I have a linear transformation, given by the following matrix $$ \begin{pmatrix} x_1\\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} 2 & 2\\ -1 & -1\\ \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \end{pmatrix} $$

How can I determine what this corresponds to geometrically, when I apply it to the $x_1,x_2$-plane. I have tried to visualise the transformation by hand, and using a fieldplot in Maple, to get an idea of what is happening. My idea was then to decompose it into scaling, rotation, reflection or some other simple transformations.

My question is: Which geometric transformation does the above linear map correspond to, and in general, what is a good strategy for solving this kind of problem?

  • $\begingroup$ It seems like a projection into a line for me $\endgroup$ – nerdy Apr 5 '13 at 10:35

The better approach is to see what happens with the elements of the canonical base. So, you have two vectors $e_1=(1,0)$ and $e_2=(0,1)$. You have the unity square generated by them and its image could be a rectangle, or parallelogram, or a line, etc....

In our case, $e_1$ is mapped to the vector $(2,-1)$. The same for $e_2$. So the image is a line, namely, generated by $v=(2,-1)$.

  • 1
    $\begingroup$ +1. This is how I tend to read (and construct) linear transformation matrices. The columns tell me where the "axis" (base) vectors go. $\endgroup$ – Blue Aug 22 '12 at 4:47
  • $\begingroup$ So the transformation is a projection, but not an orthogonal one. $\endgroup$ – utdiscant Aug 22 '12 at 5:36
  • $\begingroup$ @utdiscant, yes. Orthogonal would be injective and your clearly isn't. $\endgroup$ – Sigur Aug 22 '12 at 11:16
  • $\begingroup$ Why would orthogonal be injective? Injective means that each vector in the codomain is hit by exactly one vector in the domain. But since multiple vectors map to the same projections, I do not get the injection argument? $\endgroup$ – utdiscant Aug 23 '12 at 2:38
  • $\begingroup$ I said that an orthogonal one would be injective. Your map is not injective so it is not orthogonal. $\endgroup$ – Sigur Aug 23 '12 at 11:51

In general, to see the geometric picture, find the Jordan canonical form.

If both eigenvalues are complex, then it is an elliptic rotation followed by dilation, similar to multiplication on the complex plane. If both are real and distinct, then it is obvious what it does in the eigenvector basis. If the Jordan form is degenerate, you will see other pictures, like, say, symplectic transvections.

In your case the matrix is obviously rank-one, so its eigenvalues are 1 and 0. It is a projection.


The determinant of the matrix of the linear transformation reveals a fair amount of the geometry. Consider $L: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ where $L(v)=Av$ for $A \in \mathbb{R}^{2 \times 2}$. Suppose $v_1,v_2$ are vectors which are the sides of a parallelogram $P$. It is a simple exercise to show that $Area(P) = det[v_1|v_2]$. Consider then that the linear transformation takes $v_1,v_2$ to the transformed vector(s) $L(v_1),L(v_2)$. The area of $L(p)$ is also given by $Area(L(P)) = det[L(v_1)|L(v_2)]$. Observe that

$$ Area(L(P)) = det[ Av_1|Av_2] = det(A[v_1|v_2])=det(A)det[v_1|v_2] = det(A)Area(P) $$

Consider the possibilities:

  1. $det(A)=0$ it follows the $L(P)$ is a point or a line ($A=0$ to get a point)
  2. $det(A) >1$ then the orientation of $P$ is maintained and it expands.
  3. $det(A) <-1$ then the orientation of $P$ is reversed and it expands.
  4. $0<det(A) <1$ then the orientation of $P$ is maintained and it shrinks.
  5. $-1<det(A)<0$ then the orientation of $P$ is reversed and it shrinks.

To describe the details of how $P$ is shrunk or expanded you can study the eigenvectors, generalized eigenvector and complex eigenvectors which arise in various cases as has been pointed towards in the other answers thus far.


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