Answer
FalseIf T : P4(R) → R7 is a linear transformation
Work Step by Step
Assume a linear transformation $T$:
$T(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4)=(a_0,a_1,a_2,a_3,a_4,0,0)$
Obtain:
$Ker T=\{ p \in P_4(R):T(p)=0\}\\
=\{ p\in P_4(R):(a_0,a_1,a_2,a_3,a_4,0,0)=(0,0,0,0,0,0,0)\\
=\{ p \in P_4(R):a_0=a_1=a_2=a_3=a_4=0\}\\
=\{ 0 \}$
Hence, $\dim [Ker (T)] =0\lt 2$
Thus, the statement is false.