2

Normally, we can diagonalize a matrix A by $A = S \Lambda S^{-1}$.

When the matrix is symmetric, we have that $A = Q \Lambda Q^{-1} = Q \Lambda Q^{T}$.

The part I'm failing to understand is: why can we just make the eigenvector matrix $S$ a set of orthonormal vectors?

Looking at the relationship $Ax = \lambda x$, if I changed the magnitude of $x$, wouldn't I then require a different $\lambda$ to satisfy the equation? Wouldn't that also have to change $\Lambda$ in the diagonalization of $A$ where $A$ is symmetric?

wheresmycookie
  • 133
  • 4
  • 1
    I'm not sure what you are asking. It's a well-known theorem that a real symmetric matrix can be diagonalized by an orthogonal matrix. Are you unacquainted with the theorem, do you not understand its proof, or are you looking for some intuition as to why it's true? – saulspatz Feb 24 '19 at 14:42
  • @saulspatz the third thing - I'm looking for insight into why I can scale the eigenvectors down to orthonormal vectors when the matrix is symmetric. I can see the theorem but I want some intuition as to why it's true. – wheresmycookie Feb 24 '19 at 14:58
  • $\mu Ax=A(\mu x)=\lambda(\mu x)=\mu(\lambda X)$. –  Feb 24 '19 at 14:58

2 Answers2

1

The part I'm failing to understand is: why can we just make the eigenvector matrix $S$ a set of orthonormal vectors?

If $A$ is symmetric and $x,y$ are eigenvectors with eigenvalues $\lambda_x,\lambda_y$ then

$$\lambda_y \langle x,y\rangle= \langle x,Ay \rangle=\langle Ax,y \rangle=\lambda_x \langle x,y\rangle.$$ If $\lambda_x\ne \lambda_y$ it must be $\langle x,y\rangle=0.$

Looking at the relationship $Ax = \lambda x$, if I changed the magnitude of $x$, wouldn't I then require a different $\lambda$ to satisfy the equation? Wouldn't that also have to change $\Lambda$ in the diagonalization of $A$ where $A$ is symmetric?

If $y=kx$ and $Ax=\lambda x$ then $Ay=\lambda y.$ Note that $Ay=A(kx)=kAx=k\lambda x=\lambda k x=\lambda y.$

mfl
  • 29,399
  • Will upvote if you can verify my understanding :) That is, $\lambda$ holds true not for a single vector, rather all scalar multiples of that vector. That's the key piece that I was missing! – wheresmycookie Feb 24 '19 at 15:05
  • 1
    @wheresmycookie Yes, all scalar multiples of an eigenvector (except the zero multiple) are also eigenvectors with the same eigenvalue. – Ian Feb 24 '19 at 15:08
  • Yes. If $\lambda$ is an eigenvalue of $A$ and $Ax=\lambda x$ then all multiples of $x$ verify the same equality. That is, if $y$ is a multilple of $x$ then $Ay=\lambda y.$ – mfl Feb 24 '19 at 15:08
  • Wow, now that I worked through the last part of your answer, I can see it clearly. For some reason that wasn't immediately obvious to me. Thanks! – wheresmycookie Feb 24 '19 at 15:09
  • You're welcome. – mfl Feb 24 '19 at 15:09
1

$x$ appears on both sides of the equation $Ax=\lambda x$, hence changing the "magnitude" of $x$ just results in another eigenvector for eigenvalue $\lambda $. That is, if $x$ is an eigenvector for eigenvalue $\lambda $, then so is $\alpha x$ for any scalar $\alpha \neq0$.

All symmetric matrices are diagonalizable. Note that for a symmetric matrix, since eigenvectors for distinct eigenvalues are orthogonal, it is possible to get an orthonormal basis of eigenvectors.

  • Thanks for the response Chris. As to your second point, why do we write $A = Q \Lambda Q^T$ only in the special case that A is symmetric then? – wheresmycookie Feb 24 '19 at 15:12
  • Sure. This can always be done when $A$ is diagonalizable, because we can take the basis of eigenvectors and apply Gram-schmidt, say. –  Feb 24 '19 at 15:15
  • I'm looking at this and seeing that "For a symmetric matrix with real number entries, the eigenvalues are real numbers and it’s possible to choose a complete set of eigenvectors that are perpendicular (or even orthonormal)". Diagonalizing a matrix involves pulling out its eigenvectors into an $S$ matrix, right? But this quote seems to imply that I can only guarantee orthonormal when symmetric? – wheresmycookie Feb 24 '19 at 15:29
  • Otherwise, wouldn't I be able to write $A = Q \Lambda Q^T$ for any matrix? (I'm probably wrong, but this is just where I'm stuck) – wheresmycookie Feb 24 '19 at 15:30
  • You're right. It can't always be done. See this https://math.stackexchange.com/a/514961. My mistake. –  Feb 24 '19 at 15:42
  • Thanks for the link Chris, clears things up! – wheresmycookie Feb 24 '19 at 16:01