LINEAR-ALGEBRA

The Hidden Equivalence: Completing the Square = Gaussian Elimination

IKey Formulas

Formula	Name	Description
$x^{T}Ax = ax_1^2 + 2bx_1x_2 + cx_2^2$	Quadratic form (2×2)	$A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$ symmetric
$ax_1^2 + 2bx_1x_2 + cx_2^2 = a\left(x_1 + \tfrac{b}{a}x_2\right)^2 + \left(c - \tfrac{b^2}{a}\right)x_2^2$	Completing the square	Sum-of-squares decomposition
pivots: $d_1 = a,\ \ d_2 = c - \tfrac{b^2}{a}$	Gaussian elimination	Outer coefficients in sum-of-squares
multiplier: $\ell_{21} = \tfrac{b}{a}$	Elimination step	Inner shift inside the square
$A \succ 0 \iff$ all pivots $> 0$	Positivity test	Definiteness from row reduction

IIVisualization 1 — Pivots Decide the Shape

The signs of the pivots $d_1, d_2$ turn the level set of $x^{T}Ax$ into an ellipse, parallel lines, or a hyperbola.

a 2 b 6 c 20

An ellipse is the level set of a bowl; two crossing lines belong to a saddle; parallel lines mark the degenerate trough.

IIIVisualization 2 — Completing the Square = Elimination

Each step of completing the square on $A = \begin{pmatrix} 2 & 6 \\ 6 & c \end{pmatrix}$ is exactly one step of Gaussian elimination.

Step 1

Step 1 — The quadratic form

x^{T}Ax = 2x_1^2 + 12x_1x_2 + c\,x_2^2

A = \begin{pmatrix} 2 & 6 \\ 6 & c \end{pmatrix}

Step 2 — Gaussian elimination

First pivot

d_1 = a = 2

; multiplier

\ell_{21} = \tfrac{b}{a} = \tfrac{6}{2} = 3

R_2 \leftarrow R_2 - 3R_1

Step 3 — Complete the square

2\big(x_1^2 + 6x_1x_2\big) + c\,x_2^2

= 2\big(x_1 + 3x_2\big)^2 - 18x_2^2 + c\,x_2^2

The inner shift

+3x_2

is the multiplier

\ell_{21}

Step 4 — The pivot test

x^{T}Ax = 2\big(x_1 + 3x_2\big)^2 + (c - 18)\,x_2^2

Pivots

d_1 = 2

d_2 = c - 18

: so

A \succ 0 \iff c > 18

. At

c = 18

a trough, below it a saddle.

The multipliers become the inner shifts of the squares; the pivots become their outer coefficients.

IVPractice Questions

Test your understanding with the questions below.

Question 1

Consider the symmetric matrix $A = \begin{pmatrix} 3 & 6 \\ 6 & 14 \end{pmatrix}.$ Apply Gaussian elimination. What are the two pivots?

✅ Correct! Both pivots are positive, so $A$ is positive definite.

❌ Not quite. You wrote down the diagonal entries of $A$ , not the pivots. The second pivot changes after elimination.

❌ Not quite. Check the elimination step: $d_2 = c - b^2/a$ , not $c - b$ .

❌ Sign error. $d_2 = 14 - 12 = +2$ , not $-2$ .

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

The first pivot is the (1,1) entry: $d_1 = 3$ .

The multiplier is $\ell_{21} = 6/3 = 2$ . Subtracting $2$ times row 1 from row 2 gives a new (2,2) entry:

$d_2 = 14 - 2 \cdot 6 = 14 - 12 = 2.$

Equivalently, $d_2 = c - b^2/a = 14 - 36/3 = 14 - 12 = 2.$

Pivots: 3 and 2. Both positive — so $A$ is positive definite.

Question 2

True or False: If every entry of a symmetric matrix $A$ is positive, then $A$ is positive definite.

✅ Correct! Positive entries do not guarantee positive definiteness — only positive pivots do.

❌ Not quite. This is a common trap. The matrix $\begin{pmatrix} 2 & 6 \\ 6 & 7 \end{pmatrix}$ has all positive entries but is indefinite.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

False. Positivity of entries is not the test — the test is positivity of the pivots.

Counter-example from the video: $A = \begin{pmatrix} 2 & 6 \\ 6 & 7 \end{pmatrix}$ has all positive entries, but

$d_2 = 7 - \frac{6^2}{2} = 7 - 18 = -11 < 0.$

So $A$ is indefinite. Indeed, plugging in $x = (-3, 1)$ :

$x^{T}Ax = 2(9) + 12(-3)(1) + 7(1) = 18 - 36 + 7 = -11.$

The correct test: $A$ is positive definite if and only if all pivots are positive.

Question 3

Completing the square on $x^{T}Ax$ for $A = \begin{pmatrix} 2 & 6 \\ 6 & 20 \end{pmatrix}$ gives the decomposition

$x^{T}Ax = 2(x_1 + 3x_2)^2 + 2 x_2^2.$

Where does the inner coefficient 3 come from in the language of Gaussian elimination?

✅ Correct! Pivots become the outer coefficients of the squares; multipliers become the inner shifts.

❌ Not quite. The first pivot is $d_1 = 2$ (the outer coefficient of the first square), not 3.

❌ Not quite. The second pivot is $d_2 = 2$ (the outer coefficient of the second square).

❌ Not quite. $\det A = 2 \cdot 20 - 36 = 4$ , not 3.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

The decomposition is

$x^{T}Ax = d_1\,(x_1 + \ell_{21} x_2)^2 + d_2\,x_2^2.$

For $A = \begin{pmatrix} 2 & 6 \\ 6 & 20 \end{pmatrix}$ :

First pivot $d_1 = 2$ (outer coefficient of the first square)
Multiplier $\ell_{21} = b/a = 6/2 = 3$ (the inner shift inside the first square)
Second pivot $d_2 = 20 - 6^2/2 = 2$ (outer coefficient of the second square)

So 3 is the elimination multiplier $\ell_{21}$ — the number you multiply row 1 by before subtracting it from row 2 to zero out the (2,1) entry.

The video's punchline: pivots → outer coefficients, multipliers → inner shifts.

Question 4

Consider $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ . Classify its quadratic form $x^{T}Ax$ .

✅ Correct! One pivot is zero, so the form is a single square $(x_1+2x_2)^2$ — non-negative, but zero along the line $x_1 = -2x_2$ .

❌ Not quite. The second pivot is $d_2 = 4 - 4 = 0$ , not positive. Try $x = (-2, 1)$ : $x^{T}Ax = 0$ , so the form is not strictly positive.

❌ Not quite. The decomposition is $(x_1 + 2x_2)^2$ , a perfect square — it is never negative.

❌ Not quite. With all entries positive and $x = (1,0)$ giving $x^{T}Ax = 1 > 0$ , the form cannot be negative definite.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

Run elimination on $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ :

First pivot: $d_1 = 1$
Multiplier: $\ell_{21} = 2/1 = 2$
Second pivot: $d_2 = 4 - 2 \cdot 2 = 0$

Sum-of-squares decomposition:

$x^{T}Ax = 1\cdot(x_1 + 2x_2)^2 + 0 \cdot x_2^2 = (x_1 + 2x_2)^2.$

This is never negative (it's a square), but it equals zero along the line $x_1 = -2x_2$ (a whole line of nonzero vectors). So the form is positive semi-definite.

The signature is the missing (zero) pivot — not a negative one.

Solved: 0 / 4