Homework Help: General interpretaion of nxn determinant

2x2 is the area of a parallelogram spanned by two vectors,
3x3 is the volume of a parallelepiped spanned by three vectors

determinant of nxn is the nth dimensional volume of a parallelepiped spanned by n vectors

 

Yes, the idea of volume generalizes in a natural way to n-dimensional space.

Suppose you're working in R^n, the n-dimensional vectors of real numbers. Then there are n standard basis vectors. For example, when n = 3, these are simply the unit vectors in the x, y, and z directions.

The "unit n-cube", also called a hypercube, is the natural extension of the unit cube to n-dimensional space: all of its sides have length 1 along each edge, and it is positioned so that all of the coordinates of all of its points are >= 0, with one corner at the origin.

Because all of the edges have length 1, the volume of the unit n-cube is 1.

Any n x n matrix M maps the unit n-cube into an n-dimensional parallelpiped. The volume of that parallelpiped is precisely |det(M)|. (The absolute value of the determinant.)

The sign of the determinant doesn't have any effect on the volume of the n-dimensional parallelpiped. It controls, loosely speaking, the "polarity" of the mapping: if you number the axes of the unit cube in a certain order, which can be thought of as clockwise or counterclockwise, does the mapping preserve that order or reverse it?

 

One key thing: if the determinant is zero, then geometrically speaking the volume of the parallelpiped is zero, meaning one or more of the dimensions of the unit cube have been "collapsed" or "flattened." This means that the image of the map has lower dimension than vector space itself. You can't "undo" this reduction of dimensionality, meaning that the map is not invertible.

In fact, for a linear map T from V to itself, where V is finite-dimensional,

T is invertible IF AND ONLY IF its determinant is nonzero

Note: My argument above gives a geometric justification for this fact, but it's not a rigorous proof.

 

No, the size (magnitude) of the determinant has no correlation with how close it is to being singular.

If I take the identity matrix, and multiply it by any nonzero real number, I can make the magnitude of the determinant as big or as small as I like, and it remains equally as invertible no matter what number I choose.

There ARE measures of how close a matrix comes to being non-invertible (singular). One such measure is the ratio of the largest and smallest singular values: this is called the "condition number." Bigger condition numbers mean "closer to being singular".

In my example above, the condition number of the identity matrix is 1, as is the condition number of any (nonzero) scalar multiple of the identity matrix.

 

It is true that a matrix is invertible if and only if its determinant is non-zero. Is that what you meant? I have no idea what you could mean by "how much a matrix is invertible". It either is or it isn't- there is no "how much"!