Proving Linear Transformation for C*A^2

[squaremeplz]

Homework Statement

C is a fixed n*n matrix. Is T(A) =CA^2 a linear transformation on R^(n*n)

Homework Equations

Linear transform requirements:

1. T(A + Y) = T(A) + T(Y)

2. T(c*A) = c*T(A)

The Attempt at a Solution

I'm confused as to how the linear transform requirements apply to matrices. i.e. should I do the proof as column vectors of A i.e.

T(a_1 + a_2 + .. + a_n) = T(a_1) + T(a_2) + ... + T(a_n)

and

T(c*a_1 + c*a_2 + .. + c* a_n) = c* (T(a_1) + T(a_2) + ... + T(a_n))


I think that this is not a linear transformation since scalar multiplication does not hold

i.e. L(c*A) =/= c*L(A)

I'm just looking for a nudge in the right direction and any input is appreciated.

Thank You

 

[Mark44]

Homework Statement

C is a fixed n*n matrix. Is T(A) =CA^2 a linear transformation on R^(n*n)

Homework Equations

Linear transform requirements:

1. T(A + Y) = T(A) + T(Y)

2. T(c*A) = c*T(A)

The Attempt at a Solution

I'm confused as to how the linear transform requirements apply to matrices. i.e. should I do the proof as column vectors of A i.e.

T(a_1 + a_2 + .. + a_n) = T(a_1) + T(a_2) + ... + T(a_n)

and

T(c*a_1 + c*a_2 + .. + c* a_n) = c* (T(a_1) + T(a_2) + ... + T(a_n))


I think that this is not a linear transformation since scalar multiplication does not hold

i.e. L(c*A) =/= c*L(A)

I'm just looking for a nudge in the right direction and any input is appreciated.

Thank You

You don't need to work with the column vectors at all, just the matrices. For arbitrary n X n matrices A and B, what is T(A + B)? Is it the same as T(A) + T(B)?

And since you believe that scalar multiplication doesn't hold for transformation, you should show why T(cA) != cT(A).

 

[squaremeplz]

Thanks for the quick reply!

1. T(A+B) = C(A+B)^2
= C(A+B)(A+B)
= C(A^2 + 2AB + B^2)
= C(A^2) + C(2AB) + C(B^2)
= T(A) + T(B) + C(2AB)
T(A+B) =/= T(A) + T(B)

2. T(b*A) = C(b*A)^2 : where b is some scalar

= b^2 * CA^2

= b^2 * T(A)

 

[grey_earl]

Small correction, otherwise it's ok: Since you deal with matrices, in general AB != BA. So you have (A+B)(A+B) = A² + AB + BA + B².

 

[squaremeplz]

thanks! got to keep an eye on the details :)