# Sketching Graphs using Eigenvalues

## Homework Statement

For the conic, 5x2+4xy+5y2=9, find the direction of the principal axes, sketch the curve.

I found the eigenvalues as
3,7 but have no idea whether the 'new' equation is
3(x')2+7(y')2
or
7(x')2+3(y')2
is there a way to determine which 'way' it goes?

I took a guess and just continued using the first formula:
I found the eigenvectors by substituting the eigenvalues and got:

λ=3, V1 = (-1,1)
λ=7, V2 = (1,1)

I then thought the principal axes would therefore be:
1/√2 (-1,1) and 1/√2 (1,1)
yet the answer seems to indicate the principal axes as, (1,-1) and (1,1), why is that? I thought you had to normalise the vectors to find the principal axes..

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haruspex
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no idea whether the 'new' equation is
3(x')2+7(y')2
or
7(x')2+3(y')2
is there a way to determine which 'way' it goes?
x' and y' are just names. They don't relate in any particular way to x and y respectively. Might as well be u and v. So the two versions you offer are really the same.
I then thought the principal axes would therefore be:
1/√2 (-1,1) and 1/√2 (1,1)
yet the answer seems to indicate the principal axes as, (1,-1) and (1,1), why is that?
They're just directions. I don't see any requirement for them to be normalised.

HallsofIvy
Homework Helper

## Homework Statement

For the conic, 5x2+4xy+5y2=9, find the direction of the principal axes, sketch the curve.

I found the eigenvalues as
3,7 but have no idea whether the 'new' equation is
3(x')2+7(y')2
or
7(x')2+3(y')2
is there a way to determine which 'way' it goes?

I took a guess and just continued using the first formula:
I found the eigenvectors by substituting the eigenvalues and got:

λ=3, V1 = (-1,1)
λ=7, V2 = (1,1)

I then thought the principal axes would therefore be:
1/√2 (-1,1) and 1/√2 (1,1)
yet the answer seems to indicate the principal axes as, (1,-1) and (1,1), why is that? I thought you had to normalise the vectors to find the principal axes..
"Axes" are lines not vectors- they do not have a specific length nor can they be written as vectors. The axes are neither "$(1/\sqrt{2})(-1, 1)$" and $(1/\sqrt{2})(1, 1)$ nor "(-1, 1) and (1, 1)". The axes are lines in the directions of those vectors: y= -x in the direction of (-1, 1) and y= x in the direction of (1, 1).

Okay, that seems to make more sense. So does order really not matter after we've found the eigenvalues?

Also, I asked my lecturer this, but she couldn't properly explain the concept of diagonlisation using nullity.
Example.
|2 1 0|
|0 2 0|
|0 0 -3|

I'm trying to determine whether that is diagonizable or not, but have trouble understanding how to do it using the rank, nullity theorem.

HallsofIvy
If $\begin{bmatrix}x \\ y \\ z \end{bmatrix}$ is an eigenvalue corresponding to eigenvalue 2, then we must have
$$\begin{bmatrix}2 & 2 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -3\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= 2\begin{bmatrix}x \\ y \\ z \end{bmatrix}$$
$$\begin{bmatrix}2x+ 2y \\ 2y \\ -3z \end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}$$
which is equivalent to the three equations 2x+ 2y= 2x, 2y= 2y, and -3z= 2z. The first equation says y= 0 and the third equation says z= 0. x is undetermined so eigenvectors corresponding to eigenvalue 2 are all multiples of $\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$. The eigenspace has dimension 1 so the matrix is NOT diagonalizable.