The minimal approximation error A-A 2achievable by a rank-1 approximation A to A is
||A - A1||₂=0 where A is 2×2 matrix.
We have given a matrix A as seen in above figure or A = [ 5 15 ; 6 18 ; -1 -3 ; -4 -12 ; 2 6]
note here that C₂= 3C₁
where Cᵢ --> iᵗʰ column
v = [ 5 ; 6 ; -1 ; -4 ; 2]
||v||² = 82 => ||v|| = √82
and A At v = 820 v
and A At = [ 82 246 ; 246 738]
AAt [1;3] = 820[1;3]
=> v1 = 1/√10(1,3)^t
Then the best rank of 1 approx
= √820 /√80√10 [ 5 ; 6 ; -1 ; -4 ; 2] [ 1 3]
= A
Since , the rank of matrix A is one so, the minimal approx. value is ||A - A1||2 = 0
Hence, the minimal Approx. ||A - A1||2 = 0 .
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Complete question:
Consider the matrix A: A = [5 15] 6 18 -1 -3 -4 -12 [26] What is the minimal approximation error A-A 2 achievable by a rank-1 approximation A to A? Hint: Can you determine this without explicitly calculating the SVD?
