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I have the quadratic form $$Q(x)=x_1^2+2x_1x_4+x_2^2 +2x_2x_3+2x_3^2+2x_3x_4+2x_4^2$$

I want to diagonalize the matrix of Q. I know I need to find the matrix of the associated bilinear form but I am unsure on how to do this.

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    You want to diagonalise the matrix of $Q$ or write $Q$ as the sum of squares of linear forms? – 2017-01-02
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    @Bernard diagonalise the matrix of Q. – 2017-01-02

2 Answers 2

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Let's call the associated matrix $A = (a_{ij})$.

Then, simply apply this:

  1. $a_{ii} = c$ where c is the coefficient of $x_i^2$
  2. $a_{ij} = \frac c 2$ where c is the coefficient of $x_ix_j$

So, your matrix will be

$$A = \begin{bmatrix} 1&0&0&1\\ 0&1&1&0\\ 0&1&2&1\\ 1&0&1&2 \end{bmatrix}.$$

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    @Leonhard can you diagonalise it or should I show you that? – 2017-01-02
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    Do I find the unit eigenvectors of A and set them as the columns of a matrix? – 2017-01-02
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    @Leonhard yeah that's it – 2017-01-02
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It is $$\begin{bmatrix} 1&0&0&1\\ 0&1&1&0\\ 0&1&2&1\\ 1&0&1&2 \end{bmatrix}.$$

Some details

The diagonal coefficients are the coefficients of the squares. The $(i,j)$ and the $(j,i)$ coefficient are half the coefficient of the $x_ix_j$ monomial. This results from the reverse computation of the quadratic form associated to a symmetric matrix $(a_{ij})$.

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    Could you add some detail as to how you got it? The method is more important to than the answer as I want to know how to do it for other cases. – 2017-01-02
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    As you have 4 variables your matrix $Q$ will be 4x4. The coefficient $(i,j)$ of your matrix will be the coefficient associated to the term $x_i x_j$. For example $(q_{11})=1$ and $(q_{14})=2$. This way you can construct the matrix $Q$ – 2017-01-02
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    @Leonhard Written in matrix form, $Q$ is $(x_1,x_2,x_3,x_4)M(x_1,x_2,x_3,x_4)^T$ for some symmetric matrix $M$. If you multiply this out and compare it to $Q$, you’ll see that the diagonal elements are the coefficients of $x_i^2$ and that the off-diagonal elements are the coefficients of $x_ix_j$ (divided by two, usually, since $x_ix_j$ and $x_jx_i$ will likely have been combined into a single term $Q$). – 2017-01-02