Can a 3x3 Matrix Represent a Quadratic, Cubic, or Quartic Function?

[Bruno Tolentino]

I have a doubt...

Look this matrix equation:
\begin{bmatrix}<br /> A\\<br /> B<br /> \end{bmatrix} = \begin{bmatrix}<br /> +\frac{1}{\sqrt{2}} &amp; +\frac{1}{\sqrt{2}}\\<br /> +\frac{1}{\sqrt{2}} &amp; -\frac{1}{\sqrt{2}}<br /> \end{bmatrix} \begin{bmatrix}<br /> X\\<br /> Y<br /> \end{bmatrix}
\begin{bmatrix}<br /> X\\<br /> Y<br /> \end{bmatrix} = \begin{bmatrix}<br /> +\frac{1}{\sqrt{2}} &amp; +\frac{1}{\sqrt{2}}\\<br /> +\frac{1}{\sqrt{2}} &amp; -\frac{1}{\sqrt{2}}<br /> \end{bmatrix} \begin{bmatrix}<br /> A\\<br /> B<br /> \end{bmatrix}
By analogy, should exist a matrix 3x3 such that:
\begin{bmatrix}<br /> A\\<br /> B\\<br /> C\\<br /> \end{bmatrix} = \begin{bmatrix}<br /> ? &amp; ? &amp; ?\\<br /> ? &amp; ? &amp; ?\\<br /> ? &amp; ? &amp; ?\\<br /> \end{bmatrix} \begin{bmatrix}<br /> X\\<br /> Y\\<br /> Z\\<br /> \end{bmatrix}
\begin{bmatrix}<br /> X\\<br /> Y\\<br /> Z\\<br /> \end{bmatrix} = \begin{bmatrix}<br /> ? &amp; ? &amp; ?\\<br /> ? &amp; ? &amp; ?\\<br /> ? &amp; ? &amp; ?\\<br /> \end{bmatrix} \begin{bmatrix}<br /> A\\<br /> B\\<br /> C\\<br /> \end{bmatrix}
So, what values need be replaced in ? for the matrix equation above be right?

I think that my doubt is related with these wikipages:
https://en.wikipedia.org/wiki/Quadratic_formula#By_Lagrange_resolvents
https://en.wikipedia.org/wiki/Cubic_function#Lagrange.27s_method
https://en.wikipedia.org/wiki/Quartic_function#Solving_by_Lagrange_resolvent

EDIT: The inverse of the 3x3 matrix need be equal to itself.

 

[Mark44]

I have a doubt...

Look this matrix equation:
\begin{bmatrix}<br /> A\\<br /> B<br /> \end{bmatrix} = \begin{bmatrix}<br /> +\frac{1}{\sqrt{2}} &amp; +\frac{1}{\sqrt{2}}\\<br /> +\frac{1}{\sqrt{2}} &amp; -\frac{1}{\sqrt{2}}<br /> \end{bmatrix} \begin{bmatrix}<br /> X\\<br /> Y<br /> \end{bmatrix}
\begin{bmatrix}<br /> X\\<br /> Y<br /> \end{bmatrix} = \begin{bmatrix}<br /> +\frac{1}{\sqrt{2}} &amp; +\frac{1}{\sqrt{2}}\\<br /> +\frac{1}{\sqrt{2}} &amp; -\frac{1}{\sqrt{2}}<br /> \end{bmatrix} \begin{bmatrix}<br /> A\\<br /> B<br /> \end{bmatrix}

Your question is not very complicated. If $\vec{x} = A\vec{b}$, and A is an invertible matrix, then $\vec{b} = A^{-1}\vec{x}$. Note that capital letters are usually used for matrix names, and lower case letters are used for vectors or the components of vectors.

By analogy, should exist a matrix 3x3 such that:
\begin{bmatrix}<br /> A\\<br /> B\\<br /> C\\<br /> \end{bmatrix} = \begin{bmatrix}<br /> ? &amp; ? &amp; ?\\<br /> ? &amp; ? &amp; ?\\<br /> ? &amp; ? &amp; ?\\<br /> \end{bmatrix} \begin{bmatrix}<br /> X\\<br /> Y\\<br /> Z\\<br /> \end{bmatrix}
\begin{bmatrix}<br /> X\\<br /> Y\\<br /> Z\\<br /> \end{bmatrix} = \begin{bmatrix}<br /> ? &amp; ? &amp; ?\\<br /> ? &amp; ? &amp; ?\\<br /> ? &amp; ? &amp; ?\\<br /> \end{bmatrix} \begin{bmatrix}<br /> A\\<br /> B\\<br /> C\\<br /> \end{bmatrix}

Yes, as long as the matrix has an inverse. The equation I wrote is applicable for any square matrix A that has an inverse.

So, what values need be replaced in ? for the matrix equation above be right?

I think that my doubt is related with these wikipages:
https://en.wikipedia.org/wiki/Quadratic_formula#By_Lagrange_resolvents
https://en.wikipedia.org/wiki/Cubic_function#Lagrange.27s_method
https://en.wikipedia.org/wiki/Quartic_function#Solving_by_Lagrange_resolvent

None of these links is helpful, as far as I can see. Any linear algebra textbook will have a section on finding the inverse of a square matrix.

EDIT: The inverse of the 3x3 matrix need be equal to itself.

Of course.

 

[mathman]

The problem in matrix form is find A (if it exist other than A=I) where A^2=I. In scalar form it is 9 equations with 9 unknowns.

 

[Hawkeye18]

You need to find matrices such that $A^2=I$, or equivalently, such that $A^{-1}=A$. Any such matrix can be represented as $A=SJS^{-1}$, where $S$ is an invertible matrix, and $J$ is a diagonal matrix with entries $\pm 1$ on the diagonal . When all diagonal entries of $J$ are $1$ you get $A=I$, when all equal $-1$ you get $A=-I$; when $J$ has both $1$ and $-1$ on the diagonal you will get a non-trivial example of a matrix $A$.

This is a complete description, meaning that any matrix $A$ such that $A^{-1}=A$ can be represented as $A=SJS^{-1}$ (but the representation is not unique). And this is true in all dimensions, not just in dimension 3.