How do I know if my eigenvectors are right?

[David Koufos]

Homework Statement

Given $M = \begin{pmatrix} 2 & 2\\ 2 & -1 \end{pmatrix}$, find the eigenvalues and eigenvectors and use them to find the matrix which gives the deformation relative to the new axes.

Relevant Equations

$CMC^{T} = D$,
C is the matrix whose columns are the unit eigenvectors derived from M,
D is a diagonal matrix which gives the deformation relative to the new axes,
Remember $C^{T} = C^{-1}$, meaning C is orthogonal.

For
$M = \begin{pmatrix} 2 & 2\\ 2 & -1 \end{pmatrix}$

I found the characteristic equation:
$( λ - 3 )( λ + 2) \therefore λ = 3,-2$Going back we multiply
$$\begin{pmatrix}
2 - \lambda & 2\\
2 & -1 - \lambda
\end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}$$

Which gives
\begin{matrix}
2x - \lambda x + 2y = 0\\
2x - y - \lambda y = 0
\end{matrix}

Now plugging in $\lambda_{1}$ and $\lambda_{2}$ we get
\begin{matrix}
-x_{1} + 2y_{1} = 0\\
2x_{1} - 4y_{1} = 0
\end{matrix}

\begin{matrix}
4x_{2} + 2y_{2} = 0\\
2x_{2} + y_{2} = 0
\end{matrix}

Using this system of equations I derived
$\left( 2, 1 \right )$ & $\left( -1, 2 \right )$ for the eigenvectors. My question comes from this.

Moving on, the unit eigenvectors are
\begin{bmatrix}
\dfrac{2}{\sqrt{5}}& , \dfrac{1}{\sqrt{5}}
\end{bmatrix} and
\begin{bmatrix}
\dfrac{-1}{\sqrt{5}}& , \dfrac{2}{\sqrt{5}}
\end{bmatrix}The matrix C which is composed of the unit eigenvectors as columns is
$$C = \begin{pmatrix}
\dfrac{2}{\sqrt{5}} & \dfrac{-1}{\sqrt{5}}\\
\dfrac{1}{\sqrt{5}} & \dfrac{2}{\sqrt{5}}
\end{pmatrix}$$

and
$$C^{T} = \begin{pmatrix}
\dfrac{2}{\sqrt{5}} & \dfrac{1}{\sqrt{5}}\\
\dfrac{-1}{\sqrt{5}} & \dfrac{2}{\sqrt{5}}
\end{pmatrix}$$

But when I perform the multiplication I get
$$CMC^{T} = \begin{pmatrix}
\dfrac{-1}{5} & \dfrac{12}{5}\\
\dfrac{12}{5} & \dfrac{6}{5}
\end{pmatrix}$$ which is not a diagonal matrix.

I used maxima to compare answers and realize why my answer wasn't right. Maxima found for eigen vectors
$[ 2, 1 ] [ 1, -2]$ whereas I have $[ 2, 1] [ -1, 2]$.

When using the eigen vectors the computer found, I get
$CMC^{T} = \begin{pmatrix} 3 & 0\\ 0 & -2 \end{pmatrix}$ which is the solution I'm looking for.
My problem is that my eigen vectors satisfy the characteristic equation so why don't I get the same answer. For future problems how do I really know if my eigen vectors are right, or if they're off by a factor of -1?

 

[FactChecker]

Check them using the basic definition of the eigenvector and the associated eigenvalue. Check if the matrix times the eigenvector equals the eigenvalue times the eigenvector.

 

[David Koufos]

Check them using the basic definition of the eigenvector and the associated eigenvalue. Check if the matrix times the eigenvector equals the eigenvalue times the eigenvector.

I just tried that with this specific example, and using my 2nd eigen vector $(-1, 2)$ I do indeed get that [M]u = λu. So now I'm just more puzzled. I don't understand why my eigen vector doesn't yield the desired diagonal matrix. I know that you can just assume the diagonalized matrix is just the eigen values along its diagonal and zeroes elsewhere, but my book says that's not true %100 of the time so I want to make sure I have the process right before I move on to higher dimensions.

 

[docnet]

I know that you can just assume the diagonalized matrix is just the eigen values along its diagonal and zeroes elsewhere, but my book says that's not true %100 of the time so I want to make sure I have the process right before I move on to higher dimensions.

Dear David,

I think what you said is correct. The diagonal matrices are matrices whose diagonal entries are the eigenvalues. This works in higher dimensions. I think when the book mentions it is not true 100% of the time, they probably mean that for some matrices

geometric multiplicity of λ ≠ algebraic multiplicity of λ

meaning these matrices are not diagonalizable.

 

[FactChecker]

Your eigenvectors are just multiples of their eigenvectors. If $Mu=\lambda u$ then clearly $M(ru)=\lambda (ru)$ for any real multiplier, $r$. Likewise, if $CMC^T = A$, then $(rC)M(rC)^T = r^2A$ for any real multiplier, $r$. If one is a diagonal, the other one should also be a diagonal.

 

[vela]

Try calculating $C^{-1}MC$.

 

[jbunniii]

So now I'm just more puzzled. I don't understand why my eigen vector doesn't yield the desired diagonal matrix. I know that you can just assume the diagonalized matrix is just the eigen values along its diagonal and zeroes elsewhere, but my book says that's not true %100 of the time so I want to make sure I have the process right before I move on to higher dimensions.

If the matrix is indeed diagonalizable, then it is 100% true that the diagonal elements of the diagonalized matrix are precisely the eigenvalues. Can you provide a verbatim quote of what your book is saying? Perhaps there is a subtlety that you are overlooking.

 

[jbunniii]

Maxima found for eigen vectors
[2,1][1,−2] whereas I have [2,1][−1,2].

That's absolutely fine. Note that $[1,-2]$ is a scalar multiple of $[-1,2]$. (The scalar is $-1$.) In general, if $v$ is an eigenvector associated with a particular eigenvalue, then so is $cv$ for any nonzero scalar $c$.

 

[jbunniii]

I think you need to simply interchange $C$ and $C^T$.

Using Matlab, I find that if $v_1 = \begin{bmatrix}2/\sqrt{5} \\ 1/\sqrt{5}\end{bmatrix}$ and $v_2 = \begin{bmatrix}-1/\sqrt{5} \\ 2/\sqrt{5}\end{bmatrix}$, and
$$C = \begin{bmatrix}v_1 & v_2\end{bmatrix} = \frac{1}{\sqrt{5}}\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}$$
then indeed
$$CDC^{T} =
\frac{1}{5}\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}
\begin{bmatrix}3 & 0 \\ 0 & -2 \end{bmatrix}
\begin{bmatrix}2 & 1 \\ -1 & 2 \end{bmatrix}
= \begin{bmatrix}2 & 2 \\ 2 & -1 \end{bmatrix} = M$$

Note that the eigenvalue-eigenvector equations are
$$Mv_1 = \lambda_1 v_1$$
and
$$Mv_2 = \lambda_2 v_2$$
If you combine these into a matrix equation with $C = \begin{bmatrix}v_1 & v_2\end{bmatrix}$, then the result should be
$$MC = CD$$
or equivalently
$$C^TMC = D \text{ or }M = CDC^T$$
whereas you have written
$$CMC^T = D$$
which is incorrect.

 

[FactChecker]

Notice that $C^TMC = \begin{pmatrix}3 & 0 \\ 0 & -2 \end{pmatrix}$.

 

[David Koufos]

Try calculating $C^{-1}MC$.

Ok. I tried that and got the right diagonal matrix. Thank you. I mixed up $CMC^{T}$ with $C^{T}MC$

 

[David Koufos]

I think you need to simply interchange $C$ and $C^T$.

Using Matlab, I find that if $v_1 = \begin{bmatrix}2/\sqrt{5} \\ 1/\sqrt{5}\end{bmatrix}$ and $v_2 = \begin{bmatrix}-1/\sqrt{5} \\ 2/\sqrt{5}\end{bmatrix}$, and
$$C = \begin{bmatrix}v_1 & v_2\end{bmatrix} = \frac{1}{\sqrt{5}}\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}$$
then indeed
$$CDC^{T} =
\frac{1}{5}\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}
\begin{bmatrix}3 & 0 \\ 0 & -2 \end{bmatrix}
\begin{bmatrix}2 & 1 \\ -1 & 2 \end{bmatrix}
= \begin{bmatrix}2 & 2 \\ 2 & -1 \end{bmatrix} = M$$

Note that the eigenvalue-eigenvector equations are
$$Mv_1 = \lambda_1 v_1$$
and
$$Mv_2 = \lambda_2 v_2$$
If you combine these into a matrix equation with $C = \begin{bmatrix}v_1 & v_2\end{bmatrix}$, then the result should be
$$MC = CD$$
or equivalently
$$C^TMC = D \text{ or }M = CDC^T$$
whereas you have written
$$CMC^T = D$$
which is incorrect.

Ok I found the diagonal now. Thank you. I mixed up $CMC^{T}$ with $C^{T}MC$

 

[David Koufos]

I think you need to simply interchange $C$ and $C^T$.

Using Matlab, I find that if $v_1 = \begin{bmatrix}2/\sqrt{5} \\ 1/\sqrt{5}\end{bmatrix}$ and $v_2 = \begin{bmatrix}-1/\sqrt{5} \\ 2/\sqrt{5}\end{bmatrix}$, and
$$C = \begin{bmatrix}v_1 & v_2\end{bmatrix} = \frac{1}{\sqrt{5}}\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}$$
then indeed
$$CDC^{T} =
\frac{1}{5}\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}
\begin{bmatrix}3 & 0 \\ 0 & -2 \end{bmatrix}
\begin{bmatrix}2 & 1 \\ -1 & 2 \end{bmatrix}
= \begin{bmatrix}2 & 2 \\ 2 & -1 \end{bmatrix} = M$$

Note that the eigenvalue-eigenvector equations are
$$Mv_1 = \lambda_1 v_1$$
and
$$Mv_2 = \lambda_2 v_2$$
If you combine these into a matrix equation with $C = \begin{bmatrix}v_1 & v_2\end{bmatrix}$, then the result should be
$$MC = CD$$
or equivalently
$$C^TMC = D \text{ or }M = CDC^T$$
whereas you have written
$$CMC^T = D$$
which is incorrect.

Ok. I tried that and got the right diagonal matrix. Thank you. I mixed up $CMC^{T}$ with $C^{T}MC$