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

1. Sep 30, 2012

### Artusartos

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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.

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?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 30, 2012

### 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 $[T]_B$?

3. Oct 1, 2012

### Artusartos

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

Last edited: Oct 1, 2012
4. Oct 1, 2012

### jbunniii

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

5. Oct 1, 2012

### Artusartos

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?

6. Oct 1, 2012

### jbunniii

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.

7. Oct 1, 2012

### Artusartos

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:

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

8. Oct 1, 2012

### jbunniii

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.

9. Oct 1, 2012

### Artusartos

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

10. Oct 1, 2012

### jbunniii

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?

11. Oct 1, 2012

### Artusartos

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?

12. Oct 1, 2012

### jbunniii

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.

13. Oct 1, 2012

### Artusartos

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]

14. Oct 1, 2012

### jbunniii

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.

15. Oct 1, 2012

### 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.

16. Oct 1, 2012

### Artusartos

Alright, thanks a lot.