Exercise 1208Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Kernel and image of a linear map

#kernel#image#linear map#rank-nullity

Statement

Let $f:\mathbb{R}^3\to\mathbb{R}^3$ be defined by
$f(x,y,z)=(x+y, y+z, x+2y+z).$
Compute $\ker f$ , $\operatorname{Im} f$ , and check the rank-nullity theorem.

Hint

Write the matrix of $f$ and compute its rank by row reduction.

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Solution

Matrix.
$A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 1 \end{pmatrix}.$

Row reduction. $L_3\leftarrow L_3-L_1$ :
$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}.$
$L_3\leftarrow L_3-L_2$ :
$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}.$
Rank is $2$ .

Kernel. Solve $Ax=0$ : $x+y=0$ and $y+z=0$ , so $y=-x$ and $z=x$ . Kernel is $\operatorname{span}((1,-1,1))$ . Dimension $1$ .

Image. Spanned by the first two columns: $(1,0,1)$ and $(1,1,2)$ . Dimension $2$ .

Rank-nullity check. $2+1=3=\dim\mathbb{R}^3$ .

Conclusion. $\ker f=\operatorname{span}((1,-1,1))$ and $\operatorname{Im} f=\operatorname{span}((1,0,1),(1,1,2))$ .

Kernel and image of a linear map

Statement

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