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.

Let$T:V \rightarrow V$ 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.