Linear Algebra Vector Space matrix help Let $M_{2\times2}$ be a vector space of all $2\times2$ matrices. If the transformation from $M_{2\times2}$ to $M_{2\times2}$ is $t(A)=A+A^T$ and $A$ is a $2\times2$ matrix with the top row $a,b$ and bottom row $c,d$.
Would the kernel of transformation $t$ be $0$? I keep getting $0$ since if I row reduce the transformation with plugged in numbers i keep getting $x1=0$ and $x2=0$.
Also, how do I show that the range of $t$  is the set of symmetric matrices? I know that means I am trying to find the set of $Q\in M_{2\times2}$ so that $Q^T=Q$
Lastly, would any matrix in the form of A mentioned in the beginning of the question be transformed into a matrix that is equal to its transpose. Whatever numbers I put in for matrix A, my transformed matrices are always equal to their respective transposes.
 A: Let's say more about this linear transformation.
First $$A\in \ker t\iff t(A)=A+A^T=0\iff A^T=-A\iff A\in\mathcal{AS_n}(\Bbb R)$$
hence $0$ is an eigenvalue of $t$ with multiplicity equal to 
$$\dim\mathcal{AS_n}(\Bbb R)=\frac{n(n-1)}2:=\alpha_n$$
and if $A\in\mathcal{M_n}(\Bbb R)$ then
$$t(A)=A+A^T\in\mathcal{S_n}(\Bbb R)$$
and by the rank-nullity theorem we have
$$\dim \operatorname{im}(t)=\dim\mathcal{M_n}(\Bbb R)-\dim\ker(t)=\dim\mathcal{S_n}(\Bbb R)$$
hence 
$$\operatorname{im}(t)=\mathcal{S_n}(\Bbb R)$$
moreover, if $A\in\mathcal{S_n}(\Bbb R)$ then 
$$t(A)=A+A^T=2A$$
hence $2$ is an eigenvalue of $t$ with multiplicity equal to
$$\dim\mathcal{S_n}(\Bbb R)=\frac{n(n+1)}2:=\beta_n$$
hence $t$ is diagonalizable and it's minimal polynomial is
$$\pi_t(x)=x(x-2)$$
and it's characteristic polynomial is
$$\chi_t(x)=x^{\alpha_n}(x-2)^{\beta_n}$$
A: The set of anti-symmetric $2\times2$ matrices, is the kernel of $t$:since if $A=-A^T$, then 
$$t(A)=0$$
A: No, the kernel is not zero: There are matrices such that $A^T = -A$, called skew-symmetric. Try a matrix with zeros on the main diagonal.

To show that the range of $t$ is the collection of symmetric matrices, you need to check two things:


*

*For any $A$, $t(A) =  A + A^T$ is symmetric.

*For any matrix $Q$ for which $Q^T = Q$, there is a matrix $A$ such that $A + A^T = Q$. After noting that
$$Q = Q^T = (A + A^T)^T = A + A^T$$
try something like $\frac 1 2 Q$.

Once you've finished the previous section, you should see that every possible output of $t$ is symmetric, so equal to its own transpose.
