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Exercise 1845Linear algebra › Quadratic formsL1Added 3 Nov 2026

Change of variables in a quadratic form

#change of variables#quadratic form#matrix

Statement

Let Q(x,y)=x2+4xy+3y2Q(x,y)=x^2+4xy+3y^2 and P=(1201)P=\begin{pmatrix}1&2\\0&1\end{pmatrix} (invertible). Express QQ in the new variables (u,v)(u,v) defined by (xy)=P(uv)\begin{pmatrix}x\\y\end{pmatrix}=P\begin{pmatrix}u\\v\end{pmatrix}.

Hint

Q=PXQ=PX gives XTAX=UT(PTAP)UX^T A X = U^T(P^TAP)U.

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Solution

New variables. x=u+2vx=u+2v, y=vy=v.

Direct substitution.
Q=(u+2v)2+4(u+2v)v+3v2=u2+4uv+4v2+4uv+8v2+3v2=u2+8uv+15v2.Q=(u+2v)^2+4(u+2v)v+3v^2=u^2+4uv+4v^2+4uv+8v^2+3v^2=u^2+8uv+15v^2.

Matrix form. The new matrix is PTAPP^TAP:
PTAP=(1021)(1223)(1201).P^TAP=\begin{pmatrix}1&0\\2&1\end{pmatrix}\begin{pmatrix}1&2\\2&3\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}.
Compute: (1223)(1201)=(1427)\begin{pmatrix}1&2\\2&3\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}=\begin{pmatrix}1&4\\2&7\end{pmatrix}. Then
(1021)(1427)=(14415).\begin{pmatrix}1&0\\2&1\end{pmatrix}\begin{pmatrix}1&4\\2&7\end{pmatrix}=\begin{pmatrix}1&4\\4&15\end{pmatrix}.

Verification. UT(14415)U=u2+8uv+15v2U^T\begin{pmatrix}1&4\\4&15\end{pmatrix}U=u^2+8uv+15v^2. Consistent.

Conclusion. In (u,v)(u,v), Q=u2+8uv+15v2Q=u^2+8uv+15v^2.