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

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Homework Statement



Let [itex]A \in M_n(F)[/itex] and [itex]v \in F^n[/itex].

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

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


Thanks in advance



Homework Equations




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

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/lin...transformation-matrix-with-respect-to-a-basis

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

Thanks in advance

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
jbunniii
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Well, if you write [itex]b_0 = v[/itex], [itex]b_1 = Av[/itex], [itex]b_2 = A^2v[/itex], ..., [itex]b_{n-1} = A^{n-1}v[/itex], then you have [itex]T(b_k) = b_{k+1}[/itex] for [itex]0 \leq k < n-1[/itex]. What does this tell you about the first [itex]n-1[/itex] columns of [itex][T]_B[/itex]?
 
  • #3
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Well, if you write [itex]b_0 = v[/itex], [itex]b_1 = Av[/itex], [itex]b_2 = A^2v[/itex], ..., [itex]b_{n-1} = A^{n-1}v[/itex], then you have [itex]T(b_k) = b_{k+1}[/itex] for [itex]0 \leq k < n-1[/itex]. What does this tell you about the first [itex]n-1[/itex] columns of [itex][T]_B[/itex]?
They are the same elements as the basis, except for the last one, since it is [itex]A^kv[/itex]...
 
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  • #4
jbunniii
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They are the same elements as the basis, except for the last one, since it is [itex]A^kv[/itex]...
So what are the first [itex]n-1[/itex] columns? You should be able to write them using actual numbers.
 
  • #5
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So what are the first [itex]n-1[/itex] columns? You should be able to write them using actual numbers.
The first n-1 columns are: [itex]Av, A^2v, ..., A^{k-1}v[/itex]. Since each element [itex]A^pv[/itex] is transformed into [itex]A^{p+1}v[/itex], right?
 
  • #6
jbunniii
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The first n-1 columns are: [itex]Av, A^2v, ..., A^{k-1}v[/itex]. Since each element [itex]A^pv[/itex] is transformed into [itex]A^{p+1}v[/itex], right?
Let's take a step back. What is the matrix representation of [itex]b_0 = v[/itex] in terms of the basis [itex]B[/itex]? Your answer should be a column vector containing 1's and 0's.
 
  • #7
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Let's take a step back. What is the matrix representation of [itex]b_0 = v[/itex] in terms of the basis [itex]B[/itex]? 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 [itex]B[/itex] is, not with numbers but with A's and v's. Do you mean something like this:

[itex][v]_B[/itex] = [itex] 1.v + 0.Av + ... + 0.A^{k-1}v [/itex]?
 
  • #8
jbunniii
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I'm not sure I understand why it can be written with numbers. The question tells us what [itex]B[/itex] is, not with numbers but with A's and v's. Do you mean something like this:

[itex][v]_B[/itex] = [itex] 1.v + 0.Av + ... + 0.A^{k-1}v [/itex]?
That's not what [itex]v_B[/itex] is. It is true that
[tex]v = 1\cdot v + 0 \cdot Av + \ldots + 0 \cdot A^{k-1}v[/tex]
And since B is a basis, this is the unique way of writing [itex]v[/itex] as a linear combination of the elements of B. So what is [itex]v_B[/itex]? It is nothing other than the array of coefficients in that linear combination: [itex]v_B = [\begin{array}{cccccc}1 & 0 & 0 & 0 & \ldots &0\end{array}]^T[/itex] where by convention this is understood to be a column vector.
 
  • #9
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That's not what [itex]v_B[/itex] is. It is true that
[tex]v = 1\cdot v + 0 \cdot Av + \ldots + 0 \cdot A^{k-1}v[/tex]
And since B is a basis, this is the unique way of writing [itex]v[/itex] as a linear combination of the elements of B. So what is [itex]v_B[/itex]? It is nothing other than the array of coefficients in that linear combination: [itex]v_B = [\begin{array}{cccccc}1 & 0 & 0 & 0 & \ldots &0\end{array}]^T[/itex] where by convention this is understood to be a column vector.
So [itex][T]_B[/itex] is just the kxk identity matrix?
 
  • #10
jbunniii
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So [itex][T]_B[/itex] is just the kxk identity matrix?
No, because T doesn't map each element of B to itself.
In fact, T maps [itex]b_0[/itex] to [itex]b_1[/itex]. Therefore the first column of [itex][T]_B[/itex] should be the [itex][b_1]_B = [Av]_B[/itex], which is what?
 
  • #11
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No, because T doesn't map each element of B to itself.
In fact, T maps [itex]b_0[/itex] to [itex]b_1[/itex]. Therefore the first column of [itex][T]_B[/itex] should be the [itex][b_1]_B = [Av]_B[/itex], which is what?
But didn't you say that

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

So...

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

So aren't these the columns of [itex][T]_B[/itex]? So why isn't it the identity matrix?
 
  • #12
jbunniii
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But didn't you say that

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

So...

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

So aren't these the columns of [itex][T]_B[/itex]? So why isn't it the identity matrix?
Well, the first column of [itex][T]_B[/itex] is [itex][Tv]_B = [Av]_B = [\begin{array}{cccccc} 0 & 1 & 0 & 0 & \ldots &0\end{array}]^T[/itex] so already it can't be the identity matrix.
 
  • #13
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Well, the first column of [itex][T]_B[/itex] is [itex][Tv]_B = [Av]_B = [\begin{array}{cccccc} 0 & 1 & 0 & 0 & \ldots &0\end{array}]^T[/itex] 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? [itex]A^kv[/itex] must be some linear combination of the basis, but we can't know exactly what...right? So we just write...

[itex][T]_B[/itex] = [\begin{array}{cccccc} c_1 & c_2 & \ldots &c_k\end{array}]^T[/itex]
 
  • #14
jbunniii
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Oh, I get...thanks. But for the last one, we won't be able to write it with numbers, right? [itex]A^kv[/itex] must be some linear combination of the basis, but we can't know exactly what...right? So we just write...

[itex][T]_B[/itex] = [\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 [itex]T(A^{k-1}v)[/itex] in terms of B. Maybe think about the fact that [itex]v[/itex], [itex]Av[/itex], [itex]A^2v[/itex], ..., [itex]A^{k-1}v[/itex] 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.
 
  • #15
jbunniii
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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.
 
  • #16
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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. :biggrin:
 

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