# Finding change of basis matrix for bilinear form

I'm having a hard time finding change of basis matrix for bilinear form.

I'm given a matrix $$A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 &-1\\ 1 & -1 & 1 \\ \end{bmatrix}$$ and I need to find matrices $$D$$ and $$C$$ such that $$D = C^TAC$$

I've managed to find $$D = \begin{bmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1 \\ \end{bmatrix}$$ using symmetric Gaussian elimination, but I don't have much clue how to move on from here. Any hint/help would be appreciated!

• I'm guessing $A$ is the metric in some basis and you want to change basis with $C$? Do you have the old and new bases by chance? Apr 2 at 17:44
• it is just expressing each symmetric row/column step as an elementary matrix, take the identity and alter just one off-diagonal position. Call it $C_j$ at step $j,$ this being the matrix that goes on the right in $A_{j} = C_j^T A_{j-1} C_j.$ Then $C = C_1 C_2 C_3..$ See my math.stackexchange.com/questions/1388421/… Apr 2 at 17:47
• well, why not. Usually the $C_j$ are upper triangular. However, if an intermediate step creates a zero on the diagonal, the next step is lower triangular; $C_1$ (1,2) set to: $-1$ then $C_2$ (1,3) set to: $-1$ then $C_3$ (3,2) set to: $1$ then $C_4$ (2,3) set to: $-1/2 \; \; \; \;$ so this time $C_3$ is lower triangular Oh... I got to diagonal $(1,-4,1)$ rather than your $(1,-1,1).$ That can be corrected to your target with $C_5$ which is diagonal, middle element $1/2$ Apr 2 at 18:02

There is a canonical unitary matrix to diagonalise a bilinear form, which I so called the completing the square tricks. The algorithm is stated below and let $$\mathfrak{B}$$ be a bilinear form.

• Step $$1$$: Separate all monomials from $$x_1$$. You can write the bilinear form in $$f(x_1,\cdots,x_n)^2+f_2(x_2,\cdots,x_n),f,g$$ are monomials.
• Step $$2$$: Apply step $$1$$ on $$f_2$$, but this time, we separate all monomials from $$x_2$$.
• Step $$3$$: We keep applying step $$1$$ on $$f_i$$ to separate all monomials from $$x_i$$ to obtain $$f_{i+1}$$ until it only contains two variable.

Remark. Sometimes you may not be able to completing the square, like you may get a variable in the form $$x_1x_2$$ and nothing else. However, we can still write $$x_1x_2=\dfrac{1}{4}(x_1+x_2)^2-\dfrac{1}{4}(x_1-x_2)^2.$$

Last but not least, you would obtain $$\mathfrak{B}$$ as a sum of square of monomials. This gives the unitary matrix we want by setting the coefficient of $$x_i$$ of $$f_j$$ as the $$(i,j)$$-th entries, and the whole coefficient of the $$f_j$$ as the diagonal entries.

Example. We use your example $$\mathfrak{B}(x,y,z)=x^2+y^2+z^2+2xy-2yz+2xz$$.

Step $$1$$: We have $$\mathfrak{B}=(x+y+z)^2-4yz$$. So $$f=(x+y+z)$$ and $$f_2=-4yz$$.

Step $$2$$: $$-4yz=(y-z)^2-(y+z)^2$$.

So $$\mathfrak{B}=(x+y+z)^2+(y-z)^2-(y+z)^2$$. This gives

$$\begin{pmatrix}1 & 1 & 1\\1 & 1 &-1\\1&-1&1\end{pmatrix}=\begin{pmatrix}1&0&0\\1&1&1\\ 1&-1&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&-1&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 1&1&1\\ 1&-1&1\end{pmatrix}^T$$

• Arh yes, this gives diagonalization of bilinear form, but you should taking back the transpose bc you want a form $D=C^TAC$. Apr 2 at 18:43