Linear algebra, basis, linear transformation and matrix representation

[fluidistic]

Homework Statement

In a given basis $\{ e_i \}$ of a vector space, a linear transformation and a given vector of this vector space are respectively determined by $\begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0\\ 0&0&5\\ \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 2 \\3 \end{pmatrix}$.
Find the matrix representations of the transformation and the vector in a new basis such that the old one is represented by $e_1 =\begin{pmatrix} 1 \\ 1 \\0 \end{pmatrix}$, $e_2=\begin{pmatrix} 1 \\ -1 \\0 \end{pmatrix}$ and $e_3=\begin{pmatrix} 0 \\ 0 \\1 \end{pmatrix}$.

Homework Equations

Not even sure.

The Attempt at a Solution

I suspect it has to do with eigenvalues since the problem in the assignments is right after an eigenvalue problem. I would not have thought of this otherwise.
It reminds me of 2 similar matrices, say A and B, then A=P^(-1)BP but nothing more than this.
I've found that the given matrix has 3 different eigenvalues, thus 3 eigenvectors (hence non 0 vectors) that are linearly independent (since the 3 eigenvalues are all different) and thus the matrix is similar to a triangular one with the eigenvalues 1, 3 and 5 on the main diagonal.
So now I know that there exist an invertible matrix P such that A=P^(-1)BP. But I'm really stuck at seeing how this can help me.
Any tip's welcome.

 

[micromass]

First you'll need to find the change-of-basis matrix. That is, a matrix A such that if x are coordinates in the old basis, then Ax are coordinates in the new basis.

You know already what $Ae_1,~Ae_2,~Ae_3$ are. So can you use this to find the matrix A?

 

[fluidistic]

First you'll need to find the change-of-basis matrix. That is, a matrix A such that if x are coordinates in the old basis, then Ax are coordinates in the new basis.

You know already what $Ae_1,~Ae_2,~Ae_3$ are. So can you use this to find the matrix A?

Thanks for your help. I'm confused, I didn't know I knew what $Ae_i$ are.
Are these the column vectors of the first matrix of my first post?

 

[micromass]

Thanks for your help. I'm confused, I didn't know I knew what $Ae_i$ are.
Are these the column vectors of the first matrix of my first post?

Oh, no, sorry. I was thinking of the $e_i$as the original basis.

So, if I rephrase it, you know what the image of (1,0,0), (0,1,0) and (0,0,1) under A are?

 

[fluidistic]

Oh, no, sorry. I was thinking of the $e_i$as the original basis.

So, if I rephrase it, you know what the image of (1,0,0), (0,1,0) and (0,0,1) under A are?

Ah ok. :)
Hmm no either. $\{ e_i \}$ isn't necessarily the canonical basis as far as I know...

 

[micromass]

The coordinates of any basis is always (1,0,0), (0,1,0), (0,0,1). Regardless of whether it is the canonical basis.

By definition, the coordinates of a vector v with respect to $\{e_1,e_2,e_3\}$ are $(\alpha_1,\alpha_2,\alpha_3)$ such that

$v=\alpha_1e_1+\alpha_2e_2+\alpha_3e_3$.

From this, we see that the coordinates of $e_1$ are always (1,0,0).

 

[I like Serena]

Hi fluidistic!

First let's define our symbols.
Let's call your linear transformation matrix A, and your vector v.

What you need is a matrix B that defines a basis transformation.
B is defined by the fact that for instance the unit vector (1,0,0) is transformed to your e₁. That is: B (1,0,0) = e₁.

Can you deduce what the matrix B is?

If you apply this B to your vector v, you'll find the represention of the v in the alternative basis.

 

[fluidistic]

Hi fluidistic!

First let's define our symbols.
Let's call your linear transformation matrix A, and your vector v.

What you need is a matrix B that defines a basis transformation.
B is defined by the fact that for instance the unit vector (1,0,0) is transformed to your e₁. That is: B (1,0,0) = e₁.

Can you deduce what the matrix B is?

If you apply this B to your vector v, you'll find the represention of the v in the alternative basis.

Hi!
Is $B= \begin{pmatrix} 1 & 1 & 0 \\ 1& -1 & 0\\ 0& 0 & 1 \end{pmatrix}$? It seems to work for me. Its columns vectors are the one given in the problem statement.

 

[I like Serena]

Hi!
Is $B= \begin{pmatrix} 1 & 1 & 0 \\ 1& -1 & 0\\ 0& 0 & 1 \end{pmatrix}$? It seems to work for me. Its columns vectors are the one given in the problem statement.

Yep, that's it!
It's simply the given unit vectors as columns of the matrix B.

Calculate Bv and you have your representation of v in the basis defined by B.

For your matrix A, you need B^-1AB.
That is, first you transform a vector to the alternate basis (wrt which A has been defined), then you apply A, and then you transform the result back to your original basis.

 

[fluidistic]

Yep, that's it!
It's simply the given unit vectors as columns of the matrix B.

Calculate Bv and you have your representation of v in the basis defined by B.

For your matrix A, you need B^-1AB.
That is, first you transform a vector to the alternate basis (wrt which A has been defined), then you apply A, and then you transform the result back to your original basis.

I reached $v'= \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}$. And A' is a diagonal matrix with the eigenvalues of A as entries... (3, 1 and 5).

 

[I like Serena]

That's it!

I'm surprised myself that A' is diagonal, but I see that it is.

In retrospect it is clear that it is, since the given basis consists of exactly the eigenvectors of A.
Note that A is diagonized by a base transition consisting of its eigenvectors.

I suspect this problem was intended to demonstrate this feature of eigenvectors.

 

[fluidistic]

That's it!

I'm surprised myself that A' is diagonal, but I see that it is.

In retrospect it is clear that it is, since the given basis consists of exactly the eigenvectors of A.
Note that A is diagonized by a base transition consisting of its eigenvectors.

I suspect this problem was intended to demonstrate this feature of eigenvectors.

Yes it is. In my first post I saw that A was diagonalizable.
So $A=P^{-1}A'P$. I didn't know that $P^{-1}=B^{-1}$.
I had calculated the eigenvalues of A, so that I think I had calculated A'. (Up to the order of the eigenvalue on the main diagonal, but I don't think it's relevant).
Is there a way I could have found P=B which is faster than the one I've done?
In other words, knowing A' from start, is there a quick way to solve the problem?

 

[I like Serena]

Well, this problem is actually not about eigenvectors.
It's about a base transformation.
You can't assume that the new basis consists of the eigenvectors!

 

[fluidistic]

Well, this problem is actually not about eigenvectors.
It's about a base transformation.
You can't assume that the new basis consists of the eigenvectors!

Ok. I think I understand.
Thanks guys!