Aligning a matrix with its eigen vectors and other questions?

[Dr Bwts]

Hi,

I have a square symmetrical matrix A (ugly I know)


321.1115, -57.5311, -33.9206
-57.5311, 296.7836, 10.8958
-33.9206, 10.8958, 382.1050

which has the eigen values,

248.8034
341.6551
409.5415

Am I right in saying that A when aligned with its eigen vectors it is,

248.8034, 0, 0
0, 341.6551, 0
0, 0, 409.5415

?

I would also like to transform the matrix so that,

A₁₁+A₂₂+A₃₃ = 1

Thanks for any help, I feel like I should know this but have been running around in circles for the past 2 hours.

 

[HallsofIvy]

"aligned with its eigenvectors" is not standard terminology, but I think what you mean is correct. Given any linear transformation, T, from a vector space of dimension n to itself, we can always represent the transformation as a vector space by using a specific ordered basis for the vector space. The idea is that we apply T to each of the basis vectors in turn, writing the result as a linear combination of the basis vectors. For example, if we have a three dimensional vector space with ordered basis $\{v_1, v_2, v_3\}$ then the vector $x_1v_2+ x_2v_2+ x_3v_3$ would be represented by the array
$$\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}$$
In particular, $v_1= 1v_1+ 0v_2+ 0v_3$ itself is represented by
$$\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$$

So if $T(v_1)= a_1v_1+ a_2v_2_+ a_3v_3$ we can write
$$\begin{a_1 & * & * \\ a_2 & * & * \\ a_3 & * & * \end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a_1 \\ a_2\\ a_3\end{bmatrix}$$
where the "*" in the second and third columns can be anything.

That is, the result result of T applied to basis vector $v_i$ gives the ith column of the matrix representation. In particular, if the basis vectors are eigenvalues of T, then $Tv_1= \lamba_1 v_1+ 0 v_2+ 0v_3$, $Tv_2= 0v_1+ \lambda_2v_2+ 0v_3$, and $Tv_3= 0v_1+ 0v_2+ \lambda_3v_3$ so the matrix representation, in that basis, is
$$\begin{bmatrix}\lambda_2 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3\end{bmatrix}$$

To reduce those diagonal elements to 1 is to divide each eigenvector by the corresponding eigenvalue. If $v_1$ is an eigenvector of T with eigenvalue $\lambda_1$ then
$$T\frac{v_1}{\lambda_1}= \frac{1}{\lambda_1}Tv_1= \frac{1}{\lambda_1}\lambda_1 v_1= v_1$$
so representing T as as matrix by using basis vectors $v_1/\lambda_1$, $v_2/\lambda_2$, and $v_3/\lambda_3$ will give the identity matrix.

 

[Dr Bwts]

Thanks for the reply.

Once the matrix (A) has been transformed as above how can I scale it such that,

trace(A)=1

?

 

[Dr Bwts]

OK panic over I just divide the aligned matrix by its trace.

Thanks for your time.

 

[blue_raver22]

Hello,

You are correct in saying that when a matrix is aligned with its eigen vectors, it will take the form of a diagonal matrix with the eigen values along the main diagonal. This is because the eigen vectors are the directions along which the matrix has the greatest variability and thus, aligning the matrix with these vectors will simplify its representation.

To transform the matrix so that A11+A22+A33 = 1, you can use a method called normalization. This involves dividing each element in the matrix by the sum of all the elements in the matrix. This will result in a matrix where the sum of all the elements is equal to 1. However, this will also change the values of the matrix, so it is important to consider the implications of this transformation on your data.

I hope this helps and good luck with your research.