Exercise 1

Let $ f:\mathbb{R}\to\mathbb{R} $ be a function. Prove that the following are equivalent: (i) There is a constant $ a\in\mathbb{R} $ such that for every$ x\in\mathbb{R} $ we have$ f(x)=ax $. (ii) For all $ x_1,x_2,c,x\in\mathbb{R} $ we have $ f(x_1+x_2)=f(x_1)+f(x_2) $ and $ f(cx)=c\,f(x) $.

(i ⇒ ii)

Assume there exists $ a\in\mathbb{R} $ such that $ f(x)=a x $ for all $ x\in\mathbb{R} $ . Then for any $ x_1,x_2,c,x\in\mathbb{R} $ ,

  • $ f(x_1+x_2)=a(x_1+x_2)=ax_1+ax_2=f(x_1)+f(x_2) $;
  • $ f(cx)=a(cx)=c(ax)=c\,f(x) $.

Hence (ii) holds.

(ii ⇒ i)

Assume (ii) hold for all real scalars. Define $ a:=f(1) $. For any $ x\in\mathbb{R} $ we can write $ x=x\cdot 1 $; by homogeneity, we have $ f(x)=f(x\cdot 1)=x\,f(1)=a x $.

Thus (i) holds with $ a=f(1) $. Therefore (i) and (ii) are equivalent.

Exercise 2

Describe geometrically (line, plane, or all of $ \mathbb{R}^3 $) all linear combinations of the given vectors.

(a) $ \begin{pmatrix}1\\2\\3\end{pmatrix} $ and $ \begin{pmatrix}3\\6\\9\end{pmatrix} $.

  • Observe that$ \begin{pmatrix}3\\6\\9\end{pmatrix}=3\begin{pmatrix}1\\2\\3\end{pmatrix} $, so the two vectors are colinear.
  • Hence their span is $ \mathrm{Span}=\{\, t\begin{pmatrix}1\\2\\3\end{pmatrix}\mid t\in\mathbb{R}\,\} $, which is a line.

(b) $ \begin{pmatrix}1\\0\\0\end{pmatrix} $ and $ \begin{pmatrix}0\\2\\3\end{pmatrix} $.

  • These two vectors are not scalar multiples of each other, hence they are linearly independent.
  • Therefore their span $ \mathrm{Span}=\{\, \alpha\begin{pmatrix}1\\0\\0\end{pmatrix}+\beta\begin{pmatrix}0\\2\\3\end{pmatrix}\mid \alpha,\beta\in\mathbb{R}\,\} $ is a plane.

(c) $ \begin{pmatrix}2\\0\\0\end{pmatrix} $, $ \begin{pmatrix}0\\2\\2\end{pmatrix} $ and $ \begin{pmatrix}2\\2\\3\end{pmatrix} $.

  • Form the matrix with these as columns and compute the determinant: $$ \det\begin{pmatrix} 2 & 0 & 2\\ 0 & 2 & 2\\ 0 & 2 & 3 \end{pmatrix} \neq 0. $$
  • Since the determinant is nonzero, the three vectors are linearly independent,therefore their span all of $ \mathbb{R}^3 $.

Answers:

  • (a) a line.
  • (b) a plane.
  • (c) all of $ \mathbb{R}^3 $.

Exercise 3

Consider $ v=(1,-2,1) $ and $ w=(0,1,-1) $. Find $ c $ and $ d $ such that $ c v+d w=(3,3,-6) $. Why is $ (3,3,6) $ impossible?

Let $ c v+d w=(3,3,-6) $. Comparing coordinates: $$ \begin{cases} c = 3 \\ -2c+d=9 \\ c-d=-6 \end{cases} $$ It can be solved that $ c=3 $ and $ d=9 $.

For $ (3,3,6) $, the first coordinate again forces $ c=3 $, and the second gives $ d=9 $, hence the third would be $ c-d=3-9=-6\neq 6 $. Therefore it is impossible. Equivalently, every linear combination satisfies $ y+z=-x $ (since $ y=-2c+d $ and $ z=c-d $), but the vector $ (3,3,6) $ has $ y+z=9\neq -3 $; hence it does not lie in $ \mathrm{Span}\{v,w\} $.

Exercise 4

In the following, tacitly assume that every matrix operation is well-defined. Prove: (i) $ A+B=B+A $. (ii) $ c(A+B)=cA+cB $. (iii) $ A+(B+C)=(A+B)+C $. (iv) $ A(B+C)=AB+AC $. (v) $ (A+B)C=AC+BC $. (vi) $ A(BC)=(AB)C $.

Write $ A=(a_{ij}) $, $ B=(b_{ij}) $, $ C=(c_{ij}) $. Use the entrywise definitions of addition and multiplication.

  • (i) Commutativity of addition: For all $ i,j $, $ (A+B)_{ij}=a_{ij}+b_{ij}=b_{ij}+a_{ij}=(B+A)_{ij} $. Hence $ A+B=B+A $.

  • (ii) Distributivity of scalar multiplication over addition: For all $ i,j $, $ \big(c(A+B)\big)_{ij}=c(a_{ij}+b_{ij})=ca_{ij}+cb_{ij}=(cA+cB)_{ij} $.

  • (iii) Associativity of addition: For all $ i,j $, $ \big(A+(B+C)\big)_{ij}=a_{ij}+(b_{ij}+c_{ij})=(a_{ij}+b_{ij})+c_{ij}=\big((A+B)+C\big)_{ij} $.

  • (iv) Left distributivity of multiplication: $$ \big(A(B+C)\big)_{ij} =\sum_k a_{ik}(b_{kj}+c_{kj}) =\sum_k a_{ik}b_{kj}+\sum_k a_{ik}c_{kj} =(AB)_{ij}+(AC)_{ij} =(AB+AC)_{ij}. $$

  • (v) Right distributivity of multiplication: $$ \big((A+B)C\big)_{ij} =\sum_k (a_{ik}+b_{ik})c_{kj} =\sum_k a_{ik}c_{kj}+\sum_k b_{ik}c_{kj} =(AC)_{ij}+(BC)_{ij} =(AC+BC)_{ij}. $$

  • (vi) Associativity of multiplication: $$ \big(A(BC)\big)_{ij} =\sum_k a_{ik}(BC)_{kj} =\sum_k a_{ik}\sum_\ell b_{k\ell}c_{\ell j} =\sum_\ell\Big(\sum_k a_{ik}b_{k\ell}\Big)c_{\ell j} =\big((AB)C\big)_{ij}. $$

Therefore all properties (i)–(vi) hold.

Exercise 5

Let$ A=\begin{pmatrix}3&1\\[2pt] 1&-3\end{pmatrix} $. Compute $ A^{50} $ and $ A^{51} $.

First compute$ A^2 $: $$ A^2 =\begin{pmatrix}3&1\\ 1&-3\end{pmatrix} \begin{pmatrix}3&1\\ 1&-3\end{pmatrix} =\begin{pmatrix} 3\cdot 3+1\cdot 1 & 3\cdot 1+1\cdot(-3)\\ 1\cdot 3+(-3)\cdot 1 & 1\cdot 1+(-3)\cdot(-3) \end{pmatrix} =\begin{pmatrix}10&0\\ 0&10\end{pmatrix} =10\,I_2. $$ Thus $ A^2=10I_2 $. It follows that for any integer $ n\ge 1 $,

  • if $ n $ is even, $ A^n=(A^2)^{n/2}=10^{n/2}I_2 $;
  • if $ n $ is odd, $ A^n=A\cdot A^{n-1}=A\cdot 10^{(n-1)/2}I_2=10^{(n-1)/2}A $.

Therefore,

  • $ A^{50}=10^{25}I_2 $,
  • $ A^{51}=10^{25}A=10^{25}\begin{pmatrix}3&1\\[2pt] 1&-3\end{pmatrix} $.