I am wondering about the geometric intuition behind real symmetric matrices and their corresponding linear transformations.

Is it possible to understand geometrically why real symmetric matrices have only real eigenvalues? That is, what do symmetric linear transformations have in common geometrically that make this true?

I am NOT after a proof of this fact; what I am curious about is whether there is a geometric argument for it. I am after the sort of intuition one gets from looking at linear transformations in $\mathbb{R}^2$ and envisioning their eigenvalues and eigenvectors. For example, it is very intuitively clear why nontrivial rotation matrices cannot have real eigenvectors. It is also geometrically clear why diagonal matrices have their eigenvalues equal to their diagonal elements.

EDIT: Upon further thought, I have realized that a simple consequence of symmetry is that in the singular value decomposition of a matrix $M = U \Sigma V^*$, we have $U = V$. Thus requiring symmetry means that the linear transformation must be accomplished by performing a rotation, performing a scaling along the axes, and then reversing the original rotation.

The missing piece for me is why defining a transformation via a symmetric matrix means that the transformation can be decomposed in this simple way.

  • $\begingroup$ You must be careful, actually, as $U$ is not the same as $V$ if $M$ has negative eigenvalues. In this case you have to absorb the extra minus sign(s) into either $U$ or $V$. $\endgroup$ – Ian Aug 29 '15 at 16:21
  • $\begingroup$ Anyway, I think you can reduce the situation to showing that a symmetric matrix has one real eigenvalue and eigenvector, for then you can proceed by an inductive argument based on projections to see that it has a full set. Basically, the action on the orthogonal complement of the eigenspace is also given by a symmetric matrix, so you get another real eigenvalue, and the process repeats. As for showing that it has one, the most geometric argument I can think of would be to consider the extrema of the Rayleigh quotient over the sphere. $\endgroup$ – Ian Aug 29 '15 at 16:23
  • $\begingroup$ An additional mystery (at least to me): Whether a linear transformation is represented by a symmetric matrix depends on the choice of basis. But whether a linear transformation has real eigenvalues is basis-independent. $\endgroup$ – Rahul Apr 16 '17 at 5:07

A symmetric matrix with real entries is orthogonally diagonalizable.

In $\Bbb{R}^2$, $A$ is orthogonally diagonalizable means that $A$ has two orthonormal eigenvectors $u_1$ and $u_2$ with eigenvalue $\lambda_1$ and $\lambda_2$ respectively.

Suppose $x=c_1 u_1+c_2 u_2$, then $Ax=\lambda_1 c_1 u_1+\lambda_2 c_2 u_2$.

If $\lambda_1>0$ and $\lambda_2>0$. then $Ax$ can be gotten by stretching the plane (like a flat balloon) along the direction $u_1$ with $\lambda_1$ scale and along the direction $u_2$ with $\lambda_2$ scale.

enter image description here

The another equivalent aspect is as you said. Since $A=PDP^T$, where $P$ is orthogonal (stand for the rotation), transforming under $A$ is rotation, then scaling and revesing the rotation. But as Ian's comment, if there is a negative eigenvalue, we should choose the opposite direction when we were scaling. For example, $$A=\begin{pmatrix} 5 & -2\\ -2& 8\\ \end{pmatrix} =PDP^T= \begin{pmatrix} \frac{2}{\sqrt{5}} & \frac{-1}{\sqrt{5}}\\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}}\\ \end{pmatrix} \begin{pmatrix} 4 & 0 \\ 0 & 9 \\ \end{pmatrix} P^T.$$ $\theta\approx 26.6^{\circ}$.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.