You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Completing the square using a matrix in quadratic form
- Thread starter PhizKid
- Start date
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
There is NOT necessarily possible to write a quadratic form as a "sum and difference". It is always possible to write a symmetric matrix as either a difference of squares (which can be written as sum and difference) or as a su of squares.
The symmetric matrix \(\displaystyle A= \begin{bmatrix}a & b \\ b & c \end{bmatrix}\) gives the quadratic form \(\displaystyle X^TAX= \begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}a & b \\ b & c \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= ax^2+ 2bxy+ cy^2\). Now, because this matrix is symmetric, we know it is "diagonalizable". There exist an invertible matrix, P, and a diagonal matrix, D, such that \(\displaystyle A= P^{-1}DP\) so that we can write \(\displaystyle X^TAX= X^T(P^{-1}DP)X= (X^TP^{-1})D(PX)= Y^TDY\) where Y= PX.
That then gives \(\displaystyle \begin{bmatrix}y_1 & y_2\end{bmatrix}\begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}\begin{bmatrix}y_1 \\ y_2 \end{bmatrix}= a_1y_1^2+ a_2y_2^2\). Whether that is a difference of squares or a sum of squares depends upon the signs of \(\displaystyle \lambda_1\) and \(\displaystyle \lambda_2\).
Of course, \(\displaystyle \lambda_1\) and \(\displaystyle \lambda_2\) are the eigenvalues of A and the matrix, P, has the eigenvectors of A as columns.
The symmetric matrix \(\displaystyle A= \begin{bmatrix}a & b \\ b & c \end{bmatrix}\) gives the quadratic form \(\displaystyle X^TAX= \begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}a & b \\ b & c \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= ax^2+ 2bxy+ cy^2\). Now, because this matrix is symmetric, we know it is "diagonalizable". There exist an invertible matrix, P, and a diagonal matrix, D, such that \(\displaystyle A= P^{-1}DP\) so that we can write \(\displaystyle X^TAX= X^T(P^{-1}DP)X= (X^TP^{-1})D(PX)= Y^TDY\) where Y= PX.
That then gives \(\displaystyle \begin{bmatrix}y_1 & y_2\end{bmatrix}\begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}\begin{bmatrix}y_1 \\ y_2 \end{bmatrix}= a_1y_1^2+ a_2y_2^2\). Whether that is a difference of squares or a sum of squares depends upon the signs of \(\displaystyle \lambda_1\) and \(\displaystyle \lambda_2\).
Of course, \(\displaystyle \lambda_1\) and \(\displaystyle \lambda_2\) are the eigenvalues of A and the matrix, P, has the eigenvectors of A as columns.