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According to theorem 2, an\(n \times n\)matrix\(A\)is orthogonally diagonalizable if it is symmetric.

So, statement (a) is true.

According to another interpretation of theorem 2, an\(n \times n\)symmetric matrix\(A\), having the non-zero eigenvectors\({\rm{u,}}\,{\rm{v}}\),is orthogonally diagonalizable.

Thus, \(\lambda = 3\)must be one of the eigenvalues of matrix\(A\).

So, the characteristic equation\(A{\rm{u}} = 3{\rm{u}}\)and\(A{\rm{v}} = 3{\rm{v}}\)must be true.

As a result, statement in (b) is true.

According to theorem 3, an\(n \times n\)symmetric matrix\(A\)have \(n\) eigenvalues, and may have counting multiplicities.

Thus, eigenvalues need not to be distinct. So, the statement in (c) is false.

If \({\rm{u}}\)is a unit vector, then ,\(\lambda {{\rm{u}}_j}{{\rm{u}}_j}^T\)is a projection matrix such that for each vector\({\rm{v}}\)in\({\mathbb{R}^n}\),\(\left( {{\rm{u}}{{\rm{u}}^T}} \right){\rm{v}}\)is an orthogonal projection of\({\rm{v}}\)in the subspace spanned by\({\rm{u}}\).

So, the statement in (d) is false.