(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.

(Hint: since you are told that u is in the span {v1,v2,v3}, you can automatically say that there are scalars c1,c2 and c3 so that u = c1v1+c2v2+c3v3. Your goal now is to find a way to write 3u as a linear combination of {v1,v2,v3}

(2) Let u, v1,v2,v3 and v4 be vectors in the Rn . If u can be written as a linear combination of v1,v2, and v3 show that u can also be written as a linear combination of v1,v2,v3 and v4 .

(Hint: Look at the previous question and remember you can use 0 as a coefficient in your equation.)

(3) Let u, v, w1, w2, and w3 be vectors in the Rn. If u and v can both be written as linear combinations of w1, w2 and w3 show that u+v can also be written as a linear combination of w1, w2, and w3.

(hint:this is similar to the previous ones. Remember to choose different letters for your coefficients for u and v)

Respuesta :

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

[tex]3u = 3c_1v_1+3c_2v_2+3c_3v_3[/tex]

3u  is in the Span of the vectors [tex]\{v_1,v_2,v_3\}[/tex].

(2)

That's not true, consider the following counter example.

[tex]v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)[/tex]

[tex]u[/tex] is a linear combination of [tex]v_1,v_2,v_3[/tex] but is NOT a linear combination of [tex]v_1,v_2,v_3,v_4.[/tex]

Step-by-step explanation:

(1)

As the hint indicates, you know that

[tex]u = c_1 v_1 + c_2v_2+c_3v_3[/tex]

Then, if you multiply both sides of the equality by 3, you get that

[tex]3u = 3c_1v_1+3c_2v_2+3c_3v_3[/tex]

And that's it. 3u  is in the Span of the vectors [tex]\{v_1,v_2,v_3\}[/tex]

(2)

That's not true, consider the following counter example.

[tex]v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)[/tex]

[tex]u[/tex] is a linear combination of [tex]v_1,v_2,v_3[/tex] but is NOT a linear combination of [tex]v_1,v_2,v_3,v_4.[/tex]