Defining the Pseudo-Determinants of a Matrix

In this blog post, we define the "pseudo-determinants" of a matrix. We will start with what we know.

Definitions

Let ${\displaystyle A}$ and ${\displaystyle B}$ denote two square matrices, respectively.

Let ${\displaystyle d=\det \left(A\right)}$ denote the determinant of matrix ${\displaystyle A}$.

What we know:

From the second definition, ${\displaystyle \det \left(A\right)=d}$.

It is a well known theorem that ${\displaystyle \det \left(A^{T}\right)=d}$ (1).

We know that ${\displaystyle \det \left(AB\right)=\det \left(A\right)\det \left(B\right)}$ (2).

If we set ${\displaystyle B = A^T }$, (2) becomes:

${\displaystyle \det \left(AA^{T}\right)=\det \left(A\right)\det \left(A^{T}\right)}$ (3).

We then substitute (1) and the definition to get:

${\displaystyle \det \left(AA^{T}\right)=d^{2}}$ (4)

Now ${\displaystyle AA^{T}}$ is guaranteed to be a square matrix; therefore it is legal to square root both sides to conclude that:

${\displaystyle d=\pm {\sqrt {\det \left(AA^{T}\right)}}}$ (5)

However, there are some cases where the following would result in ${\displaystyle d = 0}$, therefore the following alternative is proposed:

${\displaystyle d=\pm {\sqrt {\det \left(A^{T}A\right)}}}$(6)

However, if both result in ${\displaystyle d = 0}$, then ${\displaystyle d = 0}$.

Example:

We will calculate the pseudo-determinants of this matrix:

${\displaystyle A={\begin{pmatrix}1&2\\3&4\\5&6\end{pmatrix}}}$

The formula dictates that we first transpose ${\displaystyle A}$, which leads to this:

${\displaystyle A^{T}={\begin{pmatrix}1&3&5\\2&4&6\end{pmatrix}}}$

Multiplying the transposed version of the matrix by ${\displaystyle A}$ (the proof that the reverse would result in ${\displaystyle d = 0}$ is left as an exercise to the blog post reader), we obtain:

${\displaystyle A^{T}A={\begin{pmatrix}35&44\\44&56\end{pmatrix}}}$

The determinant of the resulting matrix can be shown to be ${\displaystyle 24}$, and the pseudo-determinants can be shown to be ${\displaystyle \pm 2{\sqrt {6}}}$.

Conclusion:

The pseudo-determinant might be a novel discovery as there is inadequate description of the topic on mathematics websites.