# What is so special about eigenvector that it behaves like a pivot of linear transformation?

What is so special about eigenvector of matrix $$A$$ such that matrix $$A$$ fails to change its direction during linear transformation?

The eigenvector (say of matrix $$A$$) is such a vector in the vector space that it will only shrink or expand but not change its direction. It is like a pivot of transformation performed by $$A$$.

What allows eigenvector to be a pivot? It behaves as if it is independent of the transformation. Let's say transformation by multiplication with a $$2 \times 2$$ matrix changes $$\hat{\imath}$$ and $$\hat{\jmath\,}$$ of cartesian $$2$$-D space, then how is it possible that eigenvector, which is composed of combination of $$\hat{\imath}$$ and $$\hat{\jmath\,}$$, manages to stay unaffected?

• It is defined that way. When $Ax = \lambda x$, the matrix behaves like a scalar and $x$ is the eigenvector. Commented May 23 at 20:17
• @CroCo the color red is red because it is defined that way. yet it shouldn't stop you from exploring the light wavelength that makes it red.
– jam
Commented May 27 at 6:28
• ^you are mixing human definitions with things observed in nature. "The eigenvector (say of matrix $A$) is such a vector in the vector space that it will only shrink or expand but not change its direction" Basically you are looking for vectors that satisfy this equation $Ax=\lambda x$. Commented May 27 at 20:34

I suppose one way to look at it is that it's the vector that has the right combination of $$\hat{\imath}$$ and $$\hat{\jmath}$$ such that the stretching and skewing caused by $$A$$ on each of those basis vectors balances out just right.

We find the eigenvector $$\mathbf{v}$$ by solving the equation $$A\mathbf{v} = \lambda\mathbf{v}$$, which can also be written as $$(A - \lambda I) \mathbf{v} = \mathbf{0}$$, so once we have an eigenvalue $$\lambda$$ this is just a set of linear equations that we can easily solve for $$\mathbf{v}$$.

You can also look at it the other way around - if we have two linearly independent eigenvectors $$\mathbf{u}$$ and $$\mathbf{v}$$ with associated eigenvalues $$\mu$$ and $$\lambda$$, then we can use them as a basis for $$\mathbb{R}^2$$, meaning that any vector $$\mathbf{w}$$ can be written as a linear combination of them. If we write one the standard basis vectors as $$\hat\imath = a_1 \mathbf{u} + a_2 \mathbf{v}$$, then $$A \hat\imath = A(a_1\mathbf{u} + a_2\mathbf{v}) = a_1 \mu \mathbf{u} + a_2 \lambda \mathbf{v}$$, so the skewing caused by $$A$$ is really a result of the different mixture of $$\mathbf{u}$$ and $$\mathbf{v}$$ getting scaled by different amounts thanks to their eigenvalues. In other words, in the $$(\mathbf{u}, \mathbf{v})$$ basis, $$A$$ is just a scaling factor of $$(\mu, \lambda)$$.

• "the vector that has the right combination of ı̂ and ȷ̂ such that the stretching and skewing caused by 𝐴 on each of those basis vectors balances out just right." can you please provide an example to illustrate this intrinsic property for real eigenvector? Another point is that lets say I perform simple 45 degree rotation of 2D plane such that new j^ is (a,a) and new i^ is (a,-a) where a is sqrt(1/2). Now eigenvector contains an element that belongs to complex plane . so real plane rotates along a pivot thats in complex 2D plane? how does it fit together
– jam
Commented May 16 at 11:53
• I'd love to provide a diagram, but I don't really have that kind of skill unfortunately. And yes, if you've got complex eigenvalues then this is happening in $\mathbb{C}^2$ which is harder to visualise. Commented May 17 at 0:10

The key is in the definition of the eigenvalues/eigenvectors. In typical linear algebra textbooks, the eigenvalues/eigenvectors are defined as follows

Applied Linear Algebra, second edition by Olver and Shakiban

Definition8.2. Let $$A$$ be an $$n\times n$$ matrix. A scalar $$\lambda$$ is called an eigenvalue of $$A$$ if there is a non-zero vector $$\boldsymbol{\mathrm{v}} \neq \boldsymbol{0}$$, called an eigenvector, such that $$A\boldsymbol{\mathrm{v}} = \lambda\boldsymbol{\mathrm{v}}.$$

LINEAR ALGEBRA A Geometric Approach, second edition by Shifrin and Adams

Definition. Let $$T : V \rightarrow V$$ be a linear transformation. A nonzero vector $$\boldsymbol{\mathrm{v}} \in V$$ is called an eigenvector of $$T$$ if there is a scalar $$\lambda$$ so that $$T(\boldsymbol{\mathrm{v}}) = \lambda \boldsymbol{\mathrm{v}}$$. The scalar $$\lambda$$ is called the associated eigenvalue of $$T$$.