The condition specified in the problem does not say anything about v.
Can you post the question exactly as it was given to you?
A is a simetric metrices nxn. so and
so for some
prove that for the same
how i tried to solve it:
i just collected data from the given.
simetric matrices is diagonizable.
we were given that
so (dot product is also v)
so v is orthogonal to
what to do now?
for any 0-vector (nx1 matrix) u (= 0), and any other row-vector (1xn matrix),
we are given that the vector for some .
so multiplying it by will also give a 0 1x1 matrix, that is, the number 0.
remember, A is symmetric so λI - A is symmetric too: .
you don't need to pass to the norm, you can just use the positive-definiteness of the inner product.
There is no answer to 'why' generally. If you did not understand 'how' any of the implications are true, then I can help you.
You can do the same proof in another way. Start with the norm step and then expand it. Is that meaningful?
in your original post, you said it was given that:
the whole expression " " can be our , and we are going to multiply it by the matrix .
then we re-arrange the terms for our matrix product, to see that it is just an inner product of with itself.
the inner product of any vector with itself is positive...unless, our vector is the 0-vector,
which is what we hope to prove.