Thread: Linear & Matrix Transformations

I am beginning to learn about linear transformations and matrix-vector products, both vast subjects.

I have three questions that I have never found addressed, and I hope somebody out there can tell me or point me to the answers.

First, do there exist linear transformations in Rn that cannot be represented as matrix-vector products, where all entries are real numbers? So far as I know, there are none.

Second, do there exist matrix-vector products in Rn, where all entries are real numbers, that do not represent linear transformations? So far as I know, there are none.

Third, do there exist linear transformations in Rn, represented by matrix-vector products, that are neither isometries, dilations, nor shears? So far as I know, there are none.

I feel like I must be missing something, but I don't know what.

Thanks!

Originally Posted by zhandele

I am beginning to learn about linear transformations and matrix-vector products, both vast subjects.

I have three questions that I have never found addressed, and I hope somebody out there can tell me or point me to the answers.

First, do there exist linear transformations in Rn that cannot be represented as matrix-vector products, where all entries are real numbers? So far as I know, there are none.

No. Every linear transfromation from Rn to Rn can be expressed as an n by n matrix. The exact matrix depends upon your choice of basis for Rn- though different matrices represent the same linear transformation in different bases if and only if they are "similar matrices".

Second, do there exist matrix-vector products in Rn, where all entries are real numbers, that do not represent linear transformations? So far as I know, there are none.

Again, no. In fact, matrices are themselves linear transformations. Since any n-dimensional vector space is isomorphic to Rn, any given n by m matrix represents a linear transformation from a m-dimensional vector space to an n-dimensional vector space.

Third, do there exist linear transformations in Rn, represented by matrix-vector products, that are neither isometries, dilations, nor shears? So far as I know, there are none.

Again, no. Every linear transformation is a combination of those three kinds of linear transformations.

I feel like I must be missing something, but I don't know what.

Thanks!

it turns out that for a given pair of vector spaces, say U, where dim(U) = n, and V, where dim(V) = m (for this to make sense, these must both be vector spaces over the same field, F), that the set of all linear transformations from U to V (often denoted Hom(U,V)) is itself a vector space.

we define, for S,T in Hom(U,V), S+T by:

(S+T)(u) = S(u) + T(u) <--note the RHS addition is in V

and:

(aS)(u) = a(S(u)) <----this is the scalar multiplication of V.

this makes Hom(U,V) into a vector space. now for Mat_mxn(F), we obviously can add matrices:

if A = (a_ij) and B = (b_ij), then we set A+B = (a_ij+b_ij).

and we also have scalar multiples: cA = (ca_ij).

and we have the following important theorem: Hom(U,V) is isomorphic to Mat_mxn(F).

we will show this by actually displaying an isomorphism (note this isomorphism is NOT unique, as it depends on a choice of bases).

let {u₁,...,u_n} be a basis for U, and let {v₁,...,v_m} be a basis for V.

define the map: T_ij in Hom(U,V) by T_ij(u_j) = v_i, T_ij(u_j) = 0.

for example, with m = 3, n = 4, we have:

T₂₁(u) = T₂₁(a₁u₁+a₂u₂+a₃u₃+a₄u₄) =

a₁T₂₁(u₁) + a₂T₂₁(u₂) + a₃T₂₁(u₃) + a₄T₂₁(u₄)

= a₁v₂ + a₂0 + a₃0 + a₄0 = a₁v₂.

lemma #1: the {T_ij} form a basis for Hom(U,V).

suppose S is in Hom(U,V).

then S(u) = S(a₁u₁+...+a_nu_n) = b₁v₁+...+b_mv_m

in particular, for each j = 1,2,...,n we have:

S(u_j) = b_1jv₁+...+b_mjv_m

= b_1jT_1j(u_j) +...+ b_mjT_mj(u_j) = (b_1jT_1j+...+b_mjT_mj)(u_j).

thus S(u) = a₁S(u₁) +...+ a_nS(u_n)

= a₁(b₁₁T₁₁(u₁)+...+b_m1T_m1(u₁)) +...+ a_n(b_1nT_1n(u_n)+...+b_mnT_mn(u_n))

= b₁₁[T₁₁(a₁u₁)+...+T₁₁(a_nu_n)] +...+ b_m1[T_m1(a₁u₁)+...+T_m1(a_nu_n)] +...+ b_1n[T_1n(a₁u₁)+...+T_1n(a_nu_n)] +...+ b_mn[T_mn(a₁u₁)+...+T_mn(a_nu_n)]

(remember T_ij(u_k) = 0 if j ≠ k, so the "extra terms" i've added are all 0's).

= b₁₁T₁₁(u)+...+b_m1T_m1(u)+...+b_1nT_1n(u)+...+b_mnT_mn(u)

= (b₁₁T₁₁+...+b_1nT_1n+...+b_m1T_m1+...+b_mnT_mn)(u)

that is, any S in Hom(U,V) is a linear combination of the T_ij, so these span Hom(U,V).

next, suppose that c₁₁T₁₁+...+c_mnT_mn = 0 (the 0-map of Hom(U,V)).

then (c₁₁T₁₁+...+c_mnT_mn)(u_j) = 0 (the 0-vector of V), so

(c_1jT_1j+...+c_mjT_mj)(u_j) = 0 (these are the only terms which survive, since T_ij(u_k) = 0, for j ≠ k) so

c_1jv₁+...+c_mjv_m = 0, for EACH j, and by the linear independence of the v_j:

c_1j = ...= c_mj = 0, for all j, so the T_ij are linearly independent, and thus a basis for Hom(U,V).

now let E_ij be the mxn matrix with the ij-th entry 1, and everything else 0. the rest of the proof is simply:

ψ:Hom(U,V)→Mat_mxn(F) given by: ψ(T_ij) = E_ij is a linear isomorphism.

it suffices to prove that ψ is bijective, since our definition of ψ automatically makes it linear.

but {T_ij} has the same cardinality as {E_ij}, mn, so it is a bijection (two finite-dimensional vector spaces are isomorphic if and only if their dimensions are equal).

(if you want to prove the E_ij form a basis for Mat_mxn(F), knock yourself out...it's essentially the same proof as for the T_ij...that is, we're using the fact that:

as vector spaces, Mat_mxn(F) ≅ F^mn <--this is called "the vectorization of matrices" you just make one long column out of a matrix, by putting the 2nd column below the first, the 3rd below the 2nd, etc)