by Qiang Gao, updated at March 26, 2018
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- $$ \boldsymbol{\beta} $$ is unknown
- overfitting
- For $$ y = a x^2 + b x + c
$$, there exists a solution only if $$ b^2 \geq 4 a c $$. - For $$ \mathbf{y} = \mathbf{A} \mathbf{x}
$$, there exists a solution only if $$ \mathbf{y} $$ lies in the column space of $$ \mathbf{A} $$.
- For $$ y = a x^2 + b x + c
$$, there exists a solution and the solution is unique only if $$ b^2 = 4 a c $$. - For $$ \mathbf{y} = \mathbf{A} \mathbf{x}
$$, there exists a solution and the solution is unique only if $$ \mathbf{y} $$ lies in the column space of $$ \mathbf{A} $$ and $$ \mathbf{A} $$ has full column rank.
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If $$ y = a x^2 + b x + c $$ and $$ b^2 \geq 4 a c
$$, then the solutions are $$ x = (-b \pm \sqrt{b^2 - 4ac}) /(2a) $$. -
If $$ \mathbf{y} = \mathbf{A} \mathbf{x} $$ and $$ \mathbf{A} $$ is (square and) invertible, then the unique solution is $$ \mathbf{x} = \mathbf{A}^{-1} \mathbf{y} $$.
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For the real-valued function $$ y = f( \mathbf{x} ) = \mathbf{a}' \mathbf{x} $$ (inner product form), $$ \frac{df(\mathbf{x})}{d\mathbf{x}} = \mathbf{a}. $$
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For the real-valued function $$ y = f( \mathbf{x} ) = \mathbf{x}' \mathbf{A} \mathbf{x} $$ (quadratic from), $$ \frac{df(\mathbf{x})}{d\mathbf{x}} = \mathbf{A} \mathbf{x}. $$
There are equivalently four ways of matrix multiplication, each is very important.
Copyright ©2018 by Qiang Gao