This is the problem I am currently having trouble with:

If $\lambda_1$ and $\lambda_2$ are eigenvalues of matrix $$ \begin{bmatrix} a & b\\ c & d\\ \end{bmatrix}$$prove that $\lambda_1^2$, $\lambda_1\lambda_2$ and $\lambda_2^2$ are eigenvalues of matrix $A$. $$A=\begin{bmatrix} a^2 & ab & b^2\\ 2ac & ad+bc &2bd\\ c^2 & cd &d^2\\ \end{bmatrix}$$

What have I considered so far:

  1. I tried to find characteristic polynomial of first matrix and then find the roots which would be $\lambda_1$ and $\lambda_2$. I was hoping I can make a connection with characteristic polynomial of matrix $A$, but it got too complicated and did not seem like the right idea.
  2. Since there are two eigenvalues for first matrix, there are two eigenvectors. I thought about somehow using $A\vec v_i=\lambda_i\vec v_i$, where $i=1,2$
  3. Also, I thought about finding a connection between these two matrices. What I see is that the sum of elements of first and second column, of first matrix, are $(a+c)$ and $(b+d)$, respectively. Sum of first, second and third column of matrix $A$ are $(a+c)^2, (a+c)(b+d)$ and $(b+d)^2$. Now, if I could assign $\lambda_1$ and $\lambda_2$ to $(a+c)$ and $(b+d)$, then I would get that sum of first, second and third column of matrix $A$ are $\lambda_1^2$, $\lambda_1\lambda_2$ and $\lambda_2^2$, but what would be next?

I thought it would be good to combine 2. and 3rd idea, but I am not making any progress.

I appreciate any help.


1 Answer 1


Note: You can use the facts that Sum of the eigen values of a matrix is equal to the trace and product of the eigen values of the matrix is equal to the determinant.

For the first matrix let $\lambda_1$ and $\lambda_2$ be the eigen values. Then $\lambda_1+\lambda_2=a+d$, the trace of the matrix and $\lambda_1 \lambda_2=ad-bc$, which is the determinant. Now $(\lambda_1+\lambda_2)^2=(a+d)^2$ and adding $\lambda_1 \lambda_2=ad-bc$ gives $\lambda_1^2+\lambda_1 \lambda_2+\lambda_2^2=a^2+bc+ad+d^2$, which is the trace of the second matrix and also $\lambda_1^3\lambda_2^3=(ad-bc)^3$ which is equal to the determinant of the second matrix. So it implies that $\lambda_1^2$, $\lambda_2^2$ and $\lambda_1\lambda_2$ are eigen values of the second matrix.

  • $\begingroup$ This looks incomplete. The eigenvalues of a 3x3 matrix in general are not uniquely determined by the trace and determinant. $\endgroup$
    – user1551
    May 27, 2016 at 3:20
  • $\begingroup$ Thank you so much. I would never consider to use (λ1+λ2)^2, but it makes perfect sense now. Thank you! $\endgroup$
    – Asleen
    May 27, 2016 at 3:39

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