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Let $\phi : \mathbb{R}^3 \to \mathbb{R}^4$ be given by $$\phi (x_1, x_2, x_3) = (3x_1 - x_2 + 2x_3, x_1 + x_2 - x_3, 4x_2-5x_3, 2x_1 - 2x_2 + 3x_3)$$ I want to find $\text{im}(\phi^*)$ and $\text{ker}(\phi^*)$, $\phi^* : (\mathbb{R}^4)^* \to (\mathbb{R}^3)^*$ where $(\mathbb{R}^4)^*$ and $(\mathbb{R}^3)^*$ are the dual bases of $\mathbb{R}^4$ and $\mathbb{R}^3$, respectively. This is the first time I encounter $\phi^*$.

If we write down the matrix of $\phi$

$$ M = \begin{bmatrix} 3 & -1 & 2 \\ 1 & 1 & -1 \\ 0 & 4 & -5 \\ 2 & -2 & 3 \\ \end{bmatrix} $$

are $\text{im}(\phi^*)$ and $\text{ker}(\phi^*)$ simply equivalent to, respectively, the column space and the nullspace of $M^T$?

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Yes, actually, the matrix associated to $\phi^*$ is $M^T$. So you can do the calculation as usual with $M^T$.

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    $\begingroup$ Note that the moment you change to a non-standard basis, it's not that simple. But in an orthonormal basis (all basis vectors of unit length and orthogonal to one another), it works. $\endgroup$
    – Arthur
    Jan 9, 2017 at 21:56
  • $\begingroup$ Actually, if $M$ is the matrix of $f$ in the usual basis of $\mathbb R^3$ and $\mathbb R^4$, $M^t$ is the matrix of $f^*$ in the dual basis. I'm supposing that's what they wanted to find. $\endgroup$
    – A. Salguero-Alarcón
    Jan 9, 2017 at 21:57

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