Exercise 1195Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Image of a basis under a linear map

#basis#image#linear map

Statement

Let $f:\mathbb{R}^2\to\mathbb{R}^3$ be the linear map defined by $f(1,0)=(1,2,3)$ and $f(0,1)=(0,1,1)$ . Compute a basis and the dimension of $\operatorname{Im} f$ .

Hint

The image of a basis spans the image. Then reduce to a basis.

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Solution

Generators. $\operatorname{Im} f=\operatorname{span}((1,2,3),(0,1,1))$ .

Independence. The two vectors are not proportional, so they are independent.

Conclusion. A basis of $\operatorname{Im} f$ is $((1,2,3),(0,1,1))$ , and $\dim\operatorname{Im} f=2$ .

Remark. $\operatorname{Im} f$ is a plane in $\mathbb{R}^3$ .

Image of a basis under a linear map

Statement

Solution

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