The solution for HW2 – Q3 was missing from the solutions. Here it is:
Answer 1
td =
0.2500 0.5300 0.7500
0.7300 0.5000 0.2300
To find the first term of the svd, we find the eigen vectors of td * td’
td * td' =
0.9059 0.6200
0.6200 0.8358
<Eigen vector calculation here>
Eigen vectors of td * td’
-0.7268
-0.6869
And
0.6869
-0.7268
The eigen values are 1.4918 and 0.2499
[Note: these might be negative also, that is a valid solution too]
Therefore, the first term of the svd is
-0.7268 0.6869
-0.6869 -0.7268
The second term of the svd is the square root of the eigen values found above
1.2214 0 0
0 0.4999 0
To find the third term of the svd, we find the eigen vectors of td’ * td
Td’ * td =
0.5954 0.4975 0.3554
0.4975 0.5309 0.5125
0.3554 0.5125 0.6154
<eigen computation>
Eigen vectors are
First eigen vector:
0.5593
0.5965
0.5756
Second eigen vector:
-0.7179
0.0013
0.6962
Third eigen vector:
-0.4146
0.8026
-0.4290
Therefore the third and last term in the svd is
-0.5593 0.7179 -0.4146
-0.5965 -0.0013 0.8026
-0.5756 -0.6962 -0.4290
<But we did not find the right signs for the matrix here, so we need to find this matrix in a different way, (see recent mail sent by Dr Rao)>
Answer 2
After removing the lesser of the two eigen values, your s matrix becomes:
1.2214 0 0
0 0 0
Then to recreate the matrix, you multiply u * s * v’
0.4965 0.5296 0.5110
0.4692 0.5005 0.4829
Does it look close to the original? In my opinion, NO, it does not. We did eliminate a significant part of the variance.
(Also accept a YES answer if it claims that if you scale it properly, you would end up sort of where the original documents were.)
Answer 3
The query vector is q = [1, 0]
In the factor space, it would be q * tf
qf = -0.7268 -0.6869
The documents in the factor space are:
D1: -0.6831 0.3589
D2: -0.7286 -0.0006
D3: -0.7031 -0.3481
The similarities are:
sim(q,D1) = 0.3239
sim(q,D2) = 0.7274
sim(q,D3) = 0.9561
Answer 4
Before the transformation:
After the transformation
Clearly, after the transformation, the documents are easily separable using only one axis (the y-axis). So we can get rid of the x-axis and still differentiate between them
Answer 5
The new values added is a linear combination of the previous keywords, (0.5 * k1 + 0.5 * k2).
It will have absolutely no impact upon the calculations at all
It tells us that svd manages to find the real dimensionality of the data
Thanks and Regards,
Sushovan De