How Does Singular Value Decomposition Transform a Unit Sphere into an Ellipsoid?

[Impo]

Hi

Suppose that [tex]A \in \mathbb{R}^{3 \times 3}[/tex] who maps the unit sphere in [tex]\mathbb{R}^3[/tex] to an ellips with the following semi-axes;
[tex]x = \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}},0\right)^{T} \mapsto Ax = (2,0,0)^{T}[/tex]
[tex]x=\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},0\right)^{T} \mapsto Ax = (0,-3,0)^{T}[/tex]
[tex]x=(0,0,1)^{T} \mapsto Ax = (0,0,6)^{T}[/tex]

What is the singular value decomposition of [tex]A[/tex]? I'm not allowed to calculate [tex]A[/tex].

Thanks in advance!

 

[I like Serena]

Hi

Suppose that [tex]A \in \mathbb{R}^{3 \times 3}[/tex] who maps the unit sphere in [tex]\mathbb{R}^3[/tex] to an ellips with the following semi-axes;
[tex]x = \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}},0\right)^{T} \mapsto Ax = (2,0,0)^{T}[/tex]
[tex]x=\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},0\right)^{T} \mapsto Ax = (0,-3,0)^{T}[/tex]
[tex]x=(0,0,1)^{T} \mapsto Ax = (0,0,6)^{T}[/tex]

What is the singular value decomposition of [tex]A[/tex]? I'm not allowed to calculate [tex]A[/tex].

Thanks in advance!

Suppose you write each of your semi-axes as the columns of a matrix we will call $B$.
Then what is $AB$ equal to?

Can you rewrite that in the form $A=U\Sigma V^*$ aka a singular value decomposition?

 

[Impo]

Suppose you write each of your semi-axes as the columns of a matrix we will call $B$.

Honestly, I don't understand what you mean ...

 

[I like Serena]

Honestly, I don't understand what you mean ...

$$B=\begin{bmatrix}
-\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$

Now what is $AB$?

And can you rewrite it to a form where $A$ is equal to an orthogonal matrix times a diagonal matrix times another orthogonal matrix?

 

[Impo]

$$B=\begin{bmatrix}
-\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$

Now what is $AB$?

And can you rewrite it to a form where $A$ is equal to an orthogonal matrix times a diagonal matrix times another orthogonal matrix?

I have no idea, I don't know anything about $A$. But I thought that semi-axes of the ellips were the vectors [tex]Ax[/tex]?

 

[I like Serena]

I have no idea, I don't know anything about $A$.

Yes you do.
You know the images of each of the column vectors in B.
So you should write that down using matrix notation.

But I thought that semi-axes of the ellips were the vectors [tex]Ax[/tex]?

Right.
I intended the vectors on the unit sphere that are mapped to the semi-axes of the ellipsoid.

 

[Impo]

If [tex]\{e_1,e_2,e_3\}[/tex] is the canonical basis for [tex]\mathbb{R}^3[/tex] then the only thing I can say is
[tex]A(Be_1)=(2,0,0)^{T}[/tex]
and so on

But I don't see how this helps me ...
I think I don't quite understand the purpose of this.

 

[I like Serena]

If [tex]\{e_1,e_2,e_3\}[/tex] is the canonical basis for [tex]\mathbb{R}^3[/tex] then the only thing I can say is
[tex]A(Be_1)=(2,0,0)^{T}[/tex]
and so on

Yes... and you can combine those to write down $AB$...

But I don't see how this helps me ...
I think I don't quite understand the purpose of this.

I guess we'll get to that once we get the matrix notation down, since that seems to be the main obstacle.

 

[Impo]

Yes... and you can combine those to write down $AB$...
I guess we'll get to that once we get the matrix notation down, since that seems to be the main obstacle.

[tex]AB = \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -3 & 0 \\ 0& 0&6 \end{array} \right)[/tex]
$\Leftrightarrow A = \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -3 & 0 \\ 0& 0&6 \end{array} \right)B^{-1}$
$\Leftrightarrow A = \mbox{I}_3 \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -3 & 0 \\ 0& 0&6 \end{array} \right)B^{-1}$

In fact, if I prove that $B$ is an orthogonal matrix, that is, the columns are orthogonal vectors then I think this is a possible singular value decomposition of $A$ with singular values: $2, -3$ and $6$.

 

[I like Serena]

Yep!
That's it.