Exercise 3.2 · Q16: Prove $A^3-6A^2+7A+2I=O$

Basics you need

Solution

1. Check the orders

   $A$ is $3\times3$⁠. For $AA$⁠, the columns of the first factor (3) equal the rows of the second (3), so $A^2$ is defined and is $3\times3$⁠. Then $A^2A$ is again $3\times3$ times $3\times3$⁠, so $A^3$ is defined and is $3\times3$⁠. Scalar multiples keep the order, and $I$ is chosen of order 3. So all four terms $A^3$⁠, $-6A^2$⁠, $7A$⁠, $2I$ are $3\times3$⁠, the sum is defined, and the result is a $3\times3$ matrix. That is why the right-hand side must be the $3\times3$ zero matrix $O$⁠, not the number 0.

   A sum of matrices is only defined when all the terms have the same order. Checking this first tells us the identity we are proving makes sense, and tells us the size of $O$⁠.
   Product $(m\times n)(n\times p) = m\times p$⁠. Lesson: matrix multiplication

2. Compute $A^2 = AA$

   $$A^2 = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$$

   Row 1 of $A$ is $(1, 0, 2)$⁠; the columns of $A$ are $(1,0,2)$⁠, $(0,2,0)$⁠, $(2,1,3)$ read downwards.

   $$\begin{aligned} c_{11} &= 1\cdot1 + 0\cdot0 + 2\cdot2 = 5\\ c_{12} &= 1\cdot0 + 0\cdot2 + 2\cdot0 = 0\\ c_{13} &= 1\cdot2 + 0\cdot1 + 2\cdot3 = 8 \end{aligned}$$

   Row 2 of $A$ is $(0, 2, 1)$⁠:

   $$\begin{aligned} c_{21} &= 0\cdot1 + 2\cdot0 + 1\cdot2 = 2\\ c_{22} &= 0\cdot0 + 2\cdot2 + 1\cdot0 = 4\\ c_{23} &= 0\cdot2 + 2\cdot1 + 1\cdot3 = 5 \end{aligned}$$

   Row 3 of $A$ is $(2, 0, 3)$⁠:

   $$\begin{aligned} c_{31} &= 2\cdot1 + 0\cdot0 + 3\cdot2 = 8\\ c_{32} &= 2\cdot0 + 0\cdot2 + 3\cdot0 = 0\\ c_{33} &= 2\cdot2 + 0\cdot1 + 3\cdot3 = 13 \end{aligned}$$

   Assembling the nine entries:

   $$A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}$$
   $A^2$ is not found by squaring each entry. Squaring entries would give a first row $(1, 0, 4)$⁠, but the true first row is $(5, 0, 8)$⁠. Each entry of a product mixes a whole row with a whole column.
   Entry $(i,k)$ of a product: row $i$ of the first factor times column $k$ of the second. Basics: pair up, multiply, add

3. Compute $A^3 = A^2A$

   $$A^3 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$$

   Row 1 of $A^2$ is $(5, 0, 8)$⁠; the columns of $A$ are still $(1,0,2)$⁠, $(0,2,0)$⁠, $(2,1,3)$⁠.

   $$\begin{aligned} d_{11} &= 5\cdot1 + 0\cdot0 + 8\cdot2 = 21\\ d_{12} &= 5\cdot0 + 0\cdot2 + 8\cdot0 = 0\\ d_{13} &= 5\cdot2 + 0\cdot1 + 8\cdot3 = 34 \end{aligned}$$

   Row 2 of $A^2$ is $(2, 4, 5)$⁠:

   $$\begin{aligned} d_{21} &= 2\cdot1 + 4\cdot0 + 5\cdot2 = 12\\ d_{22} &= 2\cdot0 + 4\cdot2 + 5\cdot0 = 8\\ d_{23} &= 2\cdot2 + 4\cdot1 + 5\cdot3 = 23 \end{aligned}$$

   Row 3 of $A^2$ is $(8, 0, 13)$⁠:

   $$\begin{aligned} d_{31} &= 8\cdot1 + 0\cdot0 + 13\cdot2 = 34\\ d_{32} &= 8\cdot0 + 0\cdot2 + 13\cdot0 = 0\\ d_{33} &= 8\cdot2 + 0\cdot1 + 13\cdot3 = 55 \end{aligned}$$ $$A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}$$
   We reuse $A^2$ from Step 2 instead of multiplying three matrices from scratch. Computing $AA^2$ instead would give the same matrix, because $(AA)A = A(AA)$ by the associative law. Notice the middle column: every product in Steps 2 and 3 has 0 in positions $(1,2)$ and $(3,2)$⁠, because column 2 of $A$ is $(0, 2, 0)$⁠, which only picks out the middle entry of each row.
   Associative law $(AB)C = A(BC)$⁠, Section 3.4.6. Lesson: properties of the operations

4. Write the other three terms

   Multiply every entry of $A^2$ by $-6$⁠, every entry of $A$ by $7$⁠, and every entry of $I$ by $2$⁠:

   $$-6A^2 = \begin{bmatrix} -30 & 0 & -48 \\ -12 & -24 & -30 \\ -48 & 0 & -78 \end{bmatrix}$$ $$7A = \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix}$$ $$2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
   Writing $A^3 - 6A^2$ as $A^3 + (-6)A^2$ means everything in the next step is a plain sum, so there is only one sign to track per entry. The constant 2 in the polynomial becomes $2I$⁠: a number cannot be added to a matrix, but $2I$ is the $3\times3$ matrix that plays the role of "2".
   Scalar multiple: $kA = [k\,a_{ij}]$⁠. Lesson: addition and scalar multiplication

5. Add the four matrices entry by entry

   Each entry of the result is (entry of $A^3$⁠) + (entry of $-6A^2$⁠) + (entry of $7A$⁠) + (entry of $2I$⁠):

   $$\begin{aligned} (1,1)&: 21 - 30 + 7 + 2 = 0\\ (1,2)&: 0 + 0 + 0 + 0 = 0\\ (1,3)&: 34 - 48 + 14 + 0 = 0\\ (2,1)&: 12 - 12 + 0 + 0 = 0\\ (2,2)&: 8 - 24 + 14 + 2 = 0\\ (2,3)&: 23 - 30 + 7 + 0 = 0\\ (3,1)&: 34 - 48 + 14 + 0 = 0\\ (3,2)&: 0 + 0 + 0 + 0 = 0\\ (3,3)&: 55 - 78 + 21 + 2 = 0 \end{aligned}$$

   So

   $$A^3 - 6A^2 + 7A + 2I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O.$$
   Two matrices are equal only when every pair of corresponding entries is equal. So to prove the sum is $O$ we must show all nine entries are 0; it is not enough to check a few. The $2I$ term only affects the three diagonal entries, which is why the 2 appears only in positions $(1,1)$⁠, $(2,2)$⁠, $(3,3)$⁠.
   Equality of matrices: same order and every corresponding entry equal (Definition 2). Lesson: equality of matrices

Every fact used, and where it comes from