What is the intent of this problem, disguised as an eigenvalue - eigenvector problem?

Question

Let $$ A= \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{bmatrix} $$ $a,b,c >0$.

Find eigenvalues and a basis of eigenvectors of the matrix $A$.

My question is: what is the "real" intent of the problem? I don't think that a direct calculation of $\det(A-\lambda I)$ and solving for its zeroes to find the eigenvalues is a smart move.

The matrix is symmetric, so it is orthogonally diagonalizable.

Also, since it is symmetric, its singular value decomposition is, in fact, its eigendecomposition, so, the singular values of $A =$ the eigenvalues of $A$.

But, the singular values of A is gotten by computing the square root of the eigenvalues of $A^*A$ (or $A^TA$ if the ground field is the set of real numbers.) So, computing the SVD doesn't appear to be any easier than direct calculation of $\det{(A-\lambda I)}$.

Finally, I'm not sure how to utilize the assumption that $a,b,c >0.$

Answer 1

Note that all row-sums are $a+b+c$. This gives you one eigenvector and eigenvalue. The other two are not as nice.

EDIT: Perhaps the question was intended to be about $$ \pmatrix{a & b & c\cr c & a & b\cr b & c & a\cr} $$

Answer 2

Proceeding "blindly" by way of finding the roots of the characteristic polynomial yields $$\lambda_{0}=a+b+c$$ and $$\lambda_{\pm}=\pm\sqrt{a^{2}+b^{2}+c^{2}-ab-ac-bc}.$$ The fact that $a,b,c>0$ was not used.

Perhaps this is just an exercise in noting that Hermitian matrices have real eigenvalues?

Answer 3

This problem is about geometry. The matrix that is given to you can represent a quadric surface. One eigenvector can be easily reasoning in terms of simmetry: (1,1,1).

Answer 4

Don't understand your reluctance. As in Robert's answer, there is a famous identity, well worth memorizing, $$ a^3 + b^3 + c^3 - 3abc = (a+b+c)\left(a^2 + b^2 + c^2 - bc - ca - ab \right). $$

Meanwhile, $$ a^2 + b^2 + c^2 - bc - ca - ab = \frac{1}{2} \left( \; (b-c)^2 + (c-a)^2 + (a-b)^2 \; \right) $$ is semidefinite positive, it cannot be negative and is only zero when $a=b=c.$

Now, as in the answer by par, your original, symmetric, matrix has trace $a+b+c,$ meaning the remaining eigenvalues come in a $\pm $ pair, product $-\left( \; (b-c)^2 + (c-a)^2 + (a-b)^2 \; \right)$

Answer 5

This is a nice problem, that leads to a lot of nice & useful theory, but a little bit of a teaser if the only math exposure you have is just linear algebra. Note that each row is a cyclic permutation of the row above, e.g. row 2 is the left rotation of row 1 where the "left over" entry, an "a" cycles back to the right. A matrix with this property is call a circulant matrix. Wiki discusses these in detail, and you'd be a better person from meditating on this page a bit. Among many other properties, for a circulant matrix of any size all of the eigenvalues and eigenvectors can be computed explicitly.