Exercise · ZW-KERNEL-AND-IMAGE-001 · ★☆☆ Basic
Let \(T(x,y,z)=(x+y,\,y+z)\). Which of these vectors is in the kernel of \(T\)?
Choose one option.
Show answer
- C. \((1,\,-1,\,1)\)
Explanation
Just evaluate: \(T(1,-1,1)=(0,0)\), while the other three give a nonzero result.
Review the concept: Kernel and image
The kernel \(\ker T\) of a linear transformation \(T\) is the set of vectors it sends to zero; the image \(\operatorname{Im}T\) is the set of all vectors it produces.
\[\ker T=\{v: T(v)=0\},\qquad \operatorname{Im}T=\{T(v)\}\]
More practice problems: kernel and image
- ZW-KERNEL-AND-IMAGE-002 Let T(x,y,z)=(x,y,0). What is the image of T?
- ZW-KERNEL-AND-IMAGE-003 Let T(x,y)=(2x, 3y). What is the kernel of T?
- ZW-KERNEL-AND-IMAGE-004 What is true of the kernel of any linear transformation?
- ZW-KERNEL-AND-IMAGE-005 Let T(x,y)=(x, y, x+y) from ℝ^2 to ℝ^3. What is its image?
- ZW-KERNEL-AND-IMAGE-006 Let T(x)=Ax with A=[1 2 0; 2 4 0]. What is the dimension of its kernel?
- ZW-KERNEL-AND-IMAGE-007 Let T(x)=Bx with B=[1 0 1; 0 1 1; 1 1 2]. What is the dimension of its image?
- ZW-KERNEL-AND-IMAGE-008 Let T(x)=Ax with A=[1 2; 2 4]. Which of these vectors is in the image of T?
- ZW-KERNEL-AND-IMAGE-009 A linear transformation T is injective if and only if…