0
$\begingroup$

I will be grateful if someone can guide me through the following problem:

So I have two matrices, $A$ and $B$:

  • Matrix $A$ is $N\times N$, Symmetric and non-invertible.
  • Matrix $B$ is $(N+1)\times(N+1)$, Symmetric and non-invertible.

I want to Find Matrix $C$ [which is $(N+1)\times N$] such that: $$CAC' = B$$

is there a way to find a closed-form solution to $C$?

Thank you very much!

$\endgroup$
2
  • 1
    $\begingroup$ I think $C$ should be $(N+1)\times N$, right? $\endgroup$ Commented Apr 4, 2016 at 21:53
  • $\begingroup$ Thank you very much for the note. I fixed it. $\endgroup$
    – M.H.A
    Commented Apr 5, 2016 at 4:19

1 Answer 1

1
$\begingroup$

Edit: I had given a first answer taking the form of a counterexample. I postpone it as a second part, because I have now a complete answer that I describe in the first part.

It gives conditions such that, matrices $A$ and $B$ (resp. $N \times N$ and $(N+1) \times (N+1)$) being given, one can find a $N \times (N+1)$ matrix $C$ such that:

$$C'AC = B \ \ \ (1)$$

with the supplementary condition that $A$, at least, is non invertible (I exchange here the rôles of $C$ and $C'$ because this notation will be more practical in the following lines)

First part : Let us use block notations

$$C=[D|V] \ \ \text{and} \ \ B=\begin{bmatrix} E&F\\F'&g\end{bmatrix} \ \ \ (2)$$

$D, E$ being $N \times N$ (and symmetrical), $V,F \in \mathbb{R}^N$, and $g \in \mathbb{R}$.

With these notations, relationship (1) becomes:

$$\begin{bmatrix} D'AD&D'AV\\V'AD&V'AV\end{bmatrix}=\begin{bmatrix} E&F\\F'&g\end{bmatrix} \ \ \ (3)$$

We are now faced to 3 relationships:

$$D'AD=E \ \ \ (4a), \ \ \ \ \ \ \ (D'A)V=F \ \ \ (4b), \ \ \ \ \ \ \ V'AV=g \ \ \ (4c)$$

i.e., 3 constraints on two unknowns, matrix $D$ and vector $V$.

Due to the law of inertia (en.wikipedia.org/wiki/Sylvester's_law_of_inertia), a necessary and sufficient condition for the existence of $D$ is that $A$ and $E$ (the $N \times N$ upper-left block of $B$ have the same signature (the same number of negative, null, and positive eigenvalues).

If this condition of identical signature is fulfilled:

  • Condition (4b) is fulfilled, as $D'A$ is non invertible (because such is the case for $A$), yields an infinite number of solutions, and among them, solutions of the form $V=V_1+\mu V_2$ (where nonzero vector $V_2 \in Null(D'A)$).

  • Condition (4c) will be fulfilled at a condition: the quadratic equation in $\mu$ obtained by developing $(V'_1+\mu V'_2)A(V_1+\mu V_2)=g$ has at least one real solution.

We have thus to check that the following discriminant: $(V'_2AV_1)^2-(V'_1AV_1-g)(V'_2AV_2)$ is $\geq 0$.

Second part :

Here is a counterexample, where the most general $C$ is taken:

$$CAC'=\begin{bmatrix} a&d\\b&e\\c&f \end{bmatrix}\begin{bmatrix} 1 & 0\\0&0\end{bmatrix}\begin{bmatrix} a & b & c\\ d & e & f\end{bmatrix}=\begin{bmatrix} a^2 & ab & ac\\ ab & b^2 & bc\\ ac& bc&c^2\end{bmatrix}$$ But if the matrix $B$ has not this particular structure, for example if

$$B=\begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\ 0& 0&0\end{bmatrix}$$

no match is possible with the preceeding product, whatever the values of $a,b,c,d,e,f$.

Why was it predictable? Because the upper-left $2 \times 2$ block of $B$ has signature (0 negative, 0 null, 2 positive), whereas, for $A$ it is (0 negative, one null, 1 positive).

$\endgroup$
4
  • $\begingroup$ Thank you for taking the time to look at my question. Kindly note that A and B will never have a zero-element or a zero-row/column. I have been looking in Linear Algebra books for something related to my question (even for A and B being the same size, then C will also be square matrix in that case), but could not find a clue that enables me to start looking at C in the right way. Thank you again! $\endgroup$
    – M.H.A
    Commented Apr 5, 2016 at 4:36
  • $\begingroup$ I think that the fact that $A$ in my example has a null row or column is unimportant : this matrix can be considered as the diagonalized version of any matrix having rank 1 (it is one of you specification : rank < n = 2 here). $\endgroup$
    – Jean Marie
    Commented Apr 5, 2016 at 6:12
  • $\begingroup$ I am going to edit my old answer because I have been thinking a lot about the question you raise. I am going to answer you in a "constructive" way, explaining (by using block matrix notations and the law of inertia en.wikipedia.org/wiki/Sylvester's_law_of_inertia ) in which cases a solution exists. $\endgroup$
    – Jean Marie
    Commented Apr 6, 2016 at 12:54
  • $\begingroup$ Thank you very much!. your Technique and presentation made me understand things I did not know about. I will look into making a general solution for my case. $\endgroup$
    – M.H.A
    Commented Apr 6, 2016 at 20:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .