Proving Linear Operators and Matrix Similarity

[hola]

1. If I: W-->W is the identity linear operator on W defined by I(w) = w for w in W, prove that the matrix of I repect with to any ordered basis T for W is a nXn I matrix, where dim W= n

2. Let L: W-->W be a linear operator defined by L(w) = bw, where b is a constant. Prove that the representation of L with respect to any ordered basis for W is a scalar matrix.

3. Let X,Y, Z be sqaure matrices. Show that: (a) X is similar to Y. (b) If X is similar to Y then Y is similar to X. (c) If X is similar to Y and Y is similar to Z, then X is similar to Z.

 

[HackaB]

1. If I: W-->W is the identity linear operator on W defined by I(w) = w for w in W, prove that the matrix of I repect with to any ordered basis T for W is a nXn I matrix, where dim W= n

Let E_{ij} be the elements of the matrix of the identity operator in some ordered basis of W, with basis vectors \vec{e}_1, \vec{e}_2, ... , \vec{e}_n. If w_j are the coordinates of any vector w in that basis, then

w_i^\prime = \sum_j E_{ij} w_j

By definition, the identity operator transforms the vector w back into itself, so that w_i^\prime = w_i. Then using the elements \delta_{ij} (kronecker delta) of the identity matrix, we have

w_i = \sum_j \delta_{ij} w_j = w_i^\prime = \sum_j E_{ij} w_j

or, after subtracting

\sum_j (E_{ij} - \delta_{ij}) w_j = 0 for each i.

Since the w_j's are arbitrary, we must have that E_{ij} = \delta_{ij} for all i and j.

edit: by the way, in the step where I set w_i^\prime = w_i for all i, I have assumed that the coordinates of a given vector w in a particular basis are unique. This is easy to prove using the fact that the elements of the basis are linearly independent, by definition.

 

[thepatientmental]

1. Proof:
Let T = {v1, v2, ..., vn} be an ordered basis for W, where dim W = n. Then, for any w in W, we can write w as a linear combination of the basis vectors:
w = a1v1 + a2v2 + ... + anvn

Applying the identity operator I to w, we get:
I(w) = a1I(v1) + a2I(v2) + ... + anI(vn)
= a1v1 + a2v2 + ... + anvn
= w

This shows that I preserves the elements of W, and hence, is a linear operator on W. Now, let A be the matrix representation of I with respect to the basis T. Then, A is an nXn matrix, where the (i,j) entry of A is the coefficient of vi in the linear combination of I(vj). Since I(vj) = vj, the (i,j) entry of A is 1 if i = j, and 0 otherwise. Therefore, A is an nXn identity matrix, which proves the statement.

2. Proof:
Let T = {v1, v2, ..., vn} be an ordered basis for W, where dim W = n. Then, for any w in W, we can write w as a linear combination of the basis vectors:
w = a1v1 + a2v2 + ... + anvn

Applying the linear operator L to w, we get:
L(w) = a1L(v1) + a2L(v2) + ... + anL(vn)
= a1(bv1) + a2(bv2) + ... + an(bvn)
= b(a1v1 + a2v2 + ... + anvn)
= bw

This shows that L multiplies each vector in W by the constant b, and hence, is a scalar operator. Now, let A be the matrix representation of L with respect to the basis T. Then, A is an nXn matrix, where the (i,j) entry of A is the coefficient of vi in the linear combination of L(vj). Since L(vj) = bvj, the (i,j) entry of A is b if i = j, and 0 otherwise. Therefore, A is an nXn diagonal matrix with