Vector Operations

Level 1 - Linear Algebra topic page in Vectors In Rn.

Principle

Vector addition and scalar multiplication are the two basic operations in \(\mathbb R^n\). Addition combines vectors component-by-component. Scalar multiplication stretches, shrinks, or reverses a vector by multiplying every component by the same real number.

These two operations are the reason \(\mathbb R^n\) is a vector space.

Notation

\(\mathbf u,\mathbf v,\mathbf w\)

vectors in the same \mathbb R^n

\(u_i,v_i,w_i\)

matching components of the vectors

\(\lambda,\mu\)

real scalars

\(\mathbf u+\mathbf v\)

componentwise vector sum

\(\lambda\mathbf v\)

scalar multiple of \mathbf v by \lambda

\(-\mathbf v\)

the vector obtained by multiplying \mathbf v by -1

The Core Method

Only add vectors that have the same number of components. If

Two vectors in Rn

\[\mathbf u=(u_1,\ldots,u_n),\qquad \mathbf v=(v_1,\ldots,v_n)\]

then

Vector addition

\[\mathbf u+\mathbf v=(u_1+v_1,\ldots,u_n+v_n)\]

and

Scalar multiplication

\[\lambda\mathbf v=(\lambda v_1,\ldots,\lambda v_n)\]

A linear combination is any vector built by adding scalar multiples:

Linear combination

\[\lambda\mathbf u+\mu\mathbf v\]

The scalars multiply the whole vector, not just one component.

Worked Cases

Question

Let \(\mathbf u=(2,-1,4)\) and \(\mathbf v=(3,0,-5)\). Compute \(2\mathbf u-\mathbf v\).

Answer

First scale \(\mathbf u\): \(2\mathbf u=2(2,-1,4)=(4,-2,8)\). Next subtract \(\mathbf v\): \(2\mathbf u-\mathbf v=(4,-2,8)-(3,0,-5)\). Subtract component-by-component: \((4-3,-2-0,8-(-5))=(1,-2,13)\). Therefore \(2\mathbf u-\mathbf v=(1,-2,13)\).

Examples

Question

For \(\mathbf a=(1,2)\) and \(\mathbf b=(-3,5)\), compute \(3\mathbf a+2\mathbf b\).

Answer

Compute each scalar multiple: \(3\mathbf a=(3,6)\) and \(2\mathbf b=(-6,10)\). Add them: \(3\mathbf a+2\mathbf b=(3,6)+(-6,10)=(-3,16)\).

Vectors In Rn

Scalar Product