How do I know if my eigenvectors are right?

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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?
 
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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.
 
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FactChecker said:
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.
 
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David Koufos said:
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.
 
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.
 
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Try calculating ##C^{-1}MC##.
 
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David Koufos said:
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.
 
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##.
 
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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.
 
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  • #10
Notice that ##C^TMC = \begin{pmatrix}3 & 0 \\ 0 & -2 \end{pmatrix}##.
 
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  • #11
vela said:
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##
 
  • #12
jbunniii said:
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##
 
  • #13
jbunniii said:
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##
 
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