Linear transformations can be represented using matrix, like $$v = Au$$, which transforms vector $u$ into $v$. And my intuitive understanding about linear transformations is that, it rotates the vector $u$ by some degrees and meanwhile stretches it by some scales. But if $u$ is the eigenvector, only stretching without rotating.

Here comes my problem,

  1. So a linear transformation has just 2 effects on a vector, that is streching and rotating, right?

  2. Generally how a vector $u$ will be rotated? Someone tells me that general vector will be rotated as much as parallel with the eigenvector, what if there're more than one eigenvector pointing to different directions?

  3. Does the determinant of $A$ reveals how much a vector will be stretched?

  • $\begingroup$ You should think about the action on more than one vector at a time (ideally on $n$ "general" vectors, where $n$ is the dimension of the space $A$ acts on). For example, two vectors in $\mathbb{R}^2$ (as long as they are not parallel) define a parallelogram, and you can look at the effect of $A$ on this parallelogram. As an example, its area will change by multiplying by $\det{A}$. The order of the two vectors determines an orientation, which is either preserved or reversed by $A$ depending on whether $\det{A}>0$ or $\det{A}<0$. If $\det{A}=0$, then stranger things can happen. $\endgroup$ – mdp Sep 24 '13 at 7:49
  • $\begingroup$ @MattPressland, does the eigenvector indicate in which way a general vector will change? $\endgroup$ – avocado Sep 24 '13 at 9:41
  • $\begingroup$ The collection of all (generalized) eigenvectors will, but there will only be so much information you can get from a single eigenvector. I'm hoping someone who can express this better than me will write an answer to your question, but maybe if that hasn't happened in a few days I'll try myself. $\endgroup$ – mdp Sep 24 '13 at 9:46
  • $\begingroup$ @MattPressland, please try now or at least show me any available tutorial or intro on the web. $\endgroup$ – avocado Sep 24 '13 at 10:52
  • $\begingroup$ I don't know a good resource off the top of my head, and I don't want to say things that are confusing or wrong, so I need some more time to work out what the best thing to say is - sorry. $\endgroup$ – mdp Sep 24 '13 at 10:56

Hopefully some examples will help. The overriding theme is that the action of a matrix is determined by its action on a basis - in case you don't know what a basis is, it's a set of vectors such that every vector in the space can be written uniquely as a linear combination of those in the set. For the examples, I'll work in $\mathbb{R}^2$, and the most useful basis will be $e_1=\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)$ and $e_2\left(\begin{smallmatrix}0\\1\end{smallmatrix}\right)$.

The matrices I will be interested in are:

$$A_1=\begin{pmatrix}2&0\\0&2\end{pmatrix}\qquad A_2=\begin{pmatrix}2&0\\0&0\end{pmatrix}\qquad A_3=\begin{pmatrix}2&1\\0&2\end{pmatrix}$$

All of these matrices have eigenvector $e_1$ with eigenvalue $2$, but I aim to show that they have very different properties, by considering their action on the basis $\{e_1,e_2\}$ (or, to take a different point of view, the square they determine - this is the $1\times 1$ square with bottom-left corner at the origin).

The matrix $A_1$ is just twice the identity, so both of our basis vectors (and indeed every vector) are eigenvectors for $A_1$ with eigenvalue $2$. So this matrix stretches the plane by a factor of $2$ away from the origin. Our square has the lengths of all of its sides doubled, and so the area is scaled by a factor of $4=\det{A_1}$.

When we apply the matrix $A_2$, it doubles the length of $e_1$, but maps $e_2$ to zero. Our square is collapsed to a line of length $2$ in the direction of $e_1$, which has no area. So the area has been multiplied by a factor of $0=\det{A_2}$. This is very different from the behaviour of $A_1$, but we wouldn't have seen this difference if we only considered the vector $e_1$.

The matrix $A_3$ acts in a slightly more complicated way, doubling the length of $e_1$ as before, but mapping $e_2$ to $e_1+2e_2$. Our square is stretched in the $e_1$ direction, but also tilted away from the $e_2$ direction, to give a parallelogram (which must have area $4=\det{A_3}$). Unlike in the other two cases, $e_2$ is not an eigenvector, and indeed there is no basis of $\mathbb{R}^2$ consisting only of eigenvectors of $A_3$. (The vector $e_2$ is however a generalized eigenvector).

You may also be interested in Perron-Frobenius theory (although the wiki article is - at the time of writing - a little technical) which explains that if all the matrix entries are positive there is a unique largest eigenvalue, and the effect of multiplying by the matrix repeatedly is quite closely linked to this eigenvalue and its eigenvectors.

  • $\begingroup$ Thank you so much, there is still something for me to figure out in your answer,:-) $\endgroup$ – avocado Sep 26 '13 at 13:42

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