For the following linear mapping, determine a transformation matrix and check for the property of injectivity.
I have the following exam question:
In the $\mathbb{F_3}$ vector space $V:= M(2,2,\mathbb{F_3})$ of $2 \times 2$ matrices, the following linear mapping $\varphi$ is given:
$\varphi : V \rightarrow V, A \mapsto A + A^T$
a) Determine a basis B of $V: M(2,2,\mathbb{F_3})$.
b) Determine the transformation matrix $M_B^B(\varphi)$
c) Determine if the matrix D $\begin{pmatrix} 1 & 2 \\ 2 & 0\end{pmatrix}$ is in the image space $im(\varphi)$ of $\varphi$.
d) Determine if the mapping is injective.
Progress so far:
Calculating $\varphi(A)$ where $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ yields:
$\varphi (\begin{pmatrix} a & b \\ c & d \end{pmatrix}) \mapsto \begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} a & b + c \\ b + c & d \end{pmatrix}$ which can be identified as the set of symmetric $2\times2$ matrices. Letting $b + c = e$, it can be seen that the following set of 3 matrices forms a spanning set and thus a basis for V:
$B = \{ b_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, b_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, b_3 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\}$
As a consequence of this, it nescessarily follows that c) will be contained in the image (Calculations have been ommited here, but to answer this question more rigorously one might solve the coeffecients $a_{1,2,3}$ such that $a_1 b_1 + a_2 b_2 + a_3 b_3 = D$.
At this point I become lost. Can anyone provide some assistance for steps b) and d)?
vector-spaces linear-transformations finite-fields
José Carlos Santos
$\begingroup$ the vector space you have is isomorphic to vectors with $4$ entries, so your basis should have $4$ elements (to see the isomorphism, just map the matrix $\begin{pmatrix} a & b\\ c & d\end{pmatrix}$ onto $(a,b,c,d)$). $\endgroup$ – Student Mar 27 '18 at 18:16
a) Your answer is wrong. A basis of a $4$-dimensional space must have $4$ elements. You can take, for instance, $b_1=\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right)$, $b_2=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$, $b_3=\left(\begin{smallmatrix}0&0\\1&0\end{smallmatrix}\right)$, and $b_4=\left(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right)$.
b) The matrix you're after will be then$$\begin{pmatrix}2&0&0&0\\0&1&1&0\\0&1&1&0\\0&0&0&2\end{pmatrix}.$$
d) No. For instance, $\varphi\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)=\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$. Or you can see that the determinant of the matrix from b) is $0$.
José Carlos SantosJosé Carlos Santos
$\begingroup$ Thanks, your feedback is very helpful. Would you mind elaborating a little bit on how you determined the matrix in part b? $\endgroup$ – Oscar Mar 27 '18 at 18:22
$\begingroup$ @Oscar It follows from the fact that $\varphi(b_1)=b_1+b_1^T=2b_1$, $\varphi(b_2)=b_2+b_2^T=b_2+b_3$, $\varphi(b_3)=b_3+b_3^T=b_3+b_2$, and $\varphi(b_4)=b_4+b_4^T=2b_4$. $\endgroup$ – José Carlos Santos Mar 27 '18 at 18:25
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