A question about linear algebra (change of basis of a linear transformation)

[Artusartos]

Homework Statement

Let A \in M_n(F) and v \in F^n.

Let v, Av, A^2v, ... , A^{k-1}v be a basis, B, of V.

LetT:V \rightarrow V be induced by multiplication by A:T(w) = Aw for w in V. Find [T]_B, the matrix of T with respect to B.


Thanks in advance

Homework Equations

[T(w)]_B = [Aw]_B = C^{-1}Aw

The Attempt at a Solution

Can anybody give me a hint please? I'm trying to do this for an hour but I'm not sure how.

From here: http://www.khanacademy.org/math/linear-algebra/v/lin-alg--transformation-matrix-with-respect-to-a-basis

I learned that [T(w)]_B = [Aw]_B = C^{-1}Aw, where C= [v| Av| A^2v| ... | A^{k-1}v]. But now I don't know what the inverse of C is?

Thanks in advance

 

[jbunniii]

Well, if you write b_0 = v, b_1 = Av, b_2 = A^2v, ..., b_{n-1} = A^{n-1}v, then you have T(b_k) = b_{k+1} for 0 \leq k < n-1. What does this tell you about the first n-1 columns of [T]_B?

 

[Artusartos]

Well, if you write b_0 = v, b_1 = Av, b_2 = A^2v, ..., b_{n-1} = A^{n-1}v, then you have T(b_k) = b_{k+1} for 0 \leq k < n-1. What does this tell you about the first n-1 columns of [T]_B?

They are the same elements as the basis, except for the last one, since it is A^kv...

 

[jbunniii]

They are the same elements as the basis, except for the last one, since it is A^kv...

So what are the first n-1 columns? You should be able to write them using actual numbers.

 

[Artusartos]

So what are the first n-1 columns? You should be able to write them using actual numbers.

The first n-1 columns are: Av, A^2v, ..., A^{k-1}v. Since each element A^pv is transformed into A^{p+1}v, right?

 

[jbunniii]

The first n-1 columns are: Av, A^2v, ..., A^{k-1}v. Since each element A^pv is transformed into A^{p+1}v, right?

Let's take a step back. What is the matrix representation of b_0 = v in terms of the basis B? Your answer should be a column vector containing 1's and 0's.

 

[Artusartos]

Let's take a step back. What is the matrix representation of b_0 = v in terms of the basis B? Your answer should be a column vector containing 1's and 0's.

I'm not sure I understand why it can be written with numbers. The question tells us what B is, not with numbers but with A's and v's. Do you mean something like this:

[v]_B = 1.v + 0.Av + ... + 0.A^{k-1}v?

 

[jbunniii]

I'm not sure I understand why it can be written with numbers. The question tells us what B is, not with numbers but with A's and v's. Do you mean something like this:

[v]_B = 1.v + 0.Av + ... + 0.A^{k-1}v?

That's not what v_B is. It is true that
v = 1\cdot v + 0 \cdot Av + \ldots + 0 \cdot A^{k-1}v
And since B is a basis, this is the unique way of writing v as a linear combination of the elements of B. So what is v_B? It is nothing other than the array of coefficients in that linear combination: v_B = [\begin{array}{cccccc}1 & 0 & 0 & 0 & \ldots &0\end{array}]^T where by convention this is understood to be a column vector.

 

[Artusartos]

That's not what v_B is. It is true that
v = 1\cdot v + 0 \cdot Av + \ldots + 0 \cdot A^{k-1}v
And since B is a basis, this is the unique way of writing v as a linear combination of the elements of B. So what is v_B? It is nothing other than the array of coefficients in that linear combination: v_B = [\begin{array}{cccccc}1 & 0 & 0 & 0 & \ldots &0\end{array}]^T where by convention this is understood to be a column vector.

So [T]_B is just the kxk identity matrix?

 

[jbunniii]

So [T]_B is just the kxk identity matrix?

No, because T doesn't map each element of B to itself.
In fact, T maps b_0 to b_1. Therefore the first column of [T]_B should be the [b_1]_B = [Av]_B, which is what?

 

[Artusartos]

No, because T doesn't map each element of B to itself.
In fact, T maps b_0 to b_1. Therefore the first column of [T]_B should be the [b_1]_B = [Av]_B, which is what?

But didn't you say that

v_B = [\begin{array}{cccccc}1 & 0 & 0 & 0 & \ldots &0\end{array}]^T?

So...

[Av]_B = [\begin{array}{cccccc} 0 & 1 & 0 & 0 & \ldots &0\end{array}]^T
[A^2v]_B = [\begin{array}{cccccc} 0 & 0 & 1 & 0 & \ldots &0\end{array}]^T
...
[A^{k-1}v]_B = [\begin{array}{cccccc} 0 & 0 & 0 & 0 & \ldots &1\end{array}]^T

So aren't these the columns of [T]_B? So why isn't it the identity matrix?

 

[jbunniii]

But didn't you say that

v_B = [\begin{array}{cccccc}1 & 0 & 0 & 0 & \ldots &0\end{array}]^T?

So...

[Av]_B = [\begin{array}{cccccc} 0 & 1 & 0 & 0 & \ldots &0\end{array}]^T
[A^2v]_B = [\begin{array}{cccccc} 0 & 0 & 1 & 0 & \ldots &0\end{array}]^T
...
[A^{k-1}v]_B = [\begin{array}{cccccc} 0 & 0 & 0 & 0 & \ldots &1\end{array}]^T

So aren't these the columns of [T]_B? So why isn't it the identity matrix?

Well, the first column of [T]_B is [Tv]_B = [Av]_B = [\begin{array}{cccccc} 0 & 1 & 0 & 0 & \ldots &0\end{array}]^T so already it can't be the identity matrix.

 

[Artusartos]

Well, the first column of [T]_B is [Tv]_B = [Av]_B = [\begin{array}{cccccc} 0 & 1 & 0 & 0 & \ldots &0\end{array}]^T so already it can't be the identity matrix.

Oh, I get...thanks. But for the last one, we won't be able to write it with numbers, right? A^kv must be some linear combination of the basis, but we can't know exactly what...right? So we just write...

[T]_B = [\begin{array}{cccccc} c_1 & c_2 & \ldots &c_k\end{array}]^T[/itex]

 

[jbunniii]

Oh, I get...thanks. But for the last one, we won't be able to write it with numbers, right? A^kv must be some linear combination of the basis, but we can't know exactly what...right? So we just write...

[T]_B = [\begin{array}{cccccc} c_1 & c_2 & \ldots &c_k\end{array}]^T[/itex]

Offhand I'm not sure about the last column. I'm not sure if there is enough information given to know T(A^{k-1}v) in terms of B. Maybe think about the fact that v, Av, A^2v, ..., A^{k-1}v form a basis. This wouldn't necessarily be true for arbitrary A and v, so this gives you some information about A and v. I'll think about this some more and let you know if I have an idea.

 

[jbunniii]

Yeah, I'm pretty sure the last column can be anything. It doesn't even have to be linearly independent of the other columns as there's nothing in the problem that requires T to be invertible.

 

[Artusartos]

Yeah, I'm pretty sure the last column can be anything. It doesn't even have to be linearly independent of the other columns as there's nothing in the problem that requires T to be invertible.

Alright, thanks a lot.