(Solved): Linear Algebra 2 2 IMAGE COMPresSiON In Lab \#4 we examined grayscale images, which can be represen ...

Linear Algebra 2

2 IMAGE COMPresSiON In Lab \#4 we examined grayscale images, which can be represented by matrices whose entries correspond to the shades of gray of the corresponding pixels. Recall the definition of the Space Savings from lab. It is: $$\text{ Space Savings }=1?\frac{\text{ count of numbers needed to represent compressed data }}{\text{ count of numbers in original data }}.$$ If the original image corresponds to a $$n\times m$$ matrix, then the denominator above is $$n?m$$. Borrowing notation from the lab, if the image matrix $$M$$ is approximated by $$M_k=U_k{?}_k{V}_k^T$$, then the set of numbers needed to represent the compressed image consists of the first $$k$$ columns of $$U$$ and $$V$$, as well as the $$k$$ largest singular values, a total of $$k(n+m+1)$$ numbers. This yields $$\text{ Space Savings }=1?\frac{k(n+m+1)}{nm}\text{. }$$ When $$k$$ is small relative to $$m$$ and $$n$$, the space savings are high and the amount of data used to approximate our image $$M$$ is small. When dealing with color images, the question of how they should be encoded as matrices is a little subtle. The purpose of this long-winded question is to figure out which method is better from the perspective of low-rank approximations. 3. To represent color, we need three numerical values: the intensities of red, green, and blue. Thus a $$n\times m$$ image can be thought of in two different ways: - three $$n\times m$$ matrices, each encoding an intensity for one of the colors, - one $$3n\times m$$ matrix, made by stacking the above three matrices above on top of each other. The methods do not seem fundamentally different; the difference in the two is only semantic. However, to perform a rank $$k$$ approximation, one can approximate each of the three $$n\times m$$ matrices individually, or just the $$3n\times m$$ matrix by itself. Supposing (and this is a "big if") the rank $$k$$ approximations for both methods yield images of the same quality, which compression method is better?