I find that at a high Bachelor's or low Master's level in physics I always have to re-explain determinants. People learn the idea and they take away "it's really complicated and has something to do with whether the matrix is invertible."
An eigenvector of a matrix M is a vector v such that M v = k v -- in this direction, the matrix only scales the vector and does not rotate it to some other direction. Some matrices do not have a complete set of eigenvectors -- a rotation matrix is a good example! -- but then there is a generalized notion of eigenvectors which you can fall back on.[1] You can prove that you always have a complete set of eigenvectors for any square matrix, in the generalized sense.
The trace is the sum of the eigenvalues; the determinant is the product of the eigenvalues. This drives in part the parallelepiped interpretation that the article is giving; in fact if you have a full set of eigenvectors then any volume can be made up out of little boxes made of the eigenvectors, and so any hyper-volume must transform under the linear map by multiplying by the determinant. This in turn tells you why det(A B) = det(A) det(B) -- B must scale volumes by some amount, then A must scale volumes by some amount, and to compose those two scalings you must take their product.
A matrix is not invertible if it projects away some axis, so that information is lost. This means that one of the eigenvalues is 0. Take the product of the eigenvalues -- if it's 0, the matrix is not invertible! Done, simple.
It tells you that the determinant of a triangular matrix is the product of the diagonal elements (since those are the eigenvalues), and it also tells you that a determinant of a block-diagonal matrix is given by the product of the determinants of the blocks. This is important because the whole "why does the determinant flip sign when you interchange two columns?" question is now answered. To flip two columns, you multiply by a matrix which looks like:
1 0 0 0 0
0 1 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 0 0 1
In other words, it's block diagonal with two blocks being identity matrices and one block being [0 1; 1 0]. That block has eigenvectors [1 1] with eigenvalue +1 and [1 -1] with eigenvalue -1, so it has determinant -1. The determinant of the two identity blocks are also 1, so the whole matrix determinant is therefore -1, and det(A B) = det(A) det(B) = - det(A), where B is the block-diagonal column-swapping matrix.
With a bit of effort you can figure out the antisymmetric form for actually computing the determinant without diagonalizing the matrix; but this comment is long enough as it is. The only thing which is much simpler about the antisymmetric symbolic form for the determinant is that you immediately see that the determinant is symmetric under transpose, which tells you something a bit surprising: that a square matrix has the same left-eigenvalues as right-eigenvalues (since an eigenvalue is det(A - λ I) = 0 and all of 0, det, and I are transpose-symmetric.
Some matrices do not have a complete set of eigenvectors -- a rotation matrix is a good example!
This is incorrect. A rotation matrix is orthogonal and therefore has a complete set of eigenvectors (all of which satisfy |z|=1). For example, the eigenvectors of:
0 1
-1 0
are z=+/-i and the eigenvectors are [1,i] and [1,-i].
The right example would be a matrix like this one:
Thanks for this very lucid comment. Even people who've been working with linear algebra for years have trouble wrapping their heads around the fact that rotations in 3D have eigenvalues and eigenvectors that aren't real.
An eigenvector of a matrix M is a vector v such that M v = k v -- in this direction, the matrix only scales the vector and does not rotate it to some other direction. Some matrices do not have a complete set of eigenvectors -- a rotation matrix is a good example! -- but then there is a generalized notion of eigenvectors which you can fall back on.[1] You can prove that you always have a complete set of eigenvectors for any square matrix, in the generalized sense.
The trace is the sum of the eigenvalues; the determinant is the product of the eigenvalues. This drives in part the parallelepiped interpretation that the article is giving; in fact if you have a full set of eigenvectors then any volume can be made up out of little boxes made of the eigenvectors, and so any hyper-volume must transform under the linear map by multiplying by the determinant. This in turn tells you why det(A B) = det(A) det(B) -- B must scale volumes by some amount, then A must scale volumes by some amount, and to compose those two scalings you must take their product.
A matrix is not invertible if it projects away some axis, so that information is lost. This means that one of the eigenvalues is 0. Take the product of the eigenvalues -- if it's 0, the matrix is not invertible! Done, simple.
It tells you that the determinant of a triangular matrix is the product of the diagonal elements (since those are the eigenvalues), and it also tells you that a determinant of a block-diagonal matrix is given by the product of the determinants of the blocks. This is important because the whole "why does the determinant flip sign when you interchange two columns?" question is now answered. To flip two columns, you multiply by a matrix which looks like:
In other words, it's block diagonal with two blocks being identity matrices and one block being [0 1; 1 0]. That block has eigenvectors [1 1] with eigenvalue +1 and [1 -1] with eigenvalue -1, so it has determinant -1. The determinant of the two identity blocks are also 1, so the whole matrix determinant is therefore -1, and det(A B) = det(A) det(B) = - det(A), where B is the block-diagonal column-swapping matrix.With a bit of effort you can figure out the antisymmetric form for actually computing the determinant without diagonalizing the matrix; but this comment is long enough as it is. The only thing which is much simpler about the antisymmetric symbolic form for the determinant is that you immediately see that the determinant is symmetric under transpose, which tells you something a bit surprising: that a square matrix has the same left-eigenvalues as right-eigenvalues (since an eigenvalue is det(A - λ I) = 0 and all of 0, det, and I are transpose-symmetric.
[1] https://en.wikipedia.org/wiki/Generalized_eigenvector