Linear-Algebra-Done-Right · Unit 27 · Video 2 · Interactive Practice

Change of Basis for Bilinear Forms: A = CᵀBC

IKey Formulas

Formula	Statement	Setting
$\alpha(u, v) = \beta(u, Tv) \implies \mathcal{M}(\alpha) = \mathcal{M}(\beta)\mathcal{M}(T)$	$T$ in the second slot multiplies on the right	$\beta$ a bilinear form on $V$ , $T \in \mathcal{L}(V)$ , any basis $e_1, \dots, e_n$
$\rho(u, v) = \beta(Tu, v) \implies \mathcal{M}(\rho) = \mathcal{M}(T)^{\mathsf{t}}\mathcal{M}(\beta)$	$T$ in the first slot multiplies on the left, transposed	same; $\bigl(\mathcal{M}(T)^{\mathsf{t}}\bigr)_{j,k} = \mathcal{M}(T)_{k,j}$
$A = C^{\mathsf{t}} B C$	Change-of-basis formula	$A = \mathcal{M}(\beta, (e_1, \dots, e_n))$ , $B = \mathcal{M}(\beta, (f_1, \dots, f_n))$ , $C = \mathcal{M}(I, (e_1, \dots, e_n), (f_1, \dots, f_n))$ : column $k$ of $C$ holds the $f$ -coordinates of $e_k$
$\beta(u, v) = x^{\mathsf{t}} A y = (Cx)^{\mathsf{t}} B (Cy)$	The same formula in coordinates	$x, y$ the $e$ -coordinates of $u, v$ ; $Cx, Cy$ their $f$ -coordinates

Key Insight: An operator's matrix changes by $C^{-1}BC$ because its output must be converted back into the $e$ basis; a bilinear form outputs a scalar, so the left factor is $C^{\mathsf{t}}$ . The two formulas agree for every $B$ only when $C^{\mathsf{t}} = C^{-1}$ .

IIWhere the Transpose Comes From

In the second slot $T$ multiplies $\mathcal{M}(\beta)$ on the right; in the first slot, on the left and transposed.

Slot for T Row j Column k

IIIThe Video's Example, Any Shifted Basis

For $\beta(p, q) = p(2)\,q'(3)$ on $\mathcal{P}_2(\mathbf{R})$ , $C^{\mathsf{t}}BC$ reproduces $A$ for every basis $1,\ x - a,\ (x - b)^2$ .

a 2 b 3 Left factor

IVTranspose Versus Inverse in the Plane

With $\beta(u, v) = u^{\mathsf{t}} B v$ on $\mathbf{R}^2$ and $B = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}$ , each basis $e_1, e_2$ gives $A = C^{\mathsf{t}} B C$ ; $C^{-1} B C$ matches it only when $C^{\mathsf{t}} = C^{-1}$ .

💡 Challenge: find a basis other than $(1, 0), (0, 1)$ for which $C^{-1}BC = C^{\mathsf{t}}BC$ , then check that its $C$ satisfies $C^{\mathsf{t}}C = I$ .

VQuiz Questions

Problem 1 · T in the Second Slot

Given: the bilinear form $\beta(x, y) = 2x_1y_1 + x_1y_2 - x_2y_2$ on $\mathbf{R}^2$ and $T \in \mathcal{L}(\mathbf{R}^2)$ defined by $T(x_1, x_2) = (x_1 + 3x_2,\ 2x_1 + x_2)$ . Let $\alpha(u, v) = \beta(u, Tv)$ — find $\mathcal{M}(\alpha)$ with respect to the standard basis.

✅ Correct!

T

sits in the second slot, so

\mathcal{M}(\alpha) = \mathcal{M}(\beta)\mathcal{M}(T)

; for instance

\alpha(e_1, e_2) = \beta(e_1, (3, 1)) = 6 + 1 = 7

❌ That is $\mathcal{M}(T)\mathcal{M}(\beta)$ . Matrix multiplication does not commute; with

T

in the second slot,

\mathcal{M}(T)

goes on the right of

\mathcal{M}(\beta)

❌ That is $\mathcal{M}(\beta)\mathcal{M}(T)^{\mathsf{t}}$ . Column

k

\mathcal{M}(T)

holds the coordinates of

Te_k

, so

\mathcal{M}(T) = \begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix}

, and no transpose appears for the second slot.

❌ That is $\mathcal{M}(T)^{\mathsf{t}}\mathcal{M}(\beta)$ , the matrix of

\beta(Tu, v)

, where

T

acts in the first slot. Here

T

acts in the second slot.

Show solution

Entry $(j, k)$ of $\mathcal{M}(\beta)$ is $\beta(e_j, e_k)$ , the coefficient of $x_jy_k$ , and column $k$ of $\mathcal{M}(T)$ holds $Te_k$ : $Te_1 = (1, 2)$ , $Te_2 = (3, 1)$ .

\mathcal{M}(\beta) = \begin{pmatrix} 2 & 1 \\ 0 & -1 \end{pmatrix}, \qquad \mathcal{M}(T) = \begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix}.

Since $T$ acts in the second slot,

\mathcal{M}(\alpha) = \mathcal{M}(\beta)\mathcal{M}(T) = \begin{pmatrix} 2 \cdot 1 + 1 \cdot 2 & 2 \cdot 3 + 1 \cdot 1 \\ 0 \cdot 1 + (-1) \cdot 2 & 0 \cdot 3 + (-1) \cdot 1 \end{pmatrix} = \begin{pmatrix} 4 & 7 \\ -2 & -1 \end{pmatrix}.

Check one entry from the definition: $\alpha(e_2, e_1) = \beta(e_2, Te_1) = \beta\bigl((0, 1), (1, 2)\bigr) = -x_2y_2 = -2$ .

Problem 2 · T in the First Slot

Given: the bilinear form $\beta(x, y) = x_1y_2 - 2x_2y_1 + x_2y_2$ on $\mathbf{R}^2$ and $T \in \mathcal{L}(\mathbf{R}^2)$ defined by $T(x_1, x_2) = (x_1 - x_2,\ 3x_2)$ . Let $\rho(u, v) = \beta(Tu, v)$ — find $\mathcal{M}(\rho)$ with respect to the standard basis.

✅ Correct!

\mathcal{M}(\rho) = \mathcal{M}(T)^{\mathsf{t}}\mathcal{M}(\beta)

; for instance

\rho(e_2, e_1) = \beta\bigl((-1, 3), (1, 0)\bigr) = -2 \cdot 3 = -6

❌ That is $\mathcal{M}(T)\mathcal{M}(\beta)$ — the transpose is missing. In

\beta(Te_j, e_k)

the index

j

is a column index of

\mathcal{M}(T)

, so the left factor must be

\mathcal{M}(T)^{\mathsf{t}}

❌ That is $\mathcal{M}(\beta)\mathcal{M}(T)$ , the matrix of

\beta(u, Tv)

. Here

T

acts in the first slot.

❌ That is $\mathcal{M}(\beta)\mathcal{M}(T)^{\mathsf{t}}$ . The transposed factor belongs on the left of

\mathcal{M}(\beta)

, not on the right.

Show solution

Reading off coefficients and images: $Te_1 = (1, 0)$ , $Te_2 = (-1, 3)$ , so

\mathcal{M}(\beta) = \begin{pmatrix} 0 & 1 \\ -2 & 1 \end{pmatrix}, \qquad \mathcal{M}(T) = \begin{pmatrix} 1 & -1 \\ 0 & 3 \end{pmatrix}, \qquad \mathcal{M}(T)^{\mathsf{t}} = \begin{pmatrix} 1 & 0 \\ -1 & 3 \end{pmatrix}.

Since $T$ acts in the first slot,

\mathcal{M}(\rho) = \mathcal{M}(T)^{\mathsf{t}}\mathcal{M}(\beta) = \begin{pmatrix} 1 \cdot 0 + 0 \cdot (-2) & 1 \cdot 1 + 0 \cdot 1 \\ (-1) \cdot 0 + 3 \cdot (-2) & (-1) \cdot 1 + 3 \cdot 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -6 & 2 \end{pmatrix}.

Check from the definition: $\rho(e_2, e_2) = \beta(Te_2, e_2) = \beta\bigl((-1, 3), (0, 1)\bigr) = (-1)(1) + 3 \cdot 1 = 2$ .

Problem 3 · Change to a New Basis

Given: the bilinear form $\beta(x, y) = x_1y_1 + 2x_1y_2 + 3x_2y_2$ on $\mathbf{R}^2$ , with matrix $B$ with respect to the standard basis, and the basis $e_1 = (1, 1)$ , $e_2 = (0, 1)$ — find $A = \mathcal{M}(\beta, (e_1, e_2))$ .

✅ Correct!

A = C^{\mathsf{t}}BC

, and every entry checks against

\beta(e_j, e_k)

\beta(e_1, e_1) = 1 + 2 + 3 = 6

❌ That is $C^{-1}BC$ , the change-of-basis formula for an operator. A bilinear form outputs a scalar, so nothing is converted back: the left factor is

C^{\mathsf{t}}

❌ That is $CBC^{\mathsf{t}}$ — the transpose is on the wrong side. Entry

(j, k)

A

\beta(e_j, e_k)

, which is (column

j

C

)

^{\mathsf{t}}\, B\,

(column

k

C

❌ That is only $BC$ , the standard-basis matrix of

\beta(u, Tv)

for the operator

T

that sends the standard basis to

e_1, e_2

. Changing the basis puts

T

in both slots, so

C^{\mathsf{t}}

must also multiply on the left.

Show solution

With $f_1, f_2$ the standard basis, $B$ holds the coefficients of $x_jy_k$ , and the columns of $C$ are the standard coordinates of $e_1, e_2$ :

B = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}, \qquad C = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}.

BC = \begin{pmatrix} 3 & 2 \\ 3 & 3 \end{pmatrix}, \qquad A = C^{\mathsf{t}}BC = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 2 \\ 3 & 3 \end{pmatrix} = \begin{pmatrix} 6 & 5 \\ 3 & 3 \end{pmatrix}.

Check from the definition $A_{j,k} = \beta(e_j, e_k)$ : $\beta(e_1, e_2) = \beta\bigl((1, 1), (0, 1)\bigr) = 0 + 2 + 3 = 5$ , $\beta(e_2, e_1) = \beta\bigl((0, 1), (1, 1)\bigr) = 3$ , $\beta(e_2, e_2) = 3$ .

Problem 4 · From a Nonstandard Basis Back to the Standard One

Given: the basis $f_1 = (1, 1)$ , $f_2 = (1, -1)$ of $\mathbf{R}^2$ and a bilinear form $\beta$ on $\mathbf{R}^2$ with $B = \mathcal{M}(\beta, (f_1, f_2)) = \begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix}$ . Let $e_1, e_2$ be the standard basis — find $C = \mathcal{M}(I, (e_1, e_2), (f_1, f_2))$ and $A = \mathcal{M}(\beta, (e_1, e_2))$ .

What is C?

What is A?

✅ Correct!

e_1 = \tfrac{1}{2}f_1 + \tfrac{1}{2}f_2

and

e_2 = \tfrac{1}{2}f_1 - \tfrac{1}{2}f_2

give

C

, and

A = C^{\mathsf{t}}BC

; for instance

\beta(e_1, e_1) = \tfrac{1}{4}(1 + 0 + 2 - 1) = \tfrac{1}{2}

❌ These columns are $f_1, f_2$ written in the $e$ basis. That matrix is

\mathcal{M}(I, (f_1, f_2), (e_1, e_2)) = C^{-1}

. Column

k

C

must hold the

f

-coordinates of

e_k

❌ Check column 2:

-\tfrac{1}{2}f_1 + \tfrac{1}{2}f_2 = (0, -1) = -e_2

, not

e_2

❌ Not quite. Solve

e_k = c_1 f_1 + c_2 f_2

for each

k

; the pair

(c_1, c_2)

is column

k

C

❌ That is $F^{\mathsf{t}}BF$ with $F = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ , built from the columns

f_1, f_2

. The formula needs

C

, whose columns are the

f

-coordinates of

e_1, e_2

❌ $B$ is the matrix in the $f$ basis. The standard basis has its own matrix; for instance

\beta(e_1, e_1) = \tfrac{1}{2} \neq 1

❌ That is $C^{-1}BC$ , the operator formula. For a bilinear form the left factor is

C^{\mathsf{t}}

❌ Not quite. Compute

BC

first, then multiply by

C^{\mathsf{t}}

on the left.

Show solution

Step 1: $C$ . Solve $e_1 = c_1 f_1 + c_2 f_2$ : $(c_1 + c_2,\ c_1 - c_2) = (1, 0)$ gives $c_1 = c_2 = \tfrac{1}{2}$ . Solve $e_2 = c_1 f_1 + c_2 f_2$ : $(c_1 + c_2,\ c_1 - c_2) = (0, 1)$ gives $c_1 = \tfrac{1}{2}$ , $c_2 = -\tfrac{1}{2}$ . So

C = \begin{pmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & -\tfrac{1}{2} \end{pmatrix}.

Step 2: $A = C^{\mathsf{t}}BC$ . Here $C^{\mathsf{t}} = C$ .

BC = \begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & -\tfrac{1}{2} \end{pmatrix} = \begin{pmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & \tfrac{3}{2} \end{pmatrix}, \qquad C^{\mathsf{t}}BC = \begin{pmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & -\tfrac{1}{2} \end{pmatrix}\begin{pmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & \tfrac{3}{2} \end{pmatrix} = \begin{pmatrix} \tfrac{1}{2} & 1 \\ 0 & -\tfrac{1}{2} \end{pmatrix}.

Check: the $f$ -coordinates of $e_2$ are $\bigl(\tfrac{1}{2}, -\tfrac{1}{2}\bigr)$ , so $\beta(e_2, e_2) = \tfrac{1}{4}\bigl(1 \cdot 1 + 0 \cdot (-1) + 2 \cdot (-1) + (-1)(1)\bigr) = \tfrac{1}{4}(1 - 2 - 1) = -\tfrac{1}{2}$ .

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