# Finding rank of linear tranformation without a matrix?

A practice problem I have asks us to find the rank of the linear transformation:

$$T(x, y, z) = (x - 2y + z, 2x + y + z)$$

What I'm confused on is, thus far in my class, as far as I can recall, we've only ever found the rank of linear transformations in the form $T(x) = Ax$, where you have a matrix $--$ and you put the matrix in reduced row-echelon form and see how many leading $1$'s there are in order to determine the rank. So since we don't have a matrix here, I'm not sure how I would solve this.

I'm getting the sense that it's possible to put ANY linear transformation into matrix form, but the problem is my book hasn't reached that point yet; I'm pretty sure that's in the next section. So I don't know how to represent this transformation with a matrix (and I don't think I'm supposed to since we haven't learned that yet), but I don't know how I would solve this problem without doing that.

The problem also says NOT to compute the kernel, so I can't find the kernel and then use the rank-nullity theorem to find the rank, either.

I'm at all loss as to how I would do this. I've flipped through the section and all the problems in my book and I don't see any examples that compute the rank WITHOUT having a matrix.

Is there another way to compute the rank of this that I'm missing?

• But you have instantaneously the matrix because you can write your function under the form: $\begin{pmatrix}1&-2&1\\2&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$... – Jean Marie Feb 2 '17 at 11:07
• Yeah but the book hasn't done anything like that yet and makes a point of saying we'll learn how to represent any transformation as a matrix in the next section, so I don't think I'm supposed to do it that way yet. Is there a way to do it without representing the transformation in a matrix? If there's not, I suppose I'll just assume we're supposed to do it your way – dagny Feb 2 '17 at 11:11
• The rank of a transformation is the dimension of its image. – Arthur Feb 2 '17 at 11:19
• I would not worry too much about what you think the book wants you to think. You are a free thinker! (And I have managed to use the verb "think" more times here than in the rest of my posts :-) ). – Giuseppe Negro Feb 2 '17 at 11:52

Quite simple: $T(-1,0,2)=(1,0)$ and $T(1,0,-1)=(0,1)$, i.e. the canonical basis of $\mathbf R^2$ is attained, so the rank is $2$.

Vectors $\begin{pmatrix}1\\2\end{pmatrix}$ and $\begin{pmatrix}-2\\1\end{pmatrix}$, being "the images of something", more precisely of : $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}$ and $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\1\\0\end{pmatrix}$ resp. are in the image space.

Moreover, these vectors are independent. Thus, the rank r, which is by definition the dimension of the image space, is at least 2 but cannot be more than that because it cannot exceed the dimension of the "ambient" space $\mathbb{R^2}.$

A confirmation is brought by the fact that $\begin{pmatrix}-3\\1\\5\end{pmatrix}$ belongs to the kernel. Thus the dimension of the kernel is $k \geq 1$ and in fact no more than $1$ because of the rank-nullity theorem ($r+k=2+1=3$.)

Edit: I place here the matrix representation of the linear transformation:

$$\begin{pmatrix}1&-2&1\\2&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$$

$(x,y,z) \in ker(T)$ iff $(x - 2y + z, 2x + y + z)=(0,0)$

Its your turn to solve the linear system

$x - 2y + z=0$

$2x + y + z=0$.

Then you should see that $\dim ker(T)=1$. From

$3= \dim \mathbb R^3= \dim ker(T)+rank(T)$ we get $rank(T)=2$.

The rank is either $0, 1$ or $2$. $\begin{cases} \operatorname{rank}(T) = 0\Rightarrow T = (0,0)\\ \operatorname{rank}(T) = 1 \Rightarrow x - 2y + z = \alpha(2x + y + z) & \text {not possible} \\ \operatorname{rank}(T) = 2 &\text{the only possibility left}\end{cases}$