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I am trying , without success , to prove that DE is a symmetric matrix according to the statements below :

D and E are Symmetric matrices 2x2.
v1  = 1/√2 [1]  , v1 is an eigenvector of D with eigenvalue = λ1
           [1]

v1 is also eigenvector of E with eigenvalue =  µ1

λ1 and µ1 not necessarily equals

I would appriciate you help here.

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    Show your attempts? – 2017-01-11

1 Answers 1

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Since both D and E are symmetric 2x2 matrices $$ \text{D}=\left( \begin{array}{cc} \text{$d_0$} & \text{$d_a$} \\ \text{$d_a$} & \text{$d_1$} \\ \end{array} \right) $$ $$ \text{E}=\left( \begin{array}{cc} \text{$e_0$} & \text{$e_a$} \\ \text{$e_a$} & \text{$e_1$} \\ \end{array} \right) $$ $DE$ is symmetric if and only if $(DE)^T=DE$ or: $$ d_a e_1 + d_0 e_a=d_a e_0 + d_1 e_a \iff \frac{d_1-d_0}{d_a}=\frac{e_1-e_0}{e_a} $$ then let's see what the Eigen Value equation implies: $$ D\binom{1}{1}=\lambda\binom{1}{1} \iff \begin{cases}d_0+d_a=\lambda\\d_a+d_1=\lambda\end{cases} \implies d_0-d_1=0 $$ exactly the same argument applies to $E$ so $$ e_0-e_1=0 $$ so by substitution the condition for $DE$ to be symmetric becomes: $$ \frac{d_1-d_0}{d_a}=\frac{e_1-e_0}{e_a} \iff \frac{0}{d_a}=\frac{0}{e_a}\iff 0=0 \rm{\,\,\,\,\,which\,\, is\,\, true} $$

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    Thank you very much! , Is there any chance that you can explain how did you get to the equation below the statement (DE)T = DE (one line below)? – 2017-01-11
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    calculate explicitly the matrix product and equate it element by element to it's transpose, you will get 4 equations of which two are always satisfied (those coming from the diagonal) and the other two are identical leading only to the equation i've written – 2017-01-11
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    Oh now I got it , Thank you!! – 2017-01-11