Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is the linear transformation $T: \mathbb{R}^3\to\mathbb{R}^3$ defined as

$$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}$$

Injective and/or surjective?

Well, for injection, the solution to

$$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

Must be

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$

In this case, it isn't true - because the solution is actually

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix}t \\ 0 \\ t \end{bmatrix}$$

For some $t \in \mathbb{R}$.

For surjection, the image of $T$ must be $\mathbb{R}^3$.

I am not sure how to determine this. I have heard something like "the matrix must span $\mathbb{R}^3$".

Well then, in that case, the matrix must be linearly independent I guess... which isn't true because clearly the first column is simply a scalar product of the third column (by $-1$). Hence the matrix only has two "useful" vectors.

... so it isn't surjective. Now, I am not quite sure about this - isn't it possible for a matrix to span $\mathbb{R}^n$ if it has a number of "useful" vectors lesser than $n$?

In other words: is a check for linear independency enough to determine linear transformation surjection?

share|cite|improve this question
For injectivity $(0,0,0)$ must be the only solution. Also, don't mess scalar product with product by a scalar. – mfl Jul 6 '14 at 23:27
Remember the dimension formula in the case of endomorphisms. – gnometorule Jul 6 '14 at 23:29
up vote 2 down vote accepted

The answer is that it depends. The dimension of the image of a linear map is the dimension of its column space. In this case, the dimension of the column space (as you have pointed out) is $2$, so it cannot span $\mathbb{R}^3$. But for example, a $3\times 4$ matrix cannot have all its columns linearly independent (it can have at most $3$ linearly independent columns), yet its image can be all of $\mathbb{R}^3$.

It's worth pointing out that in the specific case you have, where the domain and range both have dimension $3$, that the fact that the map is not injective also implies that it cannot be surjective.

share|cite|improve this answer

If an endomorphism isn't injective then it isn't also surjective by the rank-nullity theorem: $$\dim \Bbb R^3=\dim \ker T+\dim \operatorname{im}T$$ so since $\dim \ker T=1$ hence $\dim \operatorname{im}T=2$ and we see from the columns of the given matrix that $$\operatorname{im}T=\operatorname{span}\left((0,0,1)^T,(1,1,1)^T\right)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.