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What is scalar data type?
Scalar data is the opposite of vector data, in that it provides a magnitude without a direction. For example, speed is a scalar quantity because it provides magnitude without a direction, whereas velocity is a vector quantity because it provides the magnitude that speed provides, but supplies us with direction.
How is linear regression used?
Linear regression can be used in statistics in order to create a model out a dependable scalar value and an explanatory variable. Linear regression has applications in finance, economics and environmental science.
What is the difference between factor analysis with multidimensional scaling?
Multidimensional scaling (MDS): Is a family of distance and scalar-product (factor) and other conjoint models. It re-scales a set of dis/similarity data into distances and produces the low-dimensional configuration that generated them. Factor Analysis / Principal Components Analysis (FA/PCA), by contrast: PCA is the full reduction of set of scalar-products to a new orthogonal set of spanning dimensions (components); FA is a dimension-reducing model (properly containing communalities and not 1 in diagonal) to orthogonal or oblique dimensions (factors). In general usage, PCA and FA are primarily dimensional and use interval-level data, whereas MDS usually uses an ordinal (non-metric) transformation of the data producing a spatial configuration where dimensions are arbitrary.
Why linear algebra is called linear - you mean what is significance of the word linear in linear algebra?
Linear algebra deals with mathematical transformations that are linear. By definition they must preserve scalar multiplication and additivity. T(u+v)= T(u) + T(v) T(R*u)=r*T(u) Where "r" is a scalar For example. T(x)=m*x where m is a scalar is a linear transform. Because T(u+v)=m(u+v) = mu + mv = T(u) + T(v) T(r*u)=m(r*u)=r*mu=r*T(u) A consequence of this is that the transformation must pass through the origin. T(x)=mx+b is not linear because it doesn't pass through the origin. Notice at x=0, the transformation is equal to "b", when it should be 0 in order to pass through the origin. This can also be seen by studying the additivity of the transformation. T(u+v)=m(u+v)+b = mu + mv +b which cannot be rearranged as T(u) + T(v) since we are missing a "b". If it was mu + mv + b + b it would work because it could be written as (mu+b) + (mv+b) which is T(u)+T(v). But it's not, so we are out of luck.
