Linear Algebra Vector Space matrix help

Question

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.

Answer 1

The set of anti-symmetric $2\times2$ matrices, is the kernel of $t$:since if $A=-A^T$, then

$$t(A)=0$$

Answer 2

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.

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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 $\frac 1 2 Q$.

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

Answer 3

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 $$\chi_t(x)=x^{\alpha_n}(x-2)^{\beta_n}$$