# Non singular matrix M such that MAM^T=F

• CAF123
In summary, the Gram-Schmidt algorithm can be used to solve the homework equations for antisymmetric matrices. The spectral theorem states that the matrix ##A^2## is symmetric and diagonalizable in an orthonormal basis. Given that ##A^2## is always symmetric, it can be diagonalized in a basis spanned by its (normalized) eigenvectors.
CAF123
Gold Member

## 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!

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

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!

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.

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##?

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 $$\begin{pmatrix} 0 & 1 \\ -1&0 \end{pmatrix}?$$

## 1. What is a non-singular matrix?

A non-singular matrix is a square matrix that has a non-zero determinant, meaning it is invertible and has a unique solution for its system of equations.

## 2. What does "MAM^T" mean in the context of a non-singular matrix?

In this context, "MAM^T" represents the product of three matrices: M, A, and the transpose of M (M^T). This is often used in linear algebra to transform a matrix A by applying a similarity transformation using matrix M.

## 3. What is the significance of a non-singular matrix in scientific research?

Non-singular matrices play a crucial role in many scientific fields, including physics, engineering, and statistics. They are used to solve systems of equations, perform transformations, and analyze data. They also have important applications in computer graphics and machine learning.

## 4. How do you determine if a matrix is non-singular?

A matrix is non-singular if its determinant is non-zero. This can be calculated by using row reduction or by using the cofactor expansion method. If the determinant is zero, the matrix is singular and is not invertible.

## 5. Can a non-singular matrix MAM^T always be written as F?

No, the matrix F is not always unique for a given non-singular matrix MAM^T. However, it is possible to find multiple matrices F that satisfy this equation. The specific values of F will depend on the dimensions and elements of the matrices M and A.

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