Singular value decomposition

Impo

New member
Hi

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

What is the singular value decomposition of $$A$$? I'm not allowed to calculate $$A$$.

Klaas van Aarsen

MHB Seeker
Staff member
Hi

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

What is the singular value decomposition of $$A$$? I'm not allowed to calculate $$A$$.

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

New member
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 ...

Klaas van Aarsen

MHB Seeker
Staff member
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

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

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 $$Ax$$?

Klaas van Aarsen

MHB Seeker
Staff member
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 $$Ax$$?
Right.
I intended the vectors on the unit sphere that are mapped to the semi-axes of the ellipsoid.

Impo

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

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

Klaas van Aarsen

MHB Seeker
Staff member
If $$\{e_1,e_2,e_3\}$$ is the canonical basis for $$\mathbb{R}^3$$ then the only thing I can say is
$$A(Be_1)=(2,0,0)^{T}$$
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

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

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 $6$.

Last edited:

Staff member
Yep!
That's it.