MHB Determinant always equal to zero?

[alexmahone]

I just finished writing a computer program that takes as input a number of matrices and computes the inverse of the product of matrices.

To test this program, I wanted to input a 3x2 matrix followed by a 2x3 matrix so that the product would be a 3x3 matrix. No matter how hard I try, the determinant of the product turns out to be zero and so the product is non-invertible. Is there a theorem in linear algebra that implies that the product of a 3x2 matrix and 2x3 matrix will always have determinant zero?

 

[Fallen Angel]

Hi,

Yes, there is a result that fit very well.

Given any two matrices $A,B$ of size $n\times m$ and $m \times r$ (this is just to ensure we can multiply them) then $rank(AB)\leq \ \min{\{ rank(A),rank(B)\}}$.

 

[alexmahone]

Hi,

Yes, there is a result that fit very well.

Given any two matrices $A,B$ of size $n\times m$ and $m \times r$ (this is just to ensure we can multiply them) then $rank(AB)\leq \ \min{\{ rank(A),rank(B)\}}$.

Sorry, I don't know about ranks. Is there a way to prove that the determinant is zero without using ranks?

 

[Fallen Angel]

Well,

You can set $A=\left(\begin{array}{cc} a_1 & a_2 \\ a_3 & a_4 \\ a_5 & a_6 \end{array}\right)$ and $B=\left( \begin{array}{ccc} b_1 & b_2 & b_3 \\ b_4 & b_5 & b_6 \end{array}\right)$ and make all the computations, but it is pretty much easier if you have some linear algebra background.

For instance, it will be enough if you know the relation between matrices and linear applications, do you know it?

 

[I like Serena]

Sorry, I don't know about ranks. Is there a way to prove that the determinant is zero without using ranks?

Hi Alexmahone,

If you put a 3-dimensional vector in a 2x3 matrix, the result is a 2-dimensional vector - a plane.
If you then put that 2-dimensional vector in a 3x2 matrix, the result is still a plane, albeit in 3 dimensions.
The consequence is that the product of those 2 matrices only has 2 independent columns - the vectors that span the plane.
Therefore the determinant is zero.

 

[Deveno]

(Multiplication by) a matrix is a LINEAR map-one consequence of this is it maps subspaces of a vector space to subspaces of another vector space.

If you have an $m \times n$ matrix with real entries, the vector spaces in question are $\Bbb R^n$ and $\Bbb R^m$, that is to say an $m \times n$ matrix when multiplied by "on the left" takes an $n$-tuple to an $m$-tuple:

$(m \times n)(n \times 1) \to m\times 1$.

So let's say you have a 3x2 matrix. Even though you are mapping INTO a space of a larger dimension, you don't wind up with the whole space "being reached", you only get (at best) a two-dimensional subspace...a plane. Which plane? The plane spanned by the two columns of our matrix:

$\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{bmatrix}\left(x\begin{bmatrix}1\\0\end{bmatrix}+y\begin{bmatrix}0\\1\end{bmatrix}\right)$

$= x\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} + y\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}$

$ = x\begin{bmatrix}a_{11}\\a_{21}\\a_{31}\end{bmatrix} + y\begin{bmatrix}a_{12}\\a_{22}\\a_{32}\end{bmatrix}$.

Now if you start with a 2x3 matrix, you take a 3-dimensional space and shrink it to a (at most) 2-dimensional space. As you can see from the above, hitting the two-dimensional space with a 3x2 matrix doesn't "blow it back up" again, it will stay at most "two-dimensional".

So the net effect is this (at best, more shrinkage could occur if we have "bad matrices"), just looking at dimensionality:

$3 \to 2 \to 2$.

We lose an entire dimension, which is to say our "combo mapping" (the composition, or matrix product) isn't 1-1, it's "many-to-one". Such a mapping cannot possibly be invertible (any inverse isn't "uniquely defined", and some triples in $\Bbb R^3$ will "never be reached", so we can't find any "pre-image").

You should have much better luck if you try "the other product": a 2x3 times a 3x2 matrix, which does this to the dimensions (if your matrices are "nice"-that is, of maximum rank):

$2 \to 2 \to 2$.