# Surjective linear transformations: is checking for linear independency enough?

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?

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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

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.
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 $$\operatorname{im}T=\operatorname{span}\left((0,0,1)^T,(1,1,1)^T\right)$$