Miscellaneous · Q11 (MCQ): $(I+A)^3-7A$ when $A^2=A$

Basics you need

Solution

1. Expand $(I + A)^2$

   A and I are both $n\times n$⁠, so $I + A$ is $n\times n$ and every product below is defined. By the distributive law (twice),

   $$\begin{aligned} (I + A)^2 &= (I + A)(I + A) \\ &= I\cdot I + I\cdot A \\ &\qquad + A\cdot I + A\cdot A \\ &= I + A + A + A^2 \\ &= I + 2A + A^2. \end{aligned}$$

   Now use $A^2 = A$⁠:

   $$(I + A)^2 = I + 2A + A = I + 3A.$$
   For general matrices X and Y, $(X + Y)^2 = X^2 + XY + YX + Y^2$⁠, and $XY$ and $YX$ can differ, so the familiar "$2XY$" is not allowed in general. Here one of the two is I, and $IA = AI = A$⁠, so the middle terms are both A and combine to $2A$⁠.
   Distributive law and multiplicative identity, Section 3.4.6. Lesson: properties of the operations

2. Multiply by $I + A$ once more

   $$\begin{aligned} (I + A)^3 &= (I + A)^2 (I + A) \\ &= (I + 3A)(I + A) \\ &= I\cdot I + I\cdot A \\ &\qquad + 3A\cdot I + 3A\cdot A \\ &= I + A + 3A + 3A^2 \\ &= I + 4A + 3A. \end{aligned}$$

   So

   $$(I + A)^3 = I + 7A.$$
   The fourth line uses $3A\cdot A = 3(AA) = 3A^2$ (a scalar factor comes out of a product), and then $A^2 = A$ turns $3A^2$ into $3A$⁠. Writing $(I + A)^3 = (I + A)^2(I + A)$ is allowed by associativity.
   Associative and distributive laws, Section 3.4.6. Lesson: properties of the operations

3. Subtract $7A$

   $$(I + A)^3 - 7A = I + 7A - 7A = I.$$

   So the expression always equals I, whatever idempotent matrix A is.

   The 7 in the question was chosen to cancel the $7A$ exactly. (A matrix with $A^2 = A$ is called idempotent: squaring it changes nothing, so neither does any higher power.)
   $7A - 7A = O$ and $I + O = I$ (Section 3.4.3). Lesson: addition and scalar multiplication

4. Check with an actual matrix

   Take the matrix F below. Its square has entries $1\cdot1 + 1\cdot0 = 1$⁠, $1\cdot1 + 1\cdot0 = 1$⁠, $0\cdot1 + 0\cdot0 = 0$ and $0\cdot1 + 0\cdot0 = 0$⁠, so $F^2 = F$⁠.

   $$F = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \quad I + F = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}$$

   Square $I + F$⁠, entry by entry:

   $$\begin{aligned} 2\cdot2 + 1\cdot0 &= 4, \quad 2\cdot1 + 1\cdot1 = 3 \\ 0\cdot2 + 1\cdot0 &= 0, \quad 0\cdot1 + 1\cdot1 = 1 \end{aligned}$$

   so $(I + F)^2$ has rows $(4, 3)$ and $(0, 1)$⁠, which is $I + 3F$ ✓. Multiply by $I + F$ again:

   $$\begin{aligned} 4\cdot2 + 3\cdot0 &= 8, \quad 4\cdot1 + 3\cdot1 = 7 \\ 0\cdot2 + 1\cdot0 &= 0, \quad 0\cdot1 + 1\cdot1 = 1 \end{aligned}$$ $$(I + F)^3 = \begin{bmatrix} 8 & 7 \\ 0 & 1 \end{bmatrix} = I + 7F$$ $$\begin{aligned} (I + F)^3 - 7F &= \begin{bmatrix} 8 - 7 & 7 - 7 \\ 0 - 0 & 1 - 0 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I. \ ✓ \end{aligned}$$
   An example cannot prove the general result (Steps 1 to 3 do that), but it confirms the algebra, and it is exactly the kind of matrix we need to rule out the wrong options in the next step.
   Row times column. Basics: pair up, multiply, add

5. Examine every option

   For the matrix F of Step 4 the expression equals I, so any option that is not I for this F is wrong.

   Option	Value for F	Verdict
   (A)	A $= F$⁠, rows $(1, 1)$⁠, $(0, 0)$	Wrong: not I. (It would also need $I + 7A - 7A = A$⁠, that is $A = I$⁠.)
   (B)	$I - F$⁠, rows $(0, -1)$⁠, $(0, 1)$	Wrong: not I. (It equals I only when $A = O$⁠.)
   (C)	I	Correct, for every A with $A^2 = A$ (Step 3).
   (D)	$3F$⁠, rows $(3, 3)$⁠, $(0, 0)$	Wrong: not I. ($3A = I$ is impossible with $A^2 = A$⁠: it would give $A = \frac13 I$⁠, and then $A^2 = \frac19 I \ne A$⁠.)

   Options (A) and (B) do coincide with I for special idempotents ($A = I$ and $A = O$ respectively), which is why a single "nice" test matrix like I or O is not enough to separate the options. F is neither I nor O, and it rules out (A), (B) and (D) at once.
   A counterexample disproves a general claim. Lesson: properties of the operations

Every fact used, and where it comes from