Hello,
1. The problem statement, all variables and given/known data
I want to show that a real symmetric matrix will have real eigenvalues and orthogonal eigenvectors.
$$
\begin{pmatrix}
A & H\\
H & B
\end{pmatrix}
$$
3. The attempt at a solution
For the matrix shown above it's clear that the charateristic equation will be
##\lambda^2-\lambda(A+B)+AB-H^2=0##
I can show that the discriminant of the quadratic equation will be greater than 0 implying that the eigenvalues must be real.
##b^2-4ac=(A+B)^2-4(AB-H^2)=A^2+2AB+B^2-4AB+4H^2##
##=(A-B)^2+4H^2##
Since ##A, B, H \in \mathbb{R}##, ##(A-B)^2+4H^2 \geq 0##
Knowing that ##\lambda## must be real for this matrix.
My only problem now is to show that the eigenvector is orthogonal.
The matrix has eigenvalues of ##\lambda_1, \lambda_2##, and hence eigenvectors ##\lambda_1v_1, \lambda_2v_2##.
How can I show,
##\lambda_1\lambda_2x_1x_2+\lambda_1\lambda_2x_1x_2=0##?
I know ##\lambda_1\lambda_2=det(M)##
It could become,
##det(M)(x_1x_2+y_1y_2)=0##
Then it's clear the vectors are orthogonal because ##det(M)## cannot be 0. But the problem is this is not a proof because I explicitly assume the dot product are 0 in the first place..
I tried substituting the complete quadratic equation into the matrix as if I know the lambda but then the matrix cannot be eliminated in simple manner and I got a mess real quick.
1. The problem statement, all variables and given/known data
I want to show that a real symmetric matrix will have real eigenvalues and orthogonal eigenvectors.
$$
\begin{pmatrix}
A & H\\
H & B
\end{pmatrix}
$$
3. The attempt at a solution
For the matrix shown above it's clear that the charateristic equation will be
##\lambda^2-\lambda(A+B)+AB-H^2=0##
I can show that the discriminant of the quadratic equation will be greater than 0 implying that the eigenvalues must be real.
##b^2-4ac=(A+B)^2-4(AB-H^2)=A^2+2AB+B^2-4AB+4H^2##
##=(A-B)^2+4H^2##
Since ##A, B, H \in \mathbb{R}##, ##(A-B)^2+4H^2 \geq 0##
Knowing that ##\lambda## must be real for this matrix.
My only problem now is to show that the eigenvector is orthogonal.
The matrix has eigenvalues of ##\lambda_1, \lambda_2##, and hence eigenvectors ##\lambda_1v_1, \lambda_2v_2##.
How can I show,
##\lambda_1\lambda_2x_1x_2+\lambda_1\lambda_2x_1x_2=0##?
I know ##\lambda_1\lambda_2=det(M)##
It could become,
##det(M)(x_1x_2+y_1y_2)=0##
Then it's clear the vectors are orthogonal because ##det(M)## cannot be 0. But the problem is this is not a proof because I explicitly assume the dot product are 0 in the first place..
I tried substituting the complete quadratic equation into the matrix as if I know the lambda but then the matrix cannot be eliminated in simple manner and I got a mess real quick.
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