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

I tried to calculate eigenvalues of matrix A and matrix B and then to compare them, but it is way to complicated to calculate, is there any easier solution to this?

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If $\left(\begin{matrix}x\\y\end{matrix}\right)$ is an eigenvector of $A$ with eigenvalue $\lambda_1$, then $$\left(\begin{matrix}a^2&ab&b^2\\2ac&ad+bc&2bd\\c^2&cd&d^2\end{matrix}\right)\left(\begin{matrix}x^2\\2xy\\y^2\end{matrix}\right)=\left(\begin{matrix}a^2x^2+2abxy+b^2y^2\\2acx^2+2\left(ad+bc\right)xy+2bdy^2\\c^2x^2+2cdxy+d^2y^2\end{matrix}\right)=\\=\left(\begin{matrix}\left(ax+by\right)^2\\2\left(ax+by\right)\left(cx+dy \right )\\\left(cx+dy\right)^2\end{matrix}\right)=\lambda_1^2\left(\begin{matrix}x^2\\2xy\\y^2\end{matrix}\right)$$ So $\lambda_1^2$ is an eigenvalue of $B$. (If you want to know how I thought of it, well, I thought we should take an eigenvector of $A$ and express with it an eigenvector of $B$; the first and third component hinted to use squares, so I tried and it worked somehow)

A similar move can be done for $\lambda_2^2$.

To find the third eigenvalue, we may use the fact that the trace of a matrix is the sum of its eigenvalues. If the last eigenvalue is $\lambda$, then $$a^2+ad+bc+d^2=\lambda_1^2+\lambda_2^2+\lambda$$ Since $ad-bc=\det A=\lambda_1\lambda_2$, we get $$a^2+2ad+d^2-\lambda_1\lambda_2=\lambda_1^2+\lambda_2^2+\lambda$$ $$\left(a+d\right)^2-\lambda_1\lambda_2=\lambda_1^2+\lambda_2^2+\lambda$$ The trace of $A$ is $\lambda_1+\lambda_2$. Putting this in the last equation yields $\lambda_3=\lambda_1\lambda_2$.

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  • $\begingroup$ You can also do a similar computation for $\lambda_1\lambda_2$. If the eigenvectors to $\lambda_1,\lambda_2$ are $(x_1,x_2),(y_1,y_2)$ then $(x_1y_1,x_1y_2+x_2y_2,x_2y_2)$ is the eigenvector to $\lambda_1\lambda_2$. I was just about to write this when your answer did pop up. But this is a nasty computation and your answer is much cleaner. Have my upvote. $\endgroup$
    – Maik Pickl
    Jun 12, 2016 at 19:30

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