Linear-Algebra-Done-Right · Unit 27 · Video 4 · Interactive Practice

Every Bilinear Form Splits: Symmetric + Alternating

IKey Formulas

Formula	Name	Hypotheses
$\alpha(v, v) = 0$ for all $v \in V$	Alternating	$\alpha \in V^{(2)}$ ; the set of these is $V^{(2)}_{\mathrm{alt}}$
$\alpha$ alternating $\iff \alpha(u, w) = -\alpha(w, u)$ for all $u, w \in V$	Alternating = antisymmetric	$\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$ , so dividing by $2$ is allowed
$\rho(u, w) = \dfrac{\beta(u, w) + \beta(w, u)}{2}$ , $\ \alpha(u, w) = \dfrac{\beta(u, w) - \beta(w, u)}{2}$	Symmetric part and alternating part; $\beta = \rho + \alpha$	$\beta \in V^{(2)}$
$V^{(2)} = V^{(2)}_{\mathrm{sym}} \oplus V^{(2)}_{\mathrm{alt}}$	Direct sum	$\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$ ; holds because $V^{(2)}_{\mathrm{sym}} \cap V^{(2)}_{\mathrm{alt}} = \{0\}$

Key Insight: Swapping the inputs of $\beta$ transposes its matrix, so with $A = \mathcal{M}(\beta)$ the split reads $A = \tfrac{1}{2}(A + A^{\mathrm{t}}) + \tfrac{1}{2}(A - A^{\mathrm{t}})$ : a symmetric matrix plus one with zero diagonal and $A_{k,j} = -A_{j,k}$ .

IIAverage and Half-Difference of a Matrix

With $A = \mathcal{M}(\beta)$ , the symmetric part has matrix $\tfrac{1}{2}(A + A^{\mathrm{t}})$ and the alternating part $\tfrac{1}{2}(A - A^{\mathrm{t}})$ .

β(e₁, e₁) 1 β(e₁, e₂) 3 β(e₂, e₁) −1 β(e₂, e₂) 2

IIIZero on Equal Inputs Forces Antisymmetry

$\beta(u + w, u + w)$ expands into four terms; if $\beta$ is alternating, the whole sum and the two terms $\beta(u, u)$ , $\beta(w, w)$ are $0$ , which forces $\beta(u, w) = -\beta(w, u)$ .

β(x, y) =

💡 The converse runs the other way: antisymmetry with $u = w = v$ gives $2\beta(v, v) = 0$ , and dividing by $2$ (allowed since $\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$ ) gives $\beta(v, v) = 0$ .

IVQuiz Questions

Problem 1 · The Alternating Part of a 3 × 3 Matrix

Given: the bilinear form $\beta$ on $\mathbf{F}^3$ with $\mathcal{M}(\beta) = \begin{pmatrix} 2 & 5 & -1 \\ 1 & 0 & 4 \\ 3 & -2 & 1 \end{pmatrix}$ (standard basis) — find the matrix of its alternating part $\alpha$ .

✅ Correct!

\mathcal{M}(\alpha) = \tfrac{1}{2}(A - A^{\mathrm{t}})

has zero diagonal and

\alpha_{k,j} = -\alpha_{j,k}

❌ You forgot the factor $\tfrac{1}{2}$ .

A - A^{\mathrm{t}}

is alternating, but adding it to the symmetric part would give

\beta

plus another copy of its alternating part, not

\beta

❌ The subtraction is reversed. That is

\tfrac{1}{2}(A^{\mathrm{t}} - A)

, the negative of the alternating part:

\alpha(u, w) = \tfrac{1}{2}\big(\beta(u, w) - \beta(w, u)\big)

starts with

\beta(u, w)

❌ The diagonal belongs to the symmetric part. An alternating form has

\alpha(e_j, e_j) = 0

, so every diagonal entry of

\mathcal{M}(\alpha)

0

Show solution

Swapping the inputs transposes the matrix: $\beta(e_k, e_j) = A_{k,j}$ , so the swapped form has matrix $A^{\mathrm{t}}$ . Then $\mathcal{M}(\alpha) = \tfrac{1}{2}(A - A^{\mathrm{t}})$ :

A^{\mathrm{t}} = \begin{pmatrix} 2 & 1 & 3 \\ 5 & 0 & -2 \\ -1 & 4 & 1 \end{pmatrix}, \qquad \tfrac{1}{2}(A - A^{\mathrm{t}}) = \tfrac{1}{2}\begin{pmatrix} 0 & 4 & -4 \\ -4 & 0 & 6 \\ 4 & -6 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 2 & -2 \\ -2 & 0 & 3 \\ 2 & -3 & 0 \end{pmatrix}.

Entry by entry: $\alpha_{1,2} = \tfrac{1}{2}(5 - 1) = 2$ , $\alpha_{1,3} = \tfrac{1}{2}(-1 - 3) = -2$ , $\alpha_{2,3} = \tfrac{1}{2}\big(4 - (-2)\big) = 3$ , and the entries below the diagonal are their negatives.

Check: the symmetric part is $\tfrac{1}{2}(A + A^{\mathrm{t}}) = \begin{pmatrix} 2 & 3 & 1 \\ 3 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ , and adding the two matrices gives back $A$ .

Problem 2 · Which Form Is Alternating?

Given: four bilinear forms on $\mathbf{F}^4$ , with $x = (x_1, x_2, x_3, x_4)$ and $y = (y_1, y_2, y_3, y_4)$ — find the one that is alternating.

✅ Correct! Its matrix has zero diagonal and opposite entries across it, so it is antisymmetric, hence alternating.

❌ Zero on each basis vector is not enough. This form is symmetric, and

\beta(e_1 + e_2, e_1 + e_2) = 1 + 1 = 2 \ne 0

❌ Put a basis vector in both slots.

\beta(e_1, e_1) = 1 \ne 0

; the term

x_1 y_1

is a nonzero diagonal entry.

❌ Only the first two terms are antisymmetric.

x_3 y_4 + x_4 y_3

gives

\beta(e_3 + e_4, e_3 + e_4) = 2 \ne 0

Show solution

Put $y = x$ in each form; a form is alternating exactly when the result is $0$ for every $x$ .

(a) $x_1 x_2 + x_2 x_1 - x_3 x_4 - x_4 x_3 = 2x_1 x_2 - 2x_3 x_4$ , not always $0$ .
(b) $x_1 x_3 - x_3 x_1 + 2x_2 x_4 - 2x_4 x_2 = 0$ for every $x$ . ✓
(c) $x_1^2 - x_2^2 + x_3 x_4 - x_4 x_3 = x_1^2 - x_2^2$ , not always $0$ .
(d) $x_1 x_2 - x_2 x_1 + x_3 x_4 + x_4 x_3 = 2x_3 x_4$ , not always $0$ .

Equivalently, by the theorem, alternating means antisymmetric: swapping $x$ and $y$ in (b) gives $x_3 y_1 - x_1 y_3 + 2x_4 y_2 - 2x_2 y_4$ , the negative of (b). Form (a) has zero diagonal entries but equal (not opposite) entries across the diagonal, which is why checking only $\beta(e_j, e_j) = 0$ misleads.

Problem 3 · Evaluate an Alternating Form on F⁴

Given: $\alpha$ is an alternating bilinear form on $\mathbf{F}^4$ with $\alpha(e_1, e_2) = 3$ , $\alpha(e_1, e_4) = -1$ , $\alpha(e_2, e_3) = 2$ , $\alpha(e_3, e_4) = 5$ , and $\alpha(e_j, e_k) = 0$ for the other pairs with $j < k$ — find $\alpha(u, w)$ for $u = (2, 0, 1, 0)$ and $w = (0, 1, 0, 1)$ .

✅ Correct!

6 - 2 - 2 + 5 = 7

, using

\alpha(e_3, e_2) = -\alpha(e_2, e_3) = -2

❌ The sign flips when the inputs swap. You used

\alpha(e_3, e_2) = +2

; antisymmetry gives

\alpha(e_3, e_2) = -\alpha(e_2, e_3) = -2

❌ $\alpha(e_3, e_2)$ is not zero. Only the unlisted pairs with

j < k

are

0

;

\alpha(e_3, e_2)

is determined by

\alpha(e_2, e_3) = 2

❌ That is $\alpha(w, u)$ . Keep

u

in the first slot:

\alpha(u, w) = -\alpha(w, u)

Show solution

Write $u = 2e_1 + e_3$ and $w = e_2 + e_4$ , and expand using linearity in each slot:

\alpha(u, w) = 2\alpha(e_1, e_2) + 2\alpha(e_1, e_4) + \alpha(e_3, e_2) + \alpha(e_3, e_4).

Alternating implies antisymmetric, so $\alpha(e_3, e_2) = -\alpha(e_2, e_3) = -2$ . Therefore

\alpha(u, w) = 2(3) + 2(-1) + (-2) + 5 = 7.

In matrix terms, $\mathcal{M}(\alpha)$ is the antisymmetric matrix $\begin{pmatrix} 0 & 3 & 0 & -1 \\ -3 & 0 & 2 & 0 \\ 0 & -2 & 0 & 5 \\ 1 & 0 & -5 & 0 \end{pmatrix}$ , and $\alpha(u, w)$ is the sum of $u_j A_{j,k} w_k$ over all $j, k$ , which again gives $7$ .

Problem 4 · Counting Dimensions

Given: $\dim V = 4$ , so $\dim V^{(2)} = 16$ — find $\dim V^{(2)}_{\mathrm{alt}}$ and $\dim V^{(2)}_{\mathrm{sym}}$ .

What is dim V⁽²⁾ alt?

What is dim V⁽²⁾ sym?

✅ Correct!

6 + 10 = 16

, exactly as

V^{(2)} = V^{(2)}_{\mathrm{sym}} \oplus V^{(2)}_{\mathrm{alt}}

requires.

❌ Count the free entries of an antisymmetric matrix. The diagonal is

0

and each entry below it is the negative of one above it.

❌ The two subspaces are not the same size. The diagonal is forced to

0

V^{(2)}_{\mathrm{alt}}

but free in

V^{(2)}_{\mathrm{sym}}

, so the

16

dimensions do not split evenly.

❌ The free entries are off the diagonal, not on it. An alternating form has every diagonal entry

0

; the

6

entries above the diagonal are the free ones.

❌ The entries below the diagonal are not free. Of the

12

off-diagonal entries, those below the diagonal are fixed by

A_{k,j} = -A_{j,k}

❌ Count the free entries of a symmetric matrix. The diagonal and the entries above it are free; those below are copies.

❌ The diagonal is free for a symmetric form. Count the

4

diagonal entries together with the

6

above it.

❌ The split is not even.

\dim V^{(2)}_{\mathrm{sym}} + \dim V^{(2)}_{\mathrm{alt}} = 16

with

\dim V^{(2)}_{\mathrm{alt}} = 6

❌ Not every bilinear form is symmetric.

V^{(2)}_{\mathrm{sym}}

is a proper subspace, since

V^{(2)}_{\mathrm{alt}} \ne \{0\}

and the two meet only in

0

Show solution

Fix a basis of $V$ ; $\beta \mapsto \mathcal{M}(\beta)$ is an isomorphism from $V^{(2)}$ onto the $4 \times 4$ matrices, so each subspace has the dimension of its set of matrices.

Alternating: $\alpha$ is alternating exactly when it is antisymmetric, that is, $A_{k,j} = -A_{j,k}$ for all $j, k$ . With $j = k$ this gives $A_{j,j} = -A_{j,j}$ , so $A_{j,j} = 0$ . The $6$ entries above the diagonal can be chosen freely and determine the rest:

\dim V^{(2)}_{\mathrm{alt}} = \tfrac{4 \cdot 3}{2} = 6.

Symmetric: $A_{k,j} = A_{j,k}$ , so the $4$ diagonal entries and the $6$ above the diagonal are free:

\dim V^{(2)}_{\mathrm{sym}} = 4 + 6 = 10.

Check: the dimension of a direct sum is the sum of the dimensions, and $10 + 6 = 16 = \dim V^{(2)}$ .

Solved: 0 / 4