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In review of linear algebra I come across this phenomenon, the Google Book link is this:

What I do not understand is Lax tried to persuade us that "there is another way of parametrizing these matrices, namely by their eigenvectors and eigenvalues.". I do not see how one can do this explicitly in symmetric real matrices.

He argue that "the first eigenvector,corresponding to the largest eigenvalue, depends on $n-1$ parameters; the second...depends on $n-2$ parameters...to the $n-1$st eigenvector that depends on one parameter. " But suppose we solved the characteristic equation and already know the eigenvalues, then solving $AX=\lambda_{1}X$ would involve $\frac{n(n+1)}{2}$ entries in $A$ right? This question is too naive to be post elsewhere, so I venture to post in here.

Similarly I could not get why Lax wrote "...all the way down to the last simple eigenvector that depends on two parameters. The remaining eigenspace is then uniquely determined". What does he meant that "depending on two parameters" and "remaining eigenspace is then uniquely determined?"

Last I also found the "avoidance of crossing" graph he gave to be ambiguous. According to his explanation, "..as $t$ approaches certain values of $t$, a pair of adjacent eigenvalues $a_{1}t$ and $a_{2}t$ appear to be on a collision course; yet at the last minute they turn aside." And the reason is "..a line or curve lying in $N$ dimensional space will in general avoid intersecting a sufrace depending on $N_2$ parameters." My questions are:

  • What does it mean "a line or curve will in general avoid intersecting a surface"? This feels totally via intuition and is not a rigorous proof, considering we could have knots, etc in the general space.
  • I tried to draw some graphs in Mathematica, but surprising I found the two eigenvalues do collide. I do not know why.

Try the command:

ParametricPlot[{Eigenvalues[{{450, 505}, {458.6, 545.3}} 
  + t {{0.325435, 0.4354334}, {0.4354353, 0.3453453}}], {t, t}}, 
  {t, 1500, -1500}, PlotPoints -> 1500]

(for unknown reason I cannot upload photos of the graph).

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1 Answer 1

This section of Lax's text deals with symmetric matrices, so the eigenvectors are orthogonal. We can assume each eigenvector has norm 1 since multiplying each one by a scalar will not change $A$.

I do not see how one can do this explicitly in symmetric real matrices.

Picking a symmetric matrix $A$ is the same as picking each of it's eigenvalues and (orthogonal) eigenvectors. Here are some details on how that goes:

When selecting an $n$-length eigenvector, you have an $n$-dimensional space to choose from. Each of the $n$ entries of the vector $(x_1,\dots,x_n)$ is a parameter you can choose. But if we assume the vector has norm 1, the last entry is determined by the others; you don't get to choose the last one because $x_n^2=1-\sum_{i=1}^{n-1}x_i^2$. So only $n-1$ parameters are really free.

If the first eigenvector is $v_1$, the second eigenvector must come from $span(v_1)^\perp$ (the space orthogonal to $v_1$). This is an $n-1$-dimensional space. The norm-1 restriction brings it down to $n-2$ parameters. The pattern continues until $v_{n-1}$ has 1 parameter and $v_n$ has 0. You wouldn't normally do this by hand, but a computer could do it explicitly.

The eigenvalues can be chosen independently from each other and from the eigenvectors. These are $n$ parameters.

Solving $AX=\Lambda X$ would involve $n(n+1)/2$ entries in $A$ right?

Yes, all the entries of $A$ are determined by the $n(n+1)/2$ in the upper triangle. Also, if you add the eigenvector and eigenvalue parameters, you have $\sum_{i=1}^n(i-1)+n=\frac{n(n+1)}{2}$ which is the same dimension as the space of $n\times n$ symmetric matrices.

If you mean $Ax=\lambda x$ when $x$ is a vector and $\lambda$ is a scalar, you can't really solve for $A$ since many matrices can share one eigenpair.

What does he meant that "depending on two parameters" and "remaining eigenspace is then uniquely determined?"

If two eigenvalues are the same (say $\lambda_{n-1}=\lambda_n$), the last simple eigenvalue is $\lambda_{n-2}$ whose eigenvector, we just argued, depends on 2 parameters (when the vectors get chosen in order).

Once $v_1,\dots,v_{n-2}$ have been chosen, the two-dimensional eigenspace associated with $\lambda_{n-1}=\lambda_n$ is determined; it's $span(v_1,v_2,\dots,v_{n-2})^\perp$. You could choose any pair of orthonormal vectors in that space for eigenvectors: $v_{n-1},v_n\in span(v_1,v_2,\dots,v_{n-2})^\perp$.

What does it mean "a line or curve will in general avoid intersecting a surface"? This feels totally via intuition and is not a rigorous proof, considering we could have knots, etc in the general space.

You're right that the last sentence in the chapter is an appeal to intuition rather than a proof of something specific. A line (or other curve) is "unlikely" to go through a particular $N-2$ dimensional space. In a plane ($n=2$), a line is unlikely to hit a particular point ($n-2=0$, and a 0-dimensional space is a point). In space ($n=3$), a line is unlikely to go through a particular line ($n-2=1$, and a 1-dimensional space is a line), and so on.

I tried to draw some graphs in Mathematica, but surprising I found the two eigenvalues do collide.

Your example actually exhibits eigenvalue avoidance, but you have an error. The Mathematica code you provided plots two lines: $(\lambda_1,\lambda_2)$ and $(t,t)$ for each $t$. What you want is $(t,\lambda_1)$ and $(t,\lambda_2)$. One way you could do this is splitting up the eigenvalues with Max and Min:

ParametricPlot[{ {t,Max[Eigenvalues[{{450,505},{458.6,545.3}}+t {{0.325435,0.4354334},{0.4354353,0.3453453}}]]}, {t,Min[Eigenvalues[{{450,505},{458.6,545.3}}+t {{0.325435,0.4354334},{0.4354353,0.3453453}}]]} },{t,-1500,-500}]

I've posted my the Matlab version of my figure below.

It's easy to invent cases where eigenvalues collide, it's just not likely if you pick "random" ones. Collision is as simple as $\left[\begin{array}{cc}1&0\\0&3\end{array}\right]+t\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]$ with $t=1$.

Matlab figure showing eigenvalue avoidance

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