Linear Algebra Basis and Dimension-1-2, guaranteed and verified 100%

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Basis and Dimension

Def. The vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 form a basis for a vector space 𝑉 , if and only
if:

i. 𝑣1 , 𝑣2 , … , 𝑣𝑛 are linearly independent
ii. 𝑆𝑝𝑎𝑛(𝑣1 , … , 𝑣𝑛 ) = 𝑉


Ex. The standard basis for ℝ𝟑 is {𝑒1 , 𝑒2 , 𝑒3 } where 𝑒1 =< 1, 0, 0 >,
𝑒2 =< 0, 1, 0 > , 𝑒3 =< 0, 0, 1 >.


i. 𝑒1 , 𝑒2 , 𝑒3 are linearly independent because

𝑐1 𝑒1 + 𝑐2 𝑒2 + 𝑐3 𝑒3 = 0

𝑐1 < 1, 0, 0 > +𝑐2 < 0, 1, 0 > +𝑐3 < 0, 0, 1 >=< 0,0,0 >
< 𝑐1 , 𝑐2 , 𝑐3 >=< 0,0,0 >
⟹ 𝑐1 = 𝑐2 = 𝑐3 = 0.


ii. Span(𝑒1 , 𝑒2 , 𝑒3 ) = ℝ3 because any vector < 𝑎1 , 𝑎2 , 𝑎3 > ∈ ℝ3 can be
written as:

< 𝑎1 , 𝑎2 , 𝑎3 >= 𝑎1 < 1, 0, 0 > + 𝑎2 < 0, 1, 0 > + 𝑎3 < 0, 0, 1 >

so Span(𝑒1 , 𝑒2 , 𝑒3 ) = ℝ3 .


Similarly, the standard basis for ℝ𝒏 is {𝑒1 , 𝑒2 , … , 𝑒𝑛 }, where
𝑒𝑖 =< 0,0,0, … ,1,0, … >, 1 in the 𝑖 𝑡ℎ component.

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There is an infinite number of sets of 3 vectors that form a basis for ℝ3 .

Ex. Show 𝑣1 =< 1, 0, 1 >, 𝑣2 =< 1, 1, 0 >, and 𝑣3 =< 0, 1, 1 > forms a
basis for ℝ3 .

i. We saw in an earlier example that 𝑣1 , 𝑣2 , 𝑣3 are linearly independent.


ii. To see that Span(𝑣1 , 𝑣2 , 𝑣3 ) = ℝ3 notice that
any vector 𝑣 ∈ ℝ3 , can be represented by 𝑣 =< 𝑎, 𝑏, 𝑐 >.
We need to show there exist 𝑐1 , 𝑐2 , 𝑐3 such that

𝑐1 𝑣1 + 𝑐2 𝑣2 + 𝑐3 𝑣3 =< 𝑎, 𝑏, 𝑐 >
𝑐1 < 1, 0, 1 > +𝑐2 < 1, 1, 0 > + 𝑐3 < 0, 0, 1 > =< 𝑎, 𝑏, 𝑐 >
< 𝑐1 + 𝑐2 , 𝑐2 + 𝑐3 , 𝑐1 + 𝑐3 > =< 𝑎, 𝑏, 𝑐 >

or 𝑐1 + 𝑐2 =𝑎
𝑐2 + 𝑐3 = 𝑏
𝑐1 + 𝑐3 = 𝑐.


We solved this system of equation when we showed that 𝑥 2 + 1,
𝑥 + 1, and 𝑥 2 + 𝑥 generated 𝑃2 (ℝ). We found that:

𝑎−𝑏+𝑐
𝑐1 =
2

𝑎+𝑏−𝑐
𝑐2 =
2

−𝑎+𝑏+𝑐
𝑐3 = .
2

Thus Span(𝑣1 , 𝑣2 , 𝑣3 ) = ℝ3 and 𝑣1 , 𝑣2 , 𝑣3 is a basis for ℝ3 .