Practise: Lists of Numbers That Do Things

Multiply entry by entry: $3u=(18,6,27)$ and $2v=(8,10,2)$. Subtract matching slots: $18-8=10$, $6-10=-4$, and $27-2=25$. The result is $(10,-4,25)$. The negative entry is legal, because a vector space allows scaling by any real number, and the list no longer describes a basket you can fill. That is the same gap between arithmetic and meaning that section 7.1 names when it averages two flats and obtains half a room. Check the first slot against the rule: three sixes minus two fours is $18-8=10$, and the other two slots follow the same pattern.

Option B fails. The list $(1,0)$ belongs to the collection, and scaling by $1/2$ produces $(1/2,0)$, whose first entry is not a whole number. So the collection leaks outside itself under a permitted move, which is exactly what section 7.1 calls a failure of closure. Options A and D are lines through the origin: scaling a list on a line through the origin keeps it on that line, and adding two lists on the same line keeps the sum there. Option C is the vertical axis, and the same argument applies. Option B is not a line through the origin; it is a grid of separated points, and halving jumps off the grid.

Matching the two entries of $a(1,2)+b(2,1)=(5,7)$ gives $a+2b=5$ and $2a+b=7$. The first equation gives $a=5-2b$. Substituting into the second gives $2(5-2b)+b=7$, so $10-3b=7$ and $b=1$. Then $a=5-2=3$. Check: $3(1,2)+1(2,1)=(3,6)+(2,1)=(5,7)$. Against the standard basis the same vector is five copies of $(1,0)$ and seven copies of $(0,1)$, so the two coordinate pairs describe one vector with two different recipes. Section 7.2 makes that the point: the numbers are not the vector, and the basis is part of the description.

The input is five copies of the first measuring list and two copies of the second: $(5,2)=5(1,0)+2(0,1)$. A linear machine sends it to $5(2,-1)+2(1,3)=(10,-5)+(2,6)=(12,1)$. The row rule agrees: the table is $\left(\begin{array}{cc}2& 1\\ -1& 3\end{array}\right)$, so the first output is $2\cdot 5+1\cdot 2=12$ and the second is $-1\cdot 5+3\cdot 2=1$. Section 7.3 calls these two readings the same arithmetic performed in two orders. The column reading explains why four numbers fix the answer for every input; the row reading is the faster way to compute one answer.

Both rows give the same equation, $x+3y=0$, because the second is twice the first. The solutions are $x=-3y$, so the kernel is the line of multiples of $(-3,1)$. Check: $1\cdot (-3)+3\cdot 1=0$ and $2\cdot (-3)+6\cdot 1=0$. The kernel has dimension 1. The columns are $(1,2)$ and $(3,6)$, and the second is three times the first, so the image is also a line, spanned by $(1,2)$, and the rank is 1. The two dimensions add to the input dimension: $1+1=2$, which is the ledger of section 7.4.

Subtract twice the first row from the second: $(2,-1,1\mid 3)-2(1,1,1\mid 6)=(0,-3,-1\mid -9)$. Subtract the first row from the third: $(1,2,-1\mid 2)-(1,1,1\mid 6)=(0,1,-2\mid -4)$. Now multiply the third row by 3 to get $(0,3,-6\mid -12)$ and add the second row: $(0,0,-7\mid -21)$. The last row reads $-7z=-21$, so $z=3$. Back substitution gives $-3y-z=-9$, so $-3y-3=-9$ and $y=2$. The first row gives $x+2+3=6$, so $x=1$. Check all three rows: $1+2+3=6$, $2-2+3=3$, and $1+4-3=2$. Every column received a pivot, so the solution is unique, as section 7.5 predicts.

The first two rows are independent and the third is their sum, so the rank is 2. The row space has dimension 2, and the column space has the same dimension, so 2. The input has four slots, so the kernel has dimension $4-2=2$. The output has three slots, so the left null space has dimension $3-2=1$. Read in order, the four dimensions are $(2,2,2,1)$. A kernel basis is $(-2,-1,1,0)$ and $(-1,-3,0,1)$: the first gives $1(-2)+0(-1)+2(1)+1(0)=0$ in row one and $0(-2)+1(-1)+1(1)+3(0)=0$ in row two, and the third row follows. A left null vector is $(-1,-1,1)$, because subtracting the first two rows from the third gives the zero row. The ledger matches section 7.6: $2+2=4$ for the inputs and $2+1=3$ for the outputs.

Option B fails the origin test: $T(0,0)=(1,0)$, while a linear map must send $(0,0)$ to $(0,0)$ because scaling the zero input by zero gives the zero output. So no table can imitate B, and section 7.3 uses exactly this shift to show that linearity is a real restriction. Option A doubles the first entry and leaves the second, which is a diagonal table. Option C returns $(x-y,0)$, whose table has rows $(1,-1)$ and $(0,0)$. Option D is the zero table, which sends everything to the origin and is linear. A map that adds a constant is affine, and an affine map is a linear map followed by a shift; the shift is what has no matrix.