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The question states:

a) Suppose that $U = V = M_{2,2} (\mathbb R)$ is the vector space of $2 × 2$ matrices over $\mathbb R $ Let $\phi$ be defined by $\phi(A) = A - A^T $ for each $A \in M_{2,2} \mathbb R$. Determine $\ker \phi $ and state its dimension.

b) Determine $\operatorname{im}\phi$ and its dimension.

So my thoughts are we use $\ker \phi = \left \{ f \in U \mid \phi(f) = 0 \right \}$ and then we follow all the steps to show that $$\ker \phi = \left \langle \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}\right \rangle $$

and we can conclude that $\dim \ker \phi = 3 $ and so by Rank and Nullity Thm, we say $\dim\operatorname{im}\phi = 1$.

My question is that is this correct, this process to get that answer, and I'm just wondering what $\operatorname{im}\phi$ would be?

Thank you in advance.

Edit: How I got to the kernel:

$$f = \begin{pmatrix} a & b\\ c & d \end{pmatrix} $$ then $$f^T = \begin{pmatrix} a & c\\ b & d \end{pmatrix}$$

then $$f-f^T = \begin{pmatrix} 0 & c-b\\ c-b & 0 \end{pmatrix}$$

So putting into a system of eqs.

$$\left \{ b-c = 0 \right. \\ \left \{ c-b = 0 \right.$$

Which implies that b=c

So $$f = \begin{pmatrix} a & b\\ c & d \end{pmatrix} = a\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} +b \begin{pmatrix} 0 & 1\\ 0& 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0\\ 0& 1 \end{pmatrix} $$

Which allows me to come to the conclusion that :

$$ker \phi = \left \langle \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}\right \rangle $$

Thank you for telling me to edit

I just realised that for kernel 2nd part, it should be $$\begin{pmatrix} 0 &1 \\ 1 & 0 \end{pmatrix}$$

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  • $\begingroup$ You need to correct the second basis vector for $\ker \phi$. Since $\dim\operatorname{im}\phi=1$ you only need to find one non-zero vector in $\operatorname{im}\phi$ which then gives you the basis. $\endgroup$
    – leoli1
    Jan 17, 2021 at 13:55
  • $\begingroup$ It would be useful if you showed us how did you get that kernel. $\endgroup$
    – Kandinskij
    Jan 17, 2021 at 13:59
  • $\begingroup$ @Eureka just added the edits, if you have the conclusion to part b, it would be appreciated $\endgroup$
    – user844410
    Jan 17, 2021 at 14:07

1 Answer 1

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The first part is correct. It is easy to see that the kernel is the set of symmetric matrices, which for dimension $n$ has dimension $n(n+1)/2$.

Since the image is one dimensional, it suffices to find one nonzero element of it, and that will form a basis. For instance, we have that $\begin{pmatrix}1&1\\0&1\end{pmatrix}\mapsto \begin{pmatrix}0&1\\-1&0\end{pmatrix}$, and the latter matrix spans the image. Notice it is antisymmetric.

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  • $\begingroup$ Wow, this exercise provides a superelegant proof that $M_n(\mathbb{K})=\text{Sym}_n(\mathbb{K})\oplus \text{Asym}_n(\mathbb{K})$ $\endgroup$
    – Kandinskij
    Jan 17, 2021 at 14:14

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