Linear Algebra Question regarding basis of a kernel

[sophiaw22]

Homework Statement

Let V be the vector space of all 2x2 matrices over Q

V= {[x₁ x₂] : x_i $$\in$$ Q}
... x₃ x₄

Let A = [ -1 0 ] and let C:V --> V be the linear map C(X) = XA + AX
.... -1 1

Find a basis for Ker(C) and a basis for Im(C)

The Attempt at a Solution

I used C(X) = XA + AX
= [-2x₁-x₂, 0 ]
...-x₁-x₄, 2x₄-x₂

but I ended up with a matrix with all independent variables.

[ -2,-1,0, 0 ]
...0, 0, 0, 0
...-1,0, 0, -1
...2,-1, 0, 0

--> ref

[1,0,0,0]
..0,1,0,0
..0,0,0,1
..0,0,0,0

Ker(C) = 0?

As for the Im(C) I am completely lost after doing the Ker(C)

Thank you very much!

 

[Outlined]

$$\left[\begin{array}{rrrr}<br /> -2 & -1 &0 & 0 \\<br /> 0 & 0 & 0 & 0 \\<br /> -1 & 0 & 0 & -1 \\<br /> 2 & -1 & 0 & 0<br /> \end{array}\right]$$

reduces to

$$\left[\begin{array}{rrrr}<br /> 0 & -1 &0 & 2 \\<br /> 0 & 0 & 0 & 0 \\<br /> -1 & 0 & 0 & -1 \\<br /> 0 & -1 & 0 & -2<br /> \end{array}\right]$$

reduces to

$$\left[\begin{array}{rrrr}<br /> 0 & 1 &0 & -2 \\<br /> 0 & 0 & 0 & 0 \\<br /> 1 & 0 & 0 & 1 \\<br /> 0 & 0 & 0 & -4<br /> \end{array}\right]$$

so kernel is
$$\mbox{Span}\left\{<br /> \left[\begin{array}{r}<br /> -1 \\<br /> 2 \\<br /> 0 \\<br /> 1 <br /> \end{array}\right], \left[\begin{array}{r}<br /> 0 \\<br /> 0 \\<br /> 1 \\<br /> 0 <br /> \end{array}\right]\right\}$$

 

[sophiaw22]

Sorry but can you expand on how you get that kernel ?

 

[Outlined]

x₂ - 2x₄ = 0
x₁ + x₄ = 0
x₃ is free
-4x₄ = 0

this reduces to

x₁ = 0
x₂ = 0
x₃ is fee
x₄ = 0

so kernel is (APOLOGIES)

$$\mbox{Span}\left\{\left[\begin{array}{r}<br /> 0 \\<br /> 0 \\<br /> 1 \\<br /> 0 <br /> \end{array}\right]\right\}$$

 

[sophiaw22]

Ah, I get it now.

Thank you so much !