Question

Let X. X; be the two eigen vectors of a real synnetric matrix A, and the
eigen vector XXe are orthogonal in pains then
a) X, X:-1, where is the identity vector
b) (X) X-1 where lis the identity vector, and I denotes the transpose.
c) X, X,-0. where is the zero vector
d) (X) X:-0, where is the identity vector and T drnote the transpone
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Answer 1

Answer:

Because equal matrices have equal dimensions, only square matrices can be symmetric.

The entries of a symmetric matrix are symmetric with respect to the main diagonal. So if {\displaystyle a_{ij}}a_{ij} denotes the entry in the {\displaystyle i}i-th row and {\displaystyle j}j-th column then

{\displaystyle A{\text{ is symmetric}}\iff {\text{ for every }}i,j,\quad a_{ji}=a_{ij}}{\displaystyle A{\text{ is symmetric}}\iff {\text{ for every }}i,j,\quad a_{ji}=a_{ij}}

for all indices {\displaystyle i}i and {\displaystyle j.}j.

Every square diagonal matrix is symmetric, since all off-diagonal elements are zero. Similarly in characteristic different from 2, each diagonal element of a skew-symmetric matrix must be zero, since each is its own negative.

In linear algebra, a real symmetric matrix represents a self-adjoint operator[1] over a real inner product space. The corresponding object for a complex inner product space is a Hermitian matrix with complex-valued entries, which is equal to its conjugate transpose. Therefore, in linear algebra over the complex numbers, it is often assumed that a symmetric matrix refers to one which has real-valued entries. Symmetric matrices appear naturally in a variety of applications, and typical numerical linear algebra software makes special accommodations for them.