Exercise I.C.10

Question

Let $\phi: \mathbb{R}^3	o\mathbb{R}$ be such that $\phi(e_1)=1, \phi(e_2)=2, \phi(e_3)=-1$. Determine $\ker \phi, \mathrm{im }\phi$ and verify the rank theorem in this case.

Toğrul Topal · Accepted Answer

We have
    \begin{align*}
      \mathrm{im } \phi = \left\{ \phi\begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} \in \mathbb{R}: x_1, x_2, x_3 \in \mathbb{R} \right\} &= \left\{ x_1 \phi\begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} + x_2 \phi\begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} + x_3 \phi\begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \in \mathbb{R}: x_1, x_2, x_3 \in \mathbb{R} \right\}\[5pt]
                                                                                                                                            &= \left\{ x_1  + 2x_2 - x_3 \in \mathbb{R}: x_1, x_2, x_3 \in \mathbb{R} \right\}
    \end{align*}
    These covers all of the real line (for instance, set $x_2 = x_3 = 0$ and modify $x_1$ as you wish), and so $\dim(\mathrm{im }\phi) = 1$. Now we look at the kernel:
    \begin{align*}
      \ker \phi &= \left\{ \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} \in \mathbb{R}^3: x_1 + 2x_2 - x_3 = 0 \right\} \[5pt]
                &= \left\{ \begin{bmatrix} x_1 \ x_2 \ x_1+2x_2 \end{bmatrix} \in \mathbb{R}^3: x_1,x_2 \in \mathbb{R} \right\} \[5pt]
                &= \left\{ x_1\begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} + x_2\begin{bmatrix} 0 \ 1 \ 2 \end{bmatrix} \in \mathbb{R}^3: x_1,x_2 \in \mathbb{R} \right\} \[5pt]
                &= \mathrm{span}\left\{ \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}, \begin{bmatrix} 0 \ 1 \ 2 \end{bmatrix} \right\}
    \end{align*}
    We have found a set of two linearly independent vectors whose span is equal to the kernel of $\phi$; i.e., we have $\dim(\ker \phi) = 2$. This verifies the dimension theorem.

Exercise I.C.10

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Exercise I.C.10

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