# Thread: Linear Algebra: Linear Transformations(1)

1. ## Linear Algebra: Linear Transformations(1)

Hello,

I am teaching myself Linear Algebra following P.R.Halmos' "Finite Dimensional Vector Spaces".

I am stuck at a couple of problems:
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Problem (1)

Prove that corresponding to every linear transformation A on a finite dimensional vector space V, there exists an invertible linear transformation P such that APA = A.(Or equivalently prove that PA is a projection)

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Problem (2)

If A,B,C are linear transformations on a finite dimensional vector space, then does (AB - BA)^2 commutes with C always?

What happens when the dimension of the vector space is 2?

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Problem (3)

From the basic definition of a determinant of a linear transformation(see below), prove that $\text{det }(A \otimes B) = \text{det }(A) ^m \text{det }(B) ^n$, where A and B are linear transformations on vector spaces of dimensions n and m respectively

(I know the proof of a particular case when A and B are matrices, using eigenvalues. But I would love to see a proof using the below definition).

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Definition of Determinant of a linear transformation:

If W is the space of all alternating n-linear forms on an n-dimensional vector space V,then we know that it is one-dimensional. Thus given a linear transformation A, we have a scalar $\delta$ such that for every w in W, $w(Ax_1,Ax_2,.....,Ax_n) = \delta w(x_1,x_2,....,x_n)$ $\forall x_1,x_2,...x_n$ vectors in V. Then the det(A) is defined to be the scalar $\delta$.

Thank you,
Srikanth

2. Originally Posted by Isomorphism
Hello,

I am teaching myself Linear Algebra following P.R.Halmos' "Finite Dimensional Vector Spaces".

I am stuck at a couple of problems:

Problem (1)

Prove that corresponding to every linear transformation A on a finite dimensional vector space V, there exists an invertible linear transformation P such that APA = A.(Or equivalently prove that PA is a projection)
the proof of this is in the book and it's quite easy! see Theorem 3, page 94.

Problem (2)

If A,B,C are linear transformations on a finite dimensional vector space, then does (AB - BA)^2 commutes with C always?
no. it's easier to work with matrices rather than transformations here. they're the same anyway: $A=e_{12}, \ B=e_{21}, \ C=e_{13}.$ (here $e_{ij}$ has 1 as its (i,j)-entry and 0 everywhere else.)

What happens when the dimension of the vector space is 2?
in this case, the claim is true: let $AB-BA=X.$ clearly $\text{tr}(X)=0.$ thus by Cayley Hamilton: $X^2 = \alpha I,$ for some scalar $\alpha.$ now it's obvious that $X^2C=CX^2,$ for all $C.$

3. in this case, the claim is true: let clearly thus by Cayley Hamilton: for some scalar now it's obvious that for all C.
Wow! This proof is fantastic .

As of now, AB and BA are linear transformations and he has not defined trace or proved Cayley-Hamilton theorem. So probably there is a elementary proof of this result.But I know both the results for matrices.

Nevertheless its a wonderful proof. Thank you NonCommAlg, you

If you have any idea on how to do the third one, can you tell me?

Thanks again,
Srikanth

4. here's a direct way for Problem 2, part 2: let $\{e_1,e_2 \}$ be a basis for our vector space. let $A(e_1)=ae_1 + be_2, \ A(e_2)=ce_1 + de_2, \ \ B(e_1)=a'e_1+b'e_2, \ B(e_2)=c'e_1 + d'e_2.$ see that:

$(AB-BA)(e_1)=re_2, \ \ (AB-BA)(e_2)=se_1,$ for some scalars $r,s.$ thus: $(AB-BA)^2(e_1)=rse_1, \ (AB-BA)^2(e_2)=rse_2.$ hence: $(AB-BA)^2=rsI$ and the result follows.