# Criterion for invertibility of a $2 \times 2$ matrix over reals

I'm revising some elementary linear algebra after a multi-decade break in which I've forgotten most of it. I've taken a look at a few introductory text books, and there seems to be a common line of argument about singular matrices that seems to me to be wrong: many of these textbooks claim that because the determinant of a singular matrix (let's say a $2 \times 2$ matrix over the reals) can be viewed as "area destroying" then the matrix maps $\mathbb{R}^2$ to $\mathbb{R}$, and it's therefore "obvious" that it cannot be inverted.

Surely this is twaddle since the cardinality of $\mathbb{R}^2$ is the same as that of $\mathbb{R}$ ? Isn't the real argument to show that a singular 2x2 matrix is not injective ?

Who's confused here? Me or the books?

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We don't measure area (or distance, or volume) by cardinality. – Arturo Magidin Sep 22 '11 at 13:55
While there are bijections from $\mathbb{R}$ to $\mathbb{R}^2$, it is impossible for them to be linear, because a basis for $\mathbb{R}$ (which has one element) would have to be in bijection with a basis for $\mathbb{R}^2$ (which has two elements). Of course, it is also true that a singular matrix is not injective, so one need not appeal to dimension to show that there can't be an inverse (in fact, I agree that arguing by non-injectivity is conceptually clearer). But the book is trying to give an intuitive explanation for why singular matrices aren't injective, not a rigorous proof. – Zev Chonoles Sep 22 '11 at 14:08
@arturo - Sorry, I don't see what point you're making - can you expand please ? Zev: I have no problem with intuitive arguments, but it seems to me that this particular intuitive argument will leave people thinking that any map from, say, $\mathbb{R}^2$ to $\mathbb{R}$ cannot be inverted (because at this level, most people will see $\mathbb{R}^2$ as "bigger" than $\mathbb{R}$)- however, surely the main point is that lack of invertibility is due to non-injectivity, and the "area destroying" property just happens also to occur, in the case of singular matrices ? – user16582 Sep 23 '11 at 10:22
@ukmaths: You say: "Surely [area destroying] is twaddle since the cardinality of $\mathbb{R}^2$ is the same as that of $\mathbb{R}$". But cardinality is not a measure of area; the fact that they have equal cardinality has nothing to do with whether or not you can consider map from $\mathbb{R}^2$ whose image is one dimensional "area destroying". The entire sentence "since the cardinality of $\mathbb{R}^2$ is the same as that of $\mathbb{R}$" is utterly irrelevant, because we don't measure area using cardinality. – Arturo Magidin Sep 23 '11 at 17:15

Let's divide the problem into smaller parts.

1. The books you are mentioning are making use of the geometric interpretation of the determinant as being the volume of the paralellepiped (tough to pronunce that one)
2. The hypothetical paralellepiped is spanned by $a v_1 + b v_2+\ldots+z v_n$ where $a,b,\ldots,z \in [0,1]$ and $v_i$ columns of the matrix.
3. Zero determinant means zero volume( area in 2x2 case). Zero determinant also means the columns of the matrix is linearly dependent since they produce a zero volume (area) with non zero vectors.
4. Since the matrix columns are linearly dependent, the image (range, columnspace ...) of this matrix is one dimensional since it is rank 1 and hence defines a line passing through the origin with some orientation and the argument defines its magnitude e.g. $$\pmatrix{2&-1\\4&-2}\pmatrix{x\\y} = \pmatrix{2x-y\\4x-2y} = \pmatrix{1\\2}(2x-y)$$ Therefore, the mapping is indeed from $\mathbb{R}\times \mathbb{R}\mapsto \mathbb{R}$ if you consider the line spanned with a unit vector multiplied with a parameter $t = (2x-y)$. The volume of a plane (area of a line) is hopefully zero. Besides, there is no unique $(x,y)$ that gives $\pmatrix{3&6}^T$ hence the non-invertibility argument. So we are not constructing a bijection between $\mathbb{R}^2$ and $\mathbb{R}$ but only parameterizing the domain and the image.

In my opinion this mapping notation is causing the problem since, we are actually mapping $\mathbb{R}\times \mathbb{R}\mapsto \mathbb{R}\times \mathbb{R}$ since the resulting vector $\pmatrix{3&6}^T$ is still in $\mathbb{R}^2$ but the information to define the resulting object uniquely requires only one parameter that is $t\in\mathbb{R}$. Therefore it is reflecting the number of copies of $\mathbb{R}$ to locate the objects in the space of points. I think you should not look at it as a construction of a bijection.

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In $\mathbb{R}^2$ at least, here's what the condition for the invertibility of a linear map means in terms of the determinant.

The following steps will lead to an enhanced understanding:

Why is that injectivity is equivalent to asking if the columns of a matrix are linearly independent?

If the determinant of a $2\times 2$ matrix $\left[\begin{array}{cc} a & c \\ b & d \end{array}\right]$ is zero, this means that $ad-bc = 0$, or that $ad = bc$, or that $b/a = d/c$ for $a,c \neq 0$. What does this mean? You can think about this geometrically in terms of which "direction" the column vectors of the matrix point to.

Note that this is not such an effective way to look at things in higher dimensions, but in $\mathbb{R}^2$ at least it helps in understanding things.

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