Matrix compression: the singular value decomposition


Posted by Diego Assencio on 2014.02.09 under Mathematics (Linear algebra)

I expect most readers of this blog to be familiar with the concept of lossy data compression. For instance, whenever you rip a song from a CD to a lossy format such as MP3 or Ogg Vorbis, you are effectively discarding some data from the song to make it smaller while still preserving most of the audio. A good compression technique yields a song which sounds almost like the original one.

Matrices can sometimes also be compressed in the sense above. For instance, given a matrix $A$, it is sometimes possible to cleverly compute another matrix $\tilde{A} \approx A$ such that $\tilde{A}$ can be represented in a way which requires much less data storage than the original matrix $A$. This approximation can be obtained from a very powerful tool in linear algebra: the singular value decomposition (SVD). This post will not present techniques for computing SVDs, but merely discuss this tool in the context of matrix compression.

The SVD of an $m \times n$ real matrix $A$ is the factorization $A = U\Sigma V^T$, where $U$ is an $m \times m$ real orthogonal matrix, $\Sigma$ is an $m \times n$ diagonal matrix with real nonnegative values along its diagonal (called the singular values of $A$) and $V$ is an $n \times n$ real orthogonal matrix. Every matrix has an SVD (for a proof of this fact, see [1], theorem 3.2).

To illustrate how one can use the SVD to approximate a matrix, I will use octave. On Ubuntu/Debian, you can install octave by opening a terminal and running the following command:

sudo apt-get install octave

Now start octave. I have generated an example matrix which is shown below (in what follows, all user input is highlighted):

octave:1> A
A =

   1.02650   0.92840   0.54947   0.98317   0.71226   0.55847
   0.92889   0.89021   0.49605   0.93776   0.62066   0.52473
   0.56184   0.49148   0.80378   0.68346   1.02731   0.64579
   0.98074   0.93973   0.69170   1.03432   0.87043   0.66371
   0.69890   0.62694   1.02294   0.87822   1.29713   0.82905
   0.56636   0.51884   0.65096   0.66109   0.82531   0.55098

You can generate the matrix above by typing 'A=[' (without the quotes), then copying and pasting the entries from $A$ above and closing it with another square bracket ']'.

Octave provides a method for computing the SVD of $A$:

octave:2> [U,Sigma,V] = svd(A)
U =

  -0.418967  -0.449082   0.754882   0.062414  -0.211638   0.065260
  -0.386832  -0.456685  -0.393449   0.517538   0.420908  -0.204913
  -0.369990   0.429218   0.277696  -0.322484   0.605310  -0.362449
  -0.455375  -0.257776  -0.427049  -0.676528  -0.289362  -0.048923
  -0.469840   0.546519  -0.099679   0.407065  -0.521257  -0.182266
  -0.331389   0.201011  -0.076997   0.029526   0.237081   0.887000

Sigma =

Diagonal Matrix

   4.6780e+00            0            0            0            0            0
            0   8.9303e-01            0            0            0            0
            0            0   4.5869e-02            0            0            0
            0            0            0   7.2919e-03            0            0
            0            0            0            0   1.5466e-16            0
            0            0            0            0            0   4.5360e-17

V =

  -0.418967  -0.449082   0.726833   0.183844  -0.240614   0.053029
  -0.386832  -0.456685  -0.363745  -0.694211  -0.126874  -0.107069
  -0.369990   0.429218  -0.101603  -0.070784  -0.356422   0.732468
  -0.455375  -0.257776  -0.373669   0.487768   0.559525   0.188602
  -0.469840   0.546519   0.309430  -0.289389   0.439192  -0.328915
  -0.331389   0.201011  -0.306111   0.396987  -0.541307  -0.552684

As the output above shows, the singular values of $A$ (the highlighted diagonal entries of $\Sigma$) are placed in decreasing order along the diagonal of $\Sigma$. Compared to the first two singular values ($\Sigma_{11}$ and $\Sigma_{22}$), the other four ($\Sigma_{33}$, $\Sigma_{44}$, $\Sigma_{55}$ and $\Sigma_{66}$) are relatively small. Being so small, and since $A = U\Sigma V^T$, the effect of discarding them should still imply $A \approx U\Sigma V^T$ for this modified $\Sigma$. Let's discard them and see what happens:

octave:3> Sigma(3,3) = Sigma(4,4) = Sigma(5,5) = Sigma(6,6) = 0
Sigma =

Diagonal Matrix

   4.67800         0         0         0         0         0
         0   0.89303         0         0         0         0
         0         0   0.00000         0         0         0
         0         0         0   0.00000         0         0
         0         0         0         0   0.00000         0
         0         0         0         0         0   0.00000

Discarding $\Sigma_{33}$ means the third column of $U$ and the third row of $V^T$ (the third column of $V$) have no effect on $U\Sigma V^T$ since they contain the only entries of $U$ and $V$ respectively which are multiplied by $\Sigma_{33}$. We can then discard these entries and also discard the third row and the third column of $\Sigma$ altogether. Similarly, we can discard the fourth, fifth and sixth columns of $U$ and $V$ and also the fourth, fifth and sixth rows and columns of $\Sigma$:

octave:4> U = U(1:6,1:2)
U =

  -0.41897  -0.44908
  -0.38683  -0.45668
  -0.36999   0.42922
  -0.45538  -0.25778
  -0.46984   0.54652
  -0.33139   0.20101

octave:5> V = V(1:6,1:2)
V =

  -0.41897  -0.44908
  -0.38683  -0.45669
  -0.36999   0.42922
  -0.45537  -0.25778
  -0.46984   0.54652
  -0.33139   0.20101

octave:6> Sigma = Sigma(1:2,1:2)
Sigma =

Diagonal Matrix

   4.67800         0
         0   0.89303

Now let's compute $U\Sigma V^T$:

octave:7> A_tilde = U*Sigma*V'
A_tilde =

   1.00125   0.94131   0.55302   0.99588   0.70167   0.56888
   0.94131   0.88626   0.49448   0.92918   0.62733   0.51770
   0.55302   0.49448   0.80491   0.68936   1.02269   0.65062
   0.99588   0.92918   0.68936   1.02940   0.87507   0.65967
   0.70168   0.62733   1.02269   0.87507   1.29940   0.82647
   0.56888   0.51770   0.65062   0.65967   0.82647   0.54982

The values of $\tilde{A}$ are very close to the values of the original matrix $A$. Let's then compute the relative error for each entry of $\tilde{A}$ (below the './' operation computes the division of each entry of $(A - \tilde{A})$ by each corresponding entry of $A$):

octave:8> (A - A_tilde) ./ A
ans =

   2.4599e-02  -1.3907e-02  -6.4608e-03  -1.2934e-02   1.4859e-02  -1.8656e-02
  -1.3375e-02   4.4309e-03   3.1586e-03   9.1550e-03  -1.0756e-02   1.3384e-02
   1.5708e-02  -6.1061e-03  -1.4028e-03  -8.6418e-03   4.4995e-03  -7.4838e-03
  -1.5442e-02   1.1226e-02   3.3826e-03   4.7511e-03  -5.3225e-03   6.0837e-03
  -3.9746e-03  -6.3450e-04   2.4899e-04   3.5946e-03  -1.7525e-03   3.1091e-03
  -4.4631e-03   2.1875e-03   5.2814e-04   2.1558e-03  -1.3991e-03   2.1170e-03

As the numbers above show, the relative errors are small, so indeed $\tilde{A} \approx A$. However, storing $A$ requires storing $N_{A} = 36$ elements, but $\tilde{A}$ needs not be stored as a $6 \times 6$ matrix. Instead, we can merely store $U$, $\Sigma$ and $V$ after many of their entries are removed. This yields: $$ N_{\tilde{A}} = \textrm{size}(U) + \textrm{size}(V) + \textrm{size}(\Sigma) = 12 + 12 + 2 = 26 $$ so we can obtain $\tilde{A}$ by storing only $N_{\tilde{A}} = 26$ instead of $N_A = 36$ elements. The reason why $\textrm{size}(\Sigma)= 2$ is because $\Sigma$ is a diagonal matrix, so its off-diagonal entries need not be stored as we know they are zero. This represents a $28\%$ reduction in the required storage space. For large matrices, the method above can lead to even better compression ratios.

It is possible to use the technique above to compress images while still preserving most of their visual properties. If you want to know how, please visit this blog again in the future :-)

References

[1]James W. Demmel, Applied Numerical Linear Algebra, SIAM; 1st edition (1997)

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