Quadratic to diagonal form

Question

In textbooks I read that it is easier for further calculation when we transform quadratic form to a diagonal form? Can any one explain what are the advantages of such transformations. Thanks for the help.

Answer 1

When we transform a quadratic form to a diagonal form, we can then take advantage of the eigenvalues and eigenvectors that are easily found once in diagonal form. After this we can use these eigenvalues and eigenvectors to deduce information about the quadratic form not immediately evident in its polynomial form. I've outlined some of the points below.

The function $f$ given by $$f(X)=X^tAX\qquad\text{for }X\in\mathbb{R}^n$$ where $X^t=(x_1,x_2,\dotsc,x_n)$ is the transpose of $X$, is termed the quadratic form associated with $A=(a_{ij})$, $A$ being a real symmetric $n\times n$ matrix, and so having a non-zero real eigenvector, moreover all eigenvalues of $A$ are real. In other words $$f(X)=\sum_{i,j=1}^{n}a_{ij}x_ix_j$$ For example if $n=2$, then $$ \begin{pmatrix} x_1&x_2 \end{pmatrix} \begin{pmatrix} a&b\\b&d \end{pmatrix} \begin{pmatrix} x_1\\x_2 \end{pmatrix} =ax_1^2+2bx_1x_2+dx_2^2 $$ Now if we wanted to seek the maximum of this function on the unit sphere, these being the set of all points for which $\Vert X\Vert=\sqrt{X\cdot X}=1$, then if a point $P$ on the sphere has $f(P)\ge f(X)$ for all $X$ with $\Vert X\Vert=1$, then $P$ is a maximum on the unit sphere.

Now it turns out that for some real symmetric matrix $A$, and $f(X)=X^tAX$ the associated quadratic form, then if $f(P)$ is a maximum on the sphere for some point $P$, then $P$ is an eigenvector of $A$, i.e., $AP=\lambda P$.

So given a quadratic form, we find the symmetric matrix associated with it, then find the eigenvectors on the unit circle where the minimum and maximum values occur by reading off the one with the smallest and highest eigenvalue respectively.

For the diagonalization of a symmetric linear map, let $V$ be a real inner product space of dimension $n$ with bilinear form $(\ast,\ast)$. Now let $$\theta\colon V\to V$$ be a symmetric linear map, that is $(\theta(x),y)=(x,\theta(y))$ for all $x,y\in V$. Hence $(x,y)=x^t\cdot y$ (matrix product). Then $x\mapsto \theta(x)=A\cdot x$, and so the map is symmetric if and only if $A^t=A$, viz. $(A\cdot x)^t\cdot y=x^t\cdot (A\cdot y)$, which holds iff $A^t=A$, i.e., symmetric as required.

(The Spectral Theorem) Let $\theta\colon V\to V$ be a symmetric linear map of the real inner product space $V$. Then $V$ has an orthonormal basis consisting of eigenvectors $\{e_1,e_2,\dotsc, e_n\}$, then the matrix of $\theta$ with respect to this basis is diagonal with the diagonal entries being the eigenvalues: $$ D= \begin{pmatrix} \lambda_1&0&\dots&0\\ 0&\lambda_2&\dots&0\\ \dots&\dots&\dots&\dots\\\tag{1} 0&0&\dots&\lambda_n\\ \end{pmatrix} $$ Hence associating a quadratic form with a real symmetric $n\times n$ matrix $A$, and then diagonalising this by a real unitary $n\times n$ matrix $U$: $$U^t A U=U^{-1} A U=D$$ where $D$ is diagonal as in (1), allows us to easily read the eigenvalues without further computation, and everything is a lot simpler and clearer.