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.