+(bx_1,\ldots,bx_n) && \text{[def. of addition in } F^n\text{]} \\ &= a(x_1,\ldots,x_n)+b(x_1,\ldots,x_n) && \text{[def. of scalar mult. in } F^n\text{]} \\ &= ax+bx && \text{[substitution]} \end{aligned} \]
\[ \therefore\ (a+b)x = ax+bx \quad \text{for all } a,b \in F,\ x \in F^n. \qquad \square \]