invertibility Archives - rweber.net

Applying Linearity

Rebecca — Mon, 24 Feb 2014 13:00:20 +0000

At the center of almost any proof involving linear transformations is to apply linearity to move between the domain and codomain, preserving the structure of linear combinations. Statements proved in such a way:

all linear transformations take 0 to 0
images of subspaces are subspaces (special example: image)
preimages of subspaces are subspaces (special example: kernel)
closing a set of vectors before or after applying a transformation gives the same vector space
we may unambiguously determine the entirety of a linear transformation from its action on a basis (linear extension) — which is what makes matrix representation possible!
having an inverse is equivalent to being bijective
the rank of a matrix is definable by its rows as well as its columns
inverses of isometries are also isometries (also uses bilinearity of inner product)

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Table of matrix possibilities

Rebecca — Mon, 11 Nov 2013 13:00:50 +0000

I like examples.

	diagonalizable	not diagonalizable
invertible	I_n,	rotation about non-axis line by, say, 60 degrees in R^3, rotation 90 degrees in R^2:
not invertible	zero matrix,	rotation 90 degrees plus projection onto yz-plane in R^3:

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Invertibility and the Determinant

Rebecca — Mon, 07 Jan 2013 13:00:16 +0000

I didn’t really like the cofactor construction of the inverse of a matrix with nonzero determinant, to prove said inverse exists. I’m willing to accept the equation det(MN) = det(M)det(N) on faith, since I am confident I could work that out if I really had to. With that in my pocket, here’s an explanation of the correspondence between nonzero determinant and invertibility.

Suppose M is invertible. Then there is some M^-1 such that MM^-1 = I, the identity matrix. Then 1 = det(I) = det(MM^-1) = det(M)det(M^-1), and 1 cannot be obtained as a product of 0 with anything, so both M and M^-1 have nonzero determinant.

Now suppose M is (square and) noninvertible. Then the kernel of the transformation T that M represents includes some nonzero vector X, and we may build a basis B for our vector space such that B includes X. The matrix M’ for T relative to B includes a column of all 0s, corresponding to the position of X in the ordering of B, so det(M’) = 0. For appropriate change of basis matrices P, P^-1 we have M = PM’P^-1, so det(M) = det(PM’P^-1) = det(P)det(M’)det(P^-1) = 0.

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