Non singular matrix M such that MAM^T=F

[CAF123]

Homework Statement

Show that there is a non-singular matrix M such that $MAM^T = F$ for any antisymmetric matrix A where the normal form F is a matrix with 2x2 blocks on its principal diagonal which are either zero or $$\begin{pmatrix} 0 &1 \\ -1&0 \end{pmatrix}$$

To do so, consider the analogue of the Gram-Schmidt method for antisymmetric matrices using the antisymmetric inner product $\langle x,y \rangle = x^TA y$

Homework Equations

Gram Schmidt equations?
Orthogonality

The Attempt at a Solution

I am just really looking for some hints to start. For vectors $x$ and $y$, $$x^T A y = (x_1 \dots x_n) \begin{pmatrix} A_{11} & A_{12} &..&A_{1n} \\ ..&.. \\ ..& ..&..&A_{nn} \end{pmatrix} \begin{pmatrix} y_1 \\ y_2\\ ..\\y_n \end{pmatrix}$$ and $A_{ii} = 0, A_{ij} = -A_{ji}$ but I am not seeing how the hint is useful.

Thanks!

 

[geoffrey159]

I feel that it is a way or another related to the spectral theorem. Try to work on $A^2$.

 

[CAF123]

In the 2x2 case, $A^2$ is proportional to the identity but this is not true for the 3x3 case and beyond so I am not sure what can be said about $A^2$ in general.

The product $x^T A x$ is the matrix multiplication of a row of $M^T$ together with $A$ and corresponding transposed row (or column) vector in $M$. The matrix $M$ is composed of the $2n$ vectors $x$ and amounts to the transition matrix from one basis to another. Is that correct? I guess Gram Schmidt comes into play when we want to work with some arbitrary basis and $M$ is the basis change of matrix that takes us to the basis in which we now have an orthogonal set of basis vectors. I just don't see a way to proceed as of yet.

Thanks!

 

[geoffrey159]

Let's say that $A$ is a real matrix.
If $A$ is anti-symmetric ($A^T = -A$), then $A^2$ is symmetric. The spectral theorem says it is diagonalizable in an orthonormal basis.
Could the diagonal matrix be a representation of $F^2$ in another orthonormal basis ? I don't know, it is just a suggestion, this is how I would start.
If yes, you will get an equality that looks like $(M A M^T)^2 = F^2$, with $M$ being a product of 2 orthogonal matrices, therefore invertible. But still, there are these squares you'll have to get rid of.

 

[CAF123]

Ok, so given that $A^2$ is always symmetric it can always be diagonalised in a basis spanned by its (normalised) eigenvectors. So, something like $$RA^2R^T = \text{diag}(\lambda_1, \dots \lambda_{2n})$$ Then multiply by $P$ on the left and by $P^T$ on the right so that $$PR A^2 R^T P^T = P\text{diag}(\lambda_1, \dots \lambda_{2n}) P^T \overset{!}{=} F^2.$$ If $M = PR$ then get $(MAM^T)^2 = F^2$

So somehow I need to prove that it is possible to go from a diagonal form $D = \text{diag}(\lambda_1 \dots \lambda_{2n})$ to $F^2$? $F^2$ is so specific though about the elements on its block diagonal. And to resolve the sign where I square root both sides i thought about considering $M=\text{Id}$ to get the sign right for det but this depends on how much I'm allowed to assume about $\text{det F}$ (I can see that it is only 1 or 0 though)

Does the hint about Gram Schmidt help me at all and in particular using the quantity $x^T A y$?

 

[CAF123]

I got the following solution from another source:

We suppose that $A$ is written in the canonical basis. Consider the product defined by $\langle x,y\rangle=x^TAy$, the bilinear antisymmetric product. There exists a basis $e_1,...e_{2n}$ such that this product in this basis is $e_{n+1}\wedge e_{1}+...+e_{2n}\wedge e_{n}$. The matrix of the bilinear for in this basis is $\pmatrix{0 & 1 \cr -1 & 0}$.

Let $M$ be the transtion matrix from the basis to $e_1,...,e_{2n}$ to the canonical basis $M^TAM =\pmatrix{ 0 &1\cr -1 & 0}$

I didn't completely understand this proof - What does the term $e_{n+1}\wedge e_{1}+...+e_{2n}\wedge e_{n}$ evaluate to and why is the bilinear in this basis in which the inner product is equal to that given by $$\begin{pmatrix} 0 & 1 \\ -1&0 \end{pmatrix}?$$