Let $\bf A$ be a real symmetric matrix of order $n\times n$ such that $\mathrm R(\mathbf A)=r(\le n)$. Then show that there exists a nonsingular matrix $\bf F$ such that $\bf F'AF=\begin{pmatrix}\bf I & \bf0 & \bf0 \\\bf0 & \bf-I & \bf0 \\\bf0 & \bf0 & \bf0 \\\end{pmatrix}$ where the orders of $\bf I$ and $\bf-I$ are respectively the number of positive and negative characteristic roots of $\bf A$, the sum of the orders being $r$. $[\mathrm R(\mathbf A)$ denotes the rank of $\bf A]$

(Attempt using hints):

Let $\mathbf A$ be nonsingular. Then as $\bf A$ is symmetric, by the Spectral decomposition theorem $\exists$ an orthogonal matrix $\bf P$ such that $\mathbf{P'AP}=\mathrm{diag}(\lambda_1,\lambda_2,...,\lambda_n)$ where $\lambda_1\ge...\lambda_q>0>\lambda_{q+1}\ge...\ge\lambda_n$ are the characteristic roots of $\bf A$.

Define $\mathbf Q=\mathrm{diag}\left(\frac{1}{\sqrt{\lambda_1}},...,\frac{1}{\sqrt{\lambda_q}},\frac{1}{\sqrt{-\lambda_{q+1}}},...,\frac{1}{\sqrt{-\lambda_n}}\right)$.

Then $\det(\mathbf Q)=\prod_{i=1}^rq_{ii}\ne0$ as the diagonal entries $q_{ii}$ of $\bf Q$ are non-zero. So $\bf Q$ is nonsingular.

So, $\mathbf{Q'(P'AP)Q}=\mathbf{(PQ)'A(PQ)}=\begin{pmatrix}\bf I_q & \bf0 \\\bf0 & \bf-I_{n-q} \\\end{pmatrix}$.

Putting $\mathbf{PQ}=\bf F$, we can say that $\bf F$ is nonsingular being the product of two nonsingular matrices.

But I am not able to use a similar argument for the general case when $\bf A$ is singular.

In this case we have $\mathbf{P'AP}=\mathrm{diag}(\lambda_1,\lambda_2,...,\lambda_r,0,...,0)$ where $\lambda_1\ge...\lambda_q>0>\lambda_{q+1}\ge...\ge\lambda_r$ are the non-zero eigenvalues of $\bf A$. But I cannot define $\bf Q$ in a similar manner as in the previous case so that $\mathbf{Q'(P'AP)Q}$ becomes $\mathrm{diag}(\mathbf I_q,\mathbf{-I_{r-q}},\mathbf0)$.

Is there a simpler approach for proving the result? A reference for the general proof would be great.

I have some further doubts regarding this result.

Let $\mathrm{R}(\mathbf A)=r(\le n)$. Then by the nonsingular transformation $\mathbf x\mapsto \mathbf y$ such that $\mathbf x=\mathbf{Py}$ $(\bf P$ is the same orthogonal matrix as before$)$, the quadratic form $\mathbf{x^\top Ax}$ is transformed to $\mathbf{y^\top (P'AP)y}=\sum_{i=1}^r \lambda_i y_i^2$, where $\mathbf y=(y_1,y_1,...,y_n)'$. This way we prove that any real quadratic form is diagonalisable.

In general we say that the quadratic form can be transformed to $\sum_{i=1}^r d_i y_i^2$ where $d_i>0$ when $\bf A$ is p.d. with full rank, and $d_i>0$ for $i=1,...,r$; $d_i=0$ for $i=r+1,...,n$ when $\bf A$ is p.s.d. with rank $r(<n)$.

But what are the $d_i$'s actually? Are they always the eigenvalues of $\bf A$? What is its connection with the matrix $\mathrm{diag}(\mathbf I,\mathbf{-I},\mathbf0)$? Does $d_i\in \{0,1,-1\}$ $\forall i=1,2,...,r$ ?


It appears that I am having trouble grasping the concept correctly.

Let $\mathbf D=\mathrm{diag}(d_1,...,d_r,0,...,0)$ be the diagonal form into which $\bf A$ is transformed.

Then $Q(\mathbf x)=\mathbf{x^\top Ax}$ is transformed to some $Q(\mathbf y)=\mathbf{y^\top Dy}=\sum_{i=1}^rd_iy_i^2$

So, if the $d_i$'s are changed to $d_i=\begin{cases}1, & \text{for } i=1,2,...,q \\-1, & \text{for } i=q+1,q+2...,r \\0, & \text{for } i=r+1,...,n\end{cases}$

then I indeed get the diagonal matrix $\mathbf D=\mathrm{diag}(\mathbf I,\mathbf{-I},\mathbf0)$ and the quadratic form $Q(\mathbf x)$ is transformed to $\mathbf{y^\top Dy}=y_1^2+...+y_q^2-y_{q+1}^2-...-y_r^2$.

My question then boils down to

how can I change the $d_i$'s in the above manner? What is the justification?

  • $\begingroup$ @Fabian Could you guide me to a reference link? $\endgroup$ – StubbornAtom Apr 15 '17 at 15:33
  • $\begingroup$ See for example here under the point Lagrange’s Reduction. $\endgroup$ – Fabian Apr 15 '17 at 17:39
  • $\begingroup$ For the singular case, the diagonal elements which are zero don’t need to be scaled, so use $1$ instead of $1/\sqrt{\pm\lambda_i}$ for $i>r$ when constructing $Q$. $\endgroup$ – amd Apr 15 '17 at 20:31

If $A$ is a real and symmetric matrix, $A$ is Hermitian and hence diagonalisable with real eigenvalues. Congruence transformations include any basis change that effects a diagonalisation of $A$, but also those that subsequently scale its eigenvalues by positive factors. We can then set positive diagonal elements to $1$ and negative ones to $-1$ while preserving those that are zero. (In your notation we need to choose the diagonal entries of $Q$ as you have for non-zero eigenvalues, but our choices for zero eigenvalues are irrelevant (as long as they are non-zero), so we may as well set these entries in $Q$ to $1$.) Finally, a rearrangement of rows and columns is also possible, thus collecting the $+1$ entries together followed by the $-1$ entries followed by the zero entries, giving the required result.


Your problem is handled by Sylvester's Law of Inertia for quadratic forms. It's proof uses just recursively completing the square to remove the nondiagonal entries of the matrix $A$.

On the other hand, if you are willing to compute the eigenvalues, you obtain an orthonormal basis with respect to which $A$ is diagonal.


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