If matrix A is symmetric and a certain condition holds then it is positive definite So the condition I am talking about is this one: $ a_{ii} > \sum_{j=1,i\neq j}^{n} {|a_{ij}|} $ 
My idea for the proof would be:
Since the matrix is symmetric then I can diagonilize it and get a matrix where all the eigenvalues are represented by the values in the diagonal. Since the operations that lead to the diaagonalized matrix are all linear, I can say that for this matrix the condition $ a_{ii} > \sum_{j=1,i\neq j}^{n}{|a_{ij}|} $ still holds. The condition then can be also rewritten as $ a_{ii} - \sum_{j=1,i\neq j}^{n}{|a_{ij}|} > 0 $. Now in the diagonalized matrix it holds that $ \sum_{j=1,i\neq j}^{n} {|a_{ij}|} = 0 $ which leads to $ a_{ii} > 0 $ which means that the eigenvalues are positive. This means that the matrix is positive definite.
Is my proof correct?
Thanks for any tips/help


Since the operations that lead to the diaagonalized matrix are all linear, I can say that for this matrix the condition $ a_{ii} > \sum_{j=1,i\neq j}^{n}{|a_{ij}|} $ still holds.

This sentence is vague and mathematically meaningless at best and false at worse:
The operation that maps $a_{ii}$ to $0$ and all other values $a_{ij}$ to themselves is linear, but it does not preserve your property.

For a better version of the proof, have you heard of Gerschgorin circles?
