- #1

- 23

- 0

Is this true? Can someone please explain why or why not??

What I think: I think it is true because I read that a linear transformation preserves the operations of vector addition and scalar multiplication.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Mola
- Start date

- #1

- 23

- 0

Is this true? Can someone please explain why or why not??

What I think: I think it is true because I read that a linear transformation preserves the operations of vector addition and scalar multiplication.

- #2

- 430

- 3

[tex]a_1v_1 + a_2v_2 + a_3v_3 = 0[/tex]

Now left-multiplying this by A you get:

[tex]A(a_1v_1+a_2v_2+a_3v_3) = A0=0[/tex]

Now use your rules for matrix arithmetic to derive:

[tex]a_1(Av_1)+a_2(Av_2)+a_3(Av_3)=0[/tex]

(HINT: Ak = kA for constants k, and A(v+w) = Av+Aw for vectors v, w where the expression makes sense).

- #3

- 23

- 0

Thanks.

That leads me to a related theory: Let's assume we are talking about {v1, v2, v3} being a linearly INDEPENDENT set now. If we multiply the vectors by the matrix A, how does it affect the independece? Would it make a differerence if the matrix A is invertible?

- #4

- 430

- 3

That leads me to a related theory: Let's assume we are talking about {v1, v2, v3} being a linearly INDEPENDENT set now. If we multiply the vectors by the matrix A, how does it affect the independece? Would it make a differerence if the matrix A is invertible?

This is actually a quite interesting little question (well in my opinion anyway). First for fixed A, v1,v2,v3 note that if we take the contrapositive of your initial result we get:

If Av1, Av2, Av3 are linearly independent, then v1,v2,v3 are linearly independent.

so for linear independence it goes backwards. For an arbitrary matrix A we can not prove your new statement since we can just let A be the 0 matrix. However if A is invertible, then we can just go backwards by noting that if,

[tex]a_1Av_1+a_2Av_2+a_3Av_3 = 0[/tex]

Then we can left-multiply by [itex]A^{-1}[/itex] to get,

[tex]a_1v_1+a_2v_2+a_3v_3 = 0[/tex]

so if v1,v2,v3 are linearly independent and A is invertible, then Av1, Av2, Av3 are linearly independent.

- #5

- 23

- 0

That was a very good help from you.

Share:

- Replies
- 3

- Views
- 6K