In particular, such a matrix is

diagonalizable, and even by an

orthogonal basis.

**Proof**:

Let

**M** be such a matrix.

For `n=1`, **M** is simply a scalar, and the statement is obvious.

So suppose the statement is true for `n-1`, and prove it for `n * n` matrices. We first wish to show **M** *has* a real eigenvalue (not all real matrices do!). To this end, consider `f(v) = (`**M**v, v), as a function from the unit sphere of R^n to R. Since the sphere is compact, f attains global extrema, say at some unit vector u. But deriving f, we see that `(grad f)(v) = 2`**M**v; by the Lagrange multiplier rule, at any extremum of f, `grad f` is in the direction of the normal to the sphere. Since the normal to the sphere at v is in the direction of v, it follows that at the extremum u, **M**u = mu for some scalar m. But that means u is an eigenvalue of **M**.

Now we shall "deflate" **M** to an `(n-1)*(n-1)` symmetric matrix **M'**, and apply the induction hypothesis. Let `V' = {v | (u,v)=0}` be the orthogonal complement of V (`dim V' = n-1`). All we must show is that **M**|V' (**M**, restricted to act only on V') is a symmetric matrix; applying the induction hypothesis, it has `n-1` distinct eigenvectors, and along with u, **M** has n distinct eigenvectors, hence it has n distinct eigenvalues counting multiplicity.

Pick any orthonormal basis `{w_1,...,w_{n-1}}` for V'. Then `{w_1,...,w_{n-1},u}` is a basis for V. Representing the operator **M** by that basis, we still have a symmetric matrix, but (since u is an eigenvector) we know that the last row/column is `(0,...,0,m)`. Dropping the last row and column, we see that **M'** is an operator from V' to V', and that it is represented by a symmetric matrix in some basis, so it is symmetric. Thus the induction hypothesis holds, and we are done.

**QED**.