What is the basis of Ker(F)?

[simpledude]

Homework Statement

Let V = M2(R) be the vector space over R of 2×2 real matrices. We consider the mapping
F : V −> V defined for all matrix M belonging to V , by F(M) = AM +MA^T where A^T denotes the transpose matrix of the matrix A given below

A =

1 2
−1 0

Question is: Determine a basis of Ker(F)

The Attempt at a Solution

So I showed that F is a linear operator, and preserves scalar addition and multiplication.
However I am lost as to how I can solve the equation:
AM +MA^T = 0

Any help appreciated, thanks :)

 

[ircdan]

AM + MA^T = 0, so AM = -MA^T. You know what A is and M is a 2x2 matrix, so pick some entres a, b, c, d for M and solve for them.

 

[jhicks]

What is zero in $M_2(\mathbb{R})$? Just apply the definition of kernel using $M = \left( \begin{array}{cc} m_{11} & m_{12}\\m_{21} & m_{22}\\\end{array} \right)$ and you will find the basis.

 

[simpledude]

I get dimension of Ker = 4,

What I did is write out the matrix and multiply it out (since we know A and M I took
as a,b,c,d). After multiplying and adding, I get a system of 4 equations, (a,b,c)
and solve them via Gauss to find how many are independent.

Is this ok?

 

[HallsofIvy]

Homework Statement

Let V = M2(R) be the vector space over R of 2×2 real matrices. We consider the mapping
F : V −> V defined for all matrix M belonging to V , by F(M) = AM +MA^T where A^T denotes the transpose matrix of the matrix A given below

A =

1 2
−1 0

Question is: Determine a basis of Ker(F)

The Attempt at a Solution

So I showed that F is a linear operator, and preserves scalar addition and multiplication.
However I am lost as to how I can solve the equation:
AM +MA^T = 0

Any help appreciated, thanks :)

You solve the matrix equation by doing the work to write it out as separate equations for the components.
Let
$$M= \left(\begin{array}{cc}a & b \\ c & d\end{array}\right)$$
Then
$$F(M)= AM+ MA^T= \left(\begin{array}{cc} 1 & 2 \\ -1 & 0\end{array}\right)\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)+ \left(\begin{array}{cc}a & b \\ c & d\end{array}\right)\left(\begin{array}{cc}1 & -1 \\ 2 & 0\end{array}\right)$$
$$= \left(\begin{array}{cc}a+2c & b+2d \\ -a & -b+d\end{array}\right)+ \left(\begin{array}{cc}a+ 2b & -a \\ c+2d & -c\end{array}\right)$$
$$= \left(\begin{array}{cc}2a+2b+2c & -a+b+2d \\ -a+ c+ 2d & -b- c+ d\end{array}\right)$$
For M be "in the kernel", that must be the 0 vector. Solve 2a+ 2b+ 2c= 0, -a+ b+ 2d= 0, -a+ c+ 2d= 0, and -b- c+ d= 0. If those equations are all independent, of course, the only solution will be a= b= c= d= 0, the 0 matrix. If not, then the kernel may have dimension 1, 2, 3, or 4. (Well, it's pretty obvious the dimension is not 4.)

 

[simpledude]

EDIT: Oh wait found a mistake in my math

 

[simpledude]

Yes did exactly that, by the way last equation should be -b-c

So after this I solve the matrix of coefficients to see how many independent columns
I have:

2 2 2 0
-1 1 0 2
-1 0 1 2
0 -1 -1 0

and simplified matrix is:

1 1 1 0
0 1 2 2
0 0 1 2
0 0 0 4
I get 4 independent columns once I solve via Gauss. So Ker F = {0}.. so is this
dim 0?