- #1

Mola

- 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.

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- Thread starter Mola
- Start date

- #1

Mola

- 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

rasmhop

- 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

Mola

- 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

rasmhop

- 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

Mola

- 23

- 0

That was a very good help from you.

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