# linear operator proof

• Oct 10th 2009, 12:37 PM
guroten
linear operator proof
Let W be the space of all n X 1 column matrices over a field F. If A is an
n X n matrix over F, then A defines a linear operator L on W through left
multiplication: L(X) = AX. Prove that every linear operator on W is left multiplication
by some n X n matrix, i.e., is L for some A.

Now suppose V is an n-dimensional vector space over the field F, and let b be an ordered basis for V. For each a in V, define Ua = [a]b (a in b-coordinates). Prove that U is an isomorphism of V onto W. If T is a linear operator on V, then UTU^(-1) is a linear operator on W. Accordingly, UTU^(-1) is left multiplication by some n X n matrix A. What is A?

Any help on these two problems would be appreciated.
• Oct 11th 2009, 08:41 AM
HallsofIvy
Quote:

Originally Posted by guroten
Let W be the space of all n X 1 column matrices over a field F. If A is an
n X n matrix over F, then A defines a linear operator L on W through left
multiplication: L(X) = AX. Prove that every linear operator on W is left multiplication
by some n X n matrix, i.e., is L for some A.

Choose a basis for W, [tex]\{M_1, M_2, ..., M_n}. Taking the column matrices with exactly one "1" and the rest of the entries "0", so that $M_1= \begin{bmatrix}1 \\ 0 \\ 0\\...\\0\end{bmatrix}$, $M_2=\begin{bmatrix}0 \\ 1 \\ 0\\...\\0\end{bmatrix}$ will work nicely. For linear transformation L, write $L(M_1)$ as a linear combination of the the basis. That coeffientes in that linear combination will give you the first column of A. Continue with $L(M_2)$, etc. to find the other columns.

This is a general method. If L is a linear transformation from vector space U to vector space V, choose bases for U and V. Apply L to each basis vector of U in turn and write the resulting vector (in V) as a linear combination of the chosen basis for V. The coefficients will give the columns of the matrix representation of L.

Quote:

Now suppose V is an n-dimensional vector space over the field F, and let a
be an ordered basis for V. For each a in V, define Ua = [a]b. Prove that U is an
isomorphism of V onto W.
You've lost me here. You say "let a be an ordered basis" but then say "for each a in V". Did you mean "let b be an ordered basis"? Does "[a]b" mean "the coefficients when a is written as a linear combination in that basis"? (I was try to think of "b" as a vector!) "W" is still the column matrices, right? Remember the basis I suggested for W above? Do you see that U maps each vector of b into the corresponding member of that basis. And any linear transformation that maps basis vector to basis vectors is an isomorphism.

[quote]
Quote:

If T is a linear operator on V, then UTU^(-1) is a linear
operator on W. Accordingly, UTU^(-1) is left multiplication by some n X n matrix A.
What is A?
Basically, what I said before. A is the matrix representing T in this basis. Its columns are the coefficients in the linear combination of each $T(v_i)$ in the basis ${v_i\}$.

Quote:

Any help on these two problems would be appreciated.
• Oct 11th 2009, 12:36 PM
guroten
Sorry, I messed up the second part. It should read:

Now suppose V is an n-dimensional vector space over the field F, and let b
be an ordered basis for V. For each a in V, define Ua = [a]b (a in b-coordinates). Prove that U is an isomorphism of V onto W. If T is a linear operator on V, then UTU^(-1) is a linear operator on W. Accordingly, UTU^(-1) is left multiplication by some n X n matrix A. What is A?

I think that corrects it.