Computational Intelligence Lab FS26

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Last updated 3:48 PM on 6/4/26
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11 Terms

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Frobenius Norm

AF=i=1nj=1maij2=tr(AA)=i=1rσi2||A||_F = \sqrt{\sum_{i=1}^{n} \sum_{j=1}^{m} |a_{ij}|^{2} } = \sqrt{tr(A^{\top} A)} = \sqrt{\sum_{i = 1}^{r} \sigma_{i}^{2}}

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Spectral Norm

A2=λmax(AHA)=σmax(A)||A||_2 = \sqrt{\lambda_{max}(A^{H}A)} = \sigma_{max}(A)

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Nuclear Norm

A=i=1rσi||A||_* = \sum_{i = 1}^{r} \sigma_{i}

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Other matrix norms (1-norm and norm\infty - \mathrm{norm})

1-norm (Maximum absolute columns sum):

A1=maxji=1naij||A||_1 = \max_{j}{\sum_{i=1}^{n} |a_{ij}|}

norm\infty - \mathrm{norm} (Maximum absolute row sum):

A=maxij=1maij||A||_\infty = \max_{i}{\sum_{j=1}^{m} |a_{ij}|}

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Eckart-Young Theorem

The Eckart-Young theorem is fundamental to many low-rank approximation problems.

It states that by pruning the singular values below σk\sigma_k in the SVD representation, we get an optimal rank k approximation of a matrix.

This means that approximations for any k can directly be read-off the SVD.

Given ARn×mA \in \mathbb{R}^{n \times m} with SVD A=UΣVA = U \Sigma V^{\top}. Then for all 1kmin{n,m}1 \le k \le min\{n,m\} we have

Ak:=Udiag(σ1,,σk)Varg min{ABF:rank(B)k}A_{k} := U diag(\sigma_{1}, … , \sigma_{k}) V^{\top} \in \text{arg min}\{||A-B||_{F}: rank(B)\le k\}

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Convex Envelope of a function

The convex envelope of a function f:RRf: R \rightarrow \mathbb{R} is the largest convex function gg for which gfg \le f on RR.

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Convex Envelope of the rank function (rank is NOT convex!)

Theorem (Fazel et al)

The convex envelope of rank(A)rank(A) on R={A21}R = \{||A||_2 \le 1\} is A||A||_*

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Sigmoid Activation Function

Sigmoid

σ(z)=11+ez\sigma(z)=\frac{1}{1+e^{-z}}

Derivative

σ(z)=σ(x)(1σ(x))\sigma’(z)=\sigma(x)(1-\sigma(x))

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Tanh Activation Function

Tanh

tanh(z)=ezezez+ez\tanh(z) = \frac{e^z - e^{-z}}{e^z + e^{-z}}

Derivative

tanh(z)=1tanh2(z)\tanh'(z) = 1 - \tanh^2(z)

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ReLU Activation function

ReLU

ReLU(z)=max(0,z)\text{ReLU}(z) = \max(0, z)

Derivative

ReLU(z)={1amp;if zgt;00amp;if z0\text{ReLU}'(z) = \begin{cases} 1 & \text{if } z > 0 \\ 0 & \text{if } z \leq 0 \end{cases}

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