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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?

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  • $\begingroup$ It is defined that way. When $Ax = \lambda x$, the matrix behaves like a scalar and $x$ is the eigenvector. $\endgroup$
    – CroCo
    Commented May 23 at 20:17
  • $\begingroup$ @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. $\endgroup$
    – jam
    Commented May 27 at 6:28
  • $\begingroup$ ^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 $Ax=\lambda x$. $\endgroup$
    – CroCo
    Commented May 27 at 20:34

2 Answers 2

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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 $(\mu, \lambda)$.

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  • $\begingroup$ "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 $\endgroup$
    – jam
    Commented May 16 at 11:53
  • $\begingroup$ 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. $\endgroup$
    – ConMan
    Commented May 17 at 0:10
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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 $T$.

The above definitions answer your questions.

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