Every square matrix has a determinant, a number that corresponds to the area, volume or hypervolume of the shape formed by the n-vectors defined by the rows of the matrix. The columns can also be used, and though the shape would be different, the determinant is unchanged.
Example: for any 2x2 matrix
a b
c d
the determinant equals ad - bc.
This is the area of a parallelogram defined by the points (0,0), (a,b), (c,d) and (a+c,b+d), or by the points (0,0), (a,c), (b,d) and (a+b,c+d).
We get a positive area if the vector (c d) is on the left hand side of the vector (a b) in the two-dimensional plane. Conversely, a negative area means the vector (c d) is on the right hand side of the vector (a b) in the two-dimensional place. A determinant of zero means there is a scalar k such that c = ak and d = bk.
There is a problem with determinants. The number of terms increases factorially, which is a growth rate even faster than exponential growth. 3x3 matrices have 6 terms in the determinant, 4x4 have 24, 5x5 have 120, etc. I will show the 3x3 formula, but I will avoid the general formula for larger cases.
Example: for any 3x3 matrix
a b c
d e f
g h i
the determinant equals aei + bfg + cdh - afh - bdi - ceg.
Why these terms and why add some while subtracting others?
Every term in an nxn determinant is the product of n entries, with exactly 1 term in every row and every column. The term aei corresponds to the main diagonal matrix
1 0 0
0 1 0
0 0 1
This is also the identity matrix. Every other term corresponds to the positions of the 1 entries in one of the permutation matrices. If the permutation consists of an even number of transpositions from the identity, the corresponding term is added to the total, and the terms that correspond to odd permutations are subtracted from the total.
Thinking in the language of group theory, the group even and odd under addition is mapped onto the group 1 and -1 under multiplication.
A matrix M is singular if det(M) = 0.
There is a standard method for finding the inverse of a matrix and one of the steps is to divide by the determinant. We can't divide by zero, which means singular matrices do not have inverses.
Stated without proof: det(MN) = det(M)det(N).
If the singular nxn matrices are removed from the set of all nxn matrices, we have a group, and determinant is a homomorphism from the group of non-singular nxn matrices under multiplication to the group (ℝ-{0}, ✕).
If a matrix M has real entries, det(M) will also be real, since the operations used to compute the determinant are multiplication, addition and subtraction. This brings us to a simply proved statement.
Lemma: If G is a finite group of nxn matrices under multiplication, the determinants of the matrices must be either 1 or -1.
Proof by contradiction: Assume one of our matrices M has a determinant x, not equal to 1 or -1. This would mean det(M²)=x² and det(M³)=x³, and so on, ad infinitum. It could not be a finite group, so instead we must assume det(M) must be either 1 or -1 for the group to be finite.
Stated without proof: If B is a matrix created by switching two rows of the matrix A, det(B)=-det(A).
Since every nxn permutation matrix can be created by multiple transpositions of rows of the identity matrix In, and the determinant of In = 1, all permutation matrices have a determinant of either 1 or -1. Having a lot of zeros in a matrix can make the computation of the determinant much simpler, since any product that includes a zero must be zero. Permutation matrices have so many zeros that there is only term that is non-zero and that term is the product 1ⁿ=1, which will be multipled by 1 or -1, depending on the parity of the number of row switches away from the idenity.
Theorem: If two matrices M and N are conjugates, det(M)=det(N).
Proof: Conjugacy means N = XMX⁻¹. If det(X)=x, then
det(X⁻¹)=1/x.
det(N)= det(XMX⁻¹) = x*det(M)*1/x= det(M).
The determinant of a matrix is invariant under conjugacy.
Stu Smith, who taught me the majority of my postgraduate math classes, said that one should always pay attention to invariance, because it usually implied something of a physical nature.
Like Tracewell, Stu Smith is someone I tried to emulate when I taught, but I admit I fell short on both counts. I was never as enthusiastic as Ted Tracewell, and I was was never as erudite as Stu Smith.
Commentary
In the middle of the 19th Century, a great number of mathematicians were working on ways to simplify the calculation of large matrix determinants, including Leopold Kronecker, who was introduced in the last post, and Charles Dodgson, better known to the general public as Lewis Carroll.
I learned a story in a math class, I forget what professor, that Queen Victoria loved Alice in Wonderland, published in 1865, after the death of her husband Prince Albert in 1861, an event that left her nearly inconsolable. She wrote to Lewis Carroll and asked him to send her the next book he published. Following her wishes, she sent him a book on determinants.
The story is true until the punch line. Dodgson knew the Queen wanted his next Lewis Carroll book Alice Through the Looking Glass, and he sent it to her as soon as it was published in 1871. He did have a quirky sense of humor, but he would never disrespect his monarch in such a fashion.
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