Recipe

Compute a new Gram matrix by congruence

Use when: you have a form's Gram matrix relative to one basis and a change-of-basis matrix $P$ , and need the Gram matrix relative to the new basis

Goal: the new Gram matrix $A'=P^{\mathsf T}AP$ , correctly distinguished from the similarity transform a linear map's matrix would use instead

Taught in Change of basis for forms: congruence, not similarity

Quick recipe

Identify $P$ , the change-of-basis matrix
Multiply $A$ by $P$ first
Left-multiply by $P^{\mathsf T}$ to get $A'$
Confirm it isn't the similarity answer

1
Build $P$ 's columns from the new basis vectors, written in the old basis's coordinates.

Example

For the new basis $\{(1,1),(1,-1)\}$ , $P=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ , one new basis vector per column.

Why: congruence needs the same change-of-basis matrix chapter 01 already builds for ordinary coordinate changes, not a new object.
2
Compute the product $AP$ before bringing in $P^{\mathsf T}$ ; matrix multiplication order matters here.

Example

For $A=\begin{pmatrix}2&0\\0&-1\end{pmatrix}$ , $AP=\begin{pmatrix}2&2\\-1&1\end{pmatrix}$ .

Why: $A'=P^{\mathsf T}AP$ is a product of three matrices, and computing $AP$ first turns the rest into a single ordinary matrix multiplication.
3
Transpose $P$ , then multiply it against the product from Step 2.

Example

$P^{\mathsf T}AP=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}2&2\\-1&1\end{pmatrix}=\begin{pmatrix}1&3\\3&1\end{pmatrix}$ .

⚡ Shortcut

If $A$ is symmetric, $A'=P^{\mathsf T}AP$ is guaranteed symmetric too, for any invertible $P$ ; checking the result's symmetry is a quick sanity test on the arithmetic.

Why: this section's derivation shows the new Gram matrix is exactly $P^{\mathsf T}AP$ , never $P^{-1}AP$ , which is the rule for a linear map's matrix instead.
4
If $P^{-1}AP$ is available for comparison, compute it too and check that it genuinely differs from $A'$ unless $P$ is orthogonal.
- $P^{-1}=P^{\mathsf T}$ (an orthogonal $P$ ) — congruence and similarity happen to give the same answer.
- $P^{-1}\ne P^{\mathsf T}$ — the two answers genuinely differ, and only $P^{\mathsf T}AP$ is correct for a form.
Why: mixing up the two formulas is this topic's single most common error, so checking they diverge, or coincide only when $P$ is orthogonal, catches the mistake immediately.

Common mistakes

Using $P^{-1}AP$ , the rule for a linear map's matrix, instead of $P^{\mathsf T}AP$ for a form's Gram matrix.
Multiplying in the wrong order and getting confused, rather than simply computing $AP$ first as a fixed habit.
Forgetting to transpose $P$ at all, and left-multiplying by $P$ itself.
Assuming $A'$ must look different from $A$ ; a symmetric form can still be congruent to one that looks similar.