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Let $\varphi: \mathbb{R}^4 \to \mathbb{R}^4$ be a linear transformation given by $$\varphi(x_1, x_2,x_3,x_4)=(x_1+x_3-x_4,x_1+x_2+x_4,x_2-x_3+x_4,-x_1-2x_2+x_3-3x_4)$$ How do I find the bases of $\text{ker}(\varphi^*)$ and $\text{im}(\varphi^*)$?

The main problem for me here is that I know how to define $\varphi^*(\psi)=\psi(\varphi)$, but not $\varphi^*$ without an explicit second transformation. I would appreciate any comments.

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    I assume by $\varphi^*$ you mean the adjoint of $\varphi$, with the standard inner product $$\left\langle\left(x_1,x_2,x_3,x_4\right),\left(y_1,y_2,y_3,y_4\right)\right\rangle\mapsto x_1y_1+x_2y_2+x_3y_3+x_4y_4\quad\mathrm ? $$ – 2017-01-27
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    @Math1000 $\varphi^*$ is the adjoint of $\varphi$. Does this mean that $M(\varphi^*)=\left(M(\varphi)\right)^T$ in the standard basis? – 2017-01-27

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