Signal Processing/Image Editing

As an image is a type of 2-D signal; instead of just time-amplitude pairs that correspond to a voice transmission, consider "time in the X domain"-"time in the Y domain"-amplitude pairs. That is, an X coordinate, a Y coordinate and an amplitude. This will give you a monochromatic image. As such, signal processing tools can be used in editing images.

Singular Value Decomposition

The Singular Value Decomposition is a matrix decomposition (another way to say factorization).

$M=U\Sigma V^{*}$

where

• U is an m×m unitary matrix over.
• Σ is an m×n diagonal matrix with nonnegative real numbers s_{n,n} on the diagonal where

$s_{1,1}>s_{2,2}>\ldots s_{n,n}$

• V*, an n×n unitary matrix. V* is the conjugate transpose (take the complex conjugate of all entries, and then perform a transpose operation on the matrix) of V.

The diagonal entries Σ_i,i of Σ are known as the singular values of M.

Image Compression

The nature of the singular values Σ is such that for a certain k,

$s_{1,1}>s_{2,2}>>s_{k,k}>\ldots s_{n,n}$

In order to transmit a 10x10 monochromatic image with 2 values ("on" or "off") it would require a matrix that has 10x10 = 100 entries. Consider the following image.

This image can be represented by the following matrix.

$M={\begin{bmatrix}1&1&1&1&1&1&1&1&1&1\\1&1&1&1&1&1&1&1&1&1\\1&1&0&0&0&0&0&0&1&1\\1&1&0&0&0&0&0&0&1&1\\1&1&0&0&1&1&0&0&1&1\\1&1&0&0&1&1&0&0&1&1\\1&1&0&0&0&0&0&0&1&1\\1&1&0&0&0&0&0&0&1&1\\1&1&1&1&1&1&1&1&1&1\\1&1&1&1&1&1&1&1&1&1\end{bmatrix}}$

The singular value decomposition of which is

$M=U\Sigma V^{*},$

Where

$\Sigma ={\begin{bmatrix}7.557&0&0&0&0&0&0&0&0&0\\0&2.979&0&0&0&0&0&0&0&0\\0&0&1.422&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\end{bmatrix}}$

Using the matrix

$\Sigma _{sm}={\begin{bmatrix}7.557&0&0\\0&2.979&0\\0&0&1.422\end{bmatrix}}$

And corresponding "truncated" versions of U and V* (Use only the first three columns of U and the first three columns of V*), we can find that

$M_{sm}={\begin{bmatrix}0.999&0.999&0.999&0.999&1.000&1.000&0.999&0.999&0.999&0.999\\0.999&0.999&0.999&0.999&1.000&1.000&0.999&0.999&0.999&0.999\\1.000&1.000&-0.001&-0.001&-0.001&-0.001&-0.001&-0.001&1.000&1.000\\1.000&1.000&-0.001&-0.001&-0.001&-0.001&-0.001&-0.001&1.000&1.000\\1.000&1.000&-0.001&-0.001&1.001&1.001&-0.001&-0.001&1.000&1.000\\1.000&1.000&-0.001&-0.001&1.001&1.001&-0.001&-0.001&1.000&1.000\\1.000&1.000&-0.001&-0.001&-0.001&-0.001&-0.001&-0.001&1.000&1.000\\1.000&1.000&-0.001&-0.001&-0.001&-0.001&-0.001&-0.001&1.000&1.000\\0.999&0.999&0.999&0.999&1.000&1.000&0.999&0.999&0.999&0.999\\0.999&0.999&0.999&0.999&1.000&1.000&0.999&0.999&0.999&0.999\end{bmatrix}}$

A cursory examination of the previous matrix will show that M_{sm} is approximately equal to M. Note that the truncated version of U and V* use 3*10 numbers each. The total number of values needed for this type of storage is 2*30+3 = 63. Which means data usage is reduced by nearly half.