[proofplan] We prove both permanence properties directly from the definition of convexity. For the product, we take two points in $C \times D$ and show that every convex combination remains in the product by checking the two coordinates separately. For the linear image, we lift two points of $T(C)$ to preimages in $C$, use convexity of $C$, and then use linearity of $T$ to identify the convex combination of the images with the image of a convex combination. [/proofplan] [step:Check convex combinations componentwise in the product] Let $z_1,z_2 \in C \times D$ and let $t \in [0,1]$. By definition of the Cartesian product, there exist $x_1,x_2 \in C$ and $y_1,y_2 \in D$ such that \begin{align*} z_1 = (x_1,y_1), \qquad z_2 = (x_2,y_2). \end{align*} The convex combination of $z_1$ and $z_2$ in $\mathbb{R}^{m+n}$ is computed componentwise: \begin{align*} (1-t)z_1 + tz_2 &= (1-t)(x_1,y_1) + t(x_2,y_2) \\ &= \big((1-t)x_1 + tx_2,\,(1-t)y_1 + ty_2\big). \end{align*} Since $C$ is convex and $x_1,x_2 \in C$, we have $(1-t)x_1 + tx_2 \in C$. Since $D$ is convex and $y_1,y_2 \in D$, we have $(1-t)y_1 + ty_2 \in D$. Therefore \begin{align*} (1-t)z_1 + tz_2 \in C \times D. \end{align*} Because $z_1,z_2 \in C \times D$ and $t \in [0,1]$ were arbitrary, $C \times D$ is convex. [/step] [step:Use linearity to move convex combinations through the map] Let $u,v \in T(C)$ and let $t \in [0,1]$. By definition of the image $T(C)$, there exist $x,y \in C$ such that \begin{align*} u = T(x), \qquad v = T(y). \end{align*} Since $C$ is convex and $x,y \in C$, the point \begin{align*} w := (1-t)x + ty \end{align*} belongs to $C$. Using linearity of the map $T: \mathbb{R}^m \to \mathbb{R}^k$, we compute \begin{align*} (1-t)u + tv &= (1-t)T(x) + tT(y) \\ &= T\big((1-t)x\big) + T(ty) \\ &= T\big((1-t)x + ty\big) \\ &= T(w). \end{align*} Since $w \in C$, the point $T(w)$ belongs to $T(C)$. Hence $(1-t)u + tv \in T(C)$. Because $u,v \in T(C)$ and $t \in [0,1]$ were arbitrary, $T(C)$ is convex. [/step]