### Inverses, determinants, column spaces, non-square matrices

PART 1

First, let's look at a graphical way of expressing a linear transformation $L:V\to W$. We choose a set of parallel lines from $V$, transform them under $L$ and graph their images in $W$. In the following example, we take the linear transformation of a rotation of $-\pi/4$ (about the origin) and use a set of lines parallel to the x or y-axes with integer y and x-values respectively.

 (Courtesy: Interactive Matrix Visualization)
If you've thought about linear transformations a bit, you might have realised that some linear transformations map distinct vectors onto the same line (and by linearity, this would imply that there are distinct vectors that are mapped to the same vector). This would only be possible if the transformation mapped the plane to a lower dimension (again, this is because we're dealing with a linear transformation). It seems that it would be useful to study such transformations in greater detail.

For one, such a transformation wouldn't have an inverse, because it is not possible to determine exactly which vector maps to a given vector in the target space.

If we have $Ax=v$ for some known $A$ and $v$, then we can find the inverse for an ordinary $A$. But if $A$ is of the form mentioned -- we call this a "singular matrix" -- this is not possible, and there is either an infinite number of solutions for $x$ (if $v$ lies in the target space/range of $A$) or no solution for $x$ (if it doesn't).

In terms of linear equations, these correspond to cases of the form $x + 2y = 4, x + 2y = 8$ and $x + 2y = 4, x + 2y = 9$ respectively. Indeed, these are the sort of linear equations that one obtains from expanding out $Ax=v$ and the equations in the components become the linear equations.

Think: functions, invertible functions, etc.

Let's think about how one can identify a singular matrix. In two dimensions, this is relatively obvious -- a matrix is singular if it maps the space to either 0 or 1 dimensions, and in both cases, the unit vectors are mapped onto the same line -- hence, the column vectors are multiples of each other.

In more than two dimensions, this is not necessary -- for example, the three basis vectors in $\mathbb{R}^3$ may be mapped onto a plane, which would cause the space to be transformed onto this plane.

Similarly, in $\mathbb{R}^4$, if four basis vectors are mapped to a 3-dimensional hypersurface, OR at least three of the basis vectors are mapped onto a 2-dimensional plane, OR at least two of the basis vectors are mapped onto a line, OR at least one of the basis vectors is mapped onto a point (the origin), the space is reduced to a space of lower dimension and the matrix is singular.

Each of these cases correspond to linear dependence between the column vectors. This is, in general, our criterion for singularity: a matrix with linearly dependent columns is singular.

Note that this statement can also be formulated based on rows, due to the connection with linear equations we mentioned earlier -- this should foreshadow a certain identity regarding the determinant that we will derive later.

Name the "kind" of singular matrices that each of the following is (i.e. what is its range?) without graphing or calculating anything. Consider where the basis vectors are mapped. $\vec a, \vec b, \vec c$ are linearly independent vectors in $\mathbb{R}^4$.

$$\left. \begin{array}{l}(a)\,\,\left[ {\begin{array}{*{20}{c}}{\vec a}&{2\vec a}&{\vec b}&{\vec c}\end{array}} \right]\\(b)\,\,\left[ {\begin{array}{*{20}{c}}{\vec a}&{\vec a + \vec b}&{\vec b}&{\vec c}\end{array}} \right]\\(c)\,\,\left[ {\begin{array}{*{20}{c}}{\vec a}&{\vec a + \vec b + 2\vec c}&{\vec b}&{\vec c}\end{array}} \right]\\(d)\,\,\left[ {\begin{array}{*{20}{c}}{\vec a}&{2\vec a}&{3\vec a}&{\vec c}\end{array}} \right]\\(e)\,\,\left[ {\begin{array}{*{20}{c}}{\vec a}&{2\vec a}&{2\vec b}&{\vec b}\end{array}} \right]\\(f)\,\,\left[ {\begin{array}{*{20}{c}}{\vec a}&{2\vec a}&{\vec a + 2\vec b}&{\vec b}\end{array}} \right]\\(g)\,\,\left[ {\begin{array}{*{20}{c}}{\vec a}&{\vec b}&0&{\vec c}\end{array}} \right]\\(h)\,\,\left[ {\begin{array}{*{20}{c}}{\vec a}&{\vec b}&0&{3\vec b}\end{array}} \right]\end{array} \right\}$$

(a) Two basis vectors are mapped onto a line, hence range is a three-dimensional subspace.
(b) Three basis vectors are mapped to the same plane, hence the range is a three-dimensional space.
(c) Four basis vectors are mapped onto a 3D subspace, hence range is a 3D subspace.
(d) Three basis vectors are mapped onto a line, hence range is a plane.
(e) Basis vectors are mapped onto one of two lines, hence range is a plane.
(f) Two basis vectors are mapped onto a line on the same plane as the other basis vectors, thus plane.
(g) A basis vector is mapped to zero, hence range is a 3D space.
(h) A basis vector is mapped to zero and two vectors to a line, hence range is range is a plane.

In general, the dimension of the range of the transformation is the number of vectors we're left with after removing the redundant ones, i.e. the range is the span of the column vectors.

Note that the zero vector is always redundant, as it doesn't span anything and is always a zero multiple of any other vector.

This "range" we've been talking about is called the column space of the matrix. The dimension of the column space is called the rank. This is true also for non-singular matrices, where the column space becomes the same space as the original, i.e. the space of all vectors the matrix can act on. Then the rank is at its maximum, and the transformation is called full-rank.

PART 2

We are interested in finding a quantity that, rather than taking discrete values and abruptly changing for singular matrices as the rank does, returns a continuous scalar value expressing how much a transformation smashes things to near zero.

It is reasonable to use the n-volume of the n-parallelepiped contained by the column vectors (i.e. by the images of the basis vectors) as this quantity. This volume is called the determinant of the transformation. It is clear that when the determinant is zero, the transformation is singular, and the reverse is true. Therefore the law to determine whether a matrix is singular applies to finding whether the determinant is zero.

Based on what we already know, it is possible to show, geometrically, that:
• Multiplying a row or column by a scalar scales the determinant by the same scalar.
• $\det(AB)=\det(A)\det(B)$ -- this essentially means that the determinant can be regarded as a "ratio" of volumes, i.e. it's a scaling factor, and measures the "amount" of the transformation, like an absolute value or magnitude.
• If $A$ and $B$ have all the same elements except one row/column, then the determinant of another matrix $C$ which also has all the same elements except that row/column which is a sum of the corresponding row/column of $A$ and $B$, is equal to $\det(A)+\det(B)$.
Here's another property: you might recall that for some triangle whose sides are given by vectors $a$, $b$ and $c$, its area can be given by $\frac12 |a\times b| = \frac12 |a\times c| = \frac12 |b\times c|$. With our knowledge of determinants, we can write $|a\times b| = \left|\det \left[ {\begin{array}{*{20}{c}} a \\ b \end{array}} \right]\right|$, and analogous expressions for $a \times c$, etc. -- these expressions are all equal. Notice that $c=a-b$ or something to that effect -- this means the determinants $\det \left[ {\begin{array}{*{20}{c}} a \\ b \end{array}} \right] = \det \left[ {\begin{array}{*{20}{c}} a \\ {b - a} \end{array}} \right]$, etc.

This can be generalised to any number of dimensions: subtracting any scalar multiple of a row or column from another row or column (respectively) of a matrix leaves the determinant unchanged.

This is very useful in computing a determinant in practice.

We now direct our attention to actually computing an inverse of a matrix.

We know that $A^{-1}A=I$, by definition. Here's our strategy to find $A^{-1}$: we multiply $A$ by a series of matrices until we get $I$. In order to keep track of the matrices, we simultaneously multiply $I$ by the same matrices. Now since the product of these matrices with $A$ is $I$, this matrix is $A^{-1}$, and multiplying them with $I$ gives us $A^{-1}$.

What are some simple linear transformations we can apply on $A$ that we know how to manipulate to eventually yield $I$ (at least for a non-singular matrix)? Well, there is a set of such linear transformations, called row operations (or column operations, which serve the same purpose), which essentially add, subtract and scalar-multiply rows and columns as if they were simultaneous equations we are trying to solve.

Think: what would be the result of doing this on a singular matrix, besides the usual "universe blows up" stuff?

As a general rule: anything we can legitimately do to linear equations, is a legitimate row operation. This makes a lot of sense, given that the solution to $Ax=v$ is $v=A^{-1}v$. It is immediately clear that $\det(A^{-1})=\det(A)^{-1}$, which generalises our earlier identity about $\det(A^n)$ very nicely to negative integers.

More on determinants and inverses in the computational guide.

PART 3

A quick note on non-square matrices: it should be clear that non-square matrices are maps between vectors of different dimensions. Their determinant is not traditionally defined for them, but some generalisations do exist -- recommended reading: Generalisations of the determinant to interdimensional transformations: a review.

PART 4

You should make a list of things equivalent to invertibility for a square matrix. Here's a list -- some of the stuff here will make sense later:
• Invertibility
• Determinant non-zero
• Linearly independent rows, columns
• Full-rank
• Zero-nullity
• Injectivity
• Surjectivity
• Bijectivity
• Image = Co-domain = Domain
• RREF is the identity matrix
• No zero eigenvalues
• No zero singular values