let V be an n-dimensional vector space over the field K and B be a basis for V.

Question

let V be an n-dimensional vector space over the field K and B be a basis for V. Let Bil(V x V, K) be the set of all bilinear maps on V x V to K.

(a) Prove that there is an isomorphism

                Bil(V x V, K)   maps to   Mat nxn (K)

(b) Let V* be the (dual) vector space of linear functionals on V, and B' be the dual basis with respect to B. Show that with respect to the bases B and B', there is an isomorphism

                Lin(V, V*) maps to Mat nxn (K)

(c) Conclude now that there is an isomorphism

                  Bil(V x V, K) maps to Lin(V, V*)

Explanation / Answer

(a)

Let B=(e1,...,en) be a basis of V

Let phi in Bil(VxV,K) then we know that by bilinearity of phi, it's determined by the values of

phi(ei,ej) for i=1..n,j=1..n.

So we can simply define the map PSI1 : Bil(V x V , K) -> Mat nxn (K ) such that :

PSI1( phi ) = the nxn matrix with elements phi(ei,ej) at position i,j for i=1..n,j=1..n

Clearly this is an homomorphism since :

PSI1(phi1+phi2) = matrix with elements (phi1+phi2)(ei,ej)

= matrix with element phi1(ei,ej) + matrix with element phi2(ei,ej)

= PSI1(phi1)+PSI1(phi2)

Now if we have PSI1(phi) = 0 then phi(ei,ej) = 0 for all i=1..n,j=1..n so phi=0 since it's determined completely by these values phi(ei,ej).

So PSI1 is one-to-one.

PSI1 is onto by construction since for any matrix A = (aij) in Mat nxn(K)

we can define phi by phi(ei,ej) = aij and we will have PSI1(phi) = A

So Bil(V x V, K) is isomorphic to Mat nxn (K)

(b) Let B=(e1,...,en) be a basis of V and B'=(e1*,...,en*) be the dual basis with respect to B.

A linear function phi : V to V* is completely define by the value phi(ei) for i=1..n

Now we can find ai1,...,ain such that for all i=1..n we have :

phi(ei) = sum(j=1..n) aij ej* ( since V* has basis B'=(e1*,...en*) and phi(ei) is in V* )

Now with this definition of aij we define PSI2 : Lin(V,V*) -> Mat nxn (K) by PSI(phi) = (aij)

Again this is trivially a homomorphism

It's one-to-one because aij=0 => phi(ei)=0 => phi=0.

It's onto by constructoin because for all A=(aij) we can define PSI2(phi)=(aij)

with phi defined as phi(ei)=sum(j=1..n) aij ej*

So Lin(V,V*) is isomorphic to Mat nxn (K)

(c) Just take PSI = PSI2^-1 o PSI1 (composition) and it's by theorem an isomorphism

from Bil(VxV, K) to Lin(V,V*) by composition of isomorphism

So Bil(VxV,K) is isomorphic to Lin(V,V*)

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