1
$\begingroup$

Let $T$ be a normal operator on a complex inner product space. Then $T$ is a self adjoint operator if and only if

1) $T$ has distinct eigen values

2) $T$ has repeated eigen values

3) All eigen values of $T$ are real

4) $T$ has atleast one real eigen value

If $T$ is a self adjoint operator then all it's eigen values are real.

This rules out 1, 2 and 4.

But how to prove the converse? i.e. If all the eigen values of a normal operator $T$ are real then it is self adjoint.

$\endgroup$

2 Answers 2

3
$\begingroup$

Since $T$ is a normal operator on a complex inner product space $V$, the Spectral theorem implies that there is an orthonormal basis $\beta$ of $V$ such that $[T]_{\beta}$ is a diagonal matrix. Since all the eigenvalues are real, we have that $[T]_{\beta}$ is a real matrix. Now,

\begin{align} [T^*]_{\beta} &= \left( [T]_{\beta} \right)^* \tag{$\beta$ is orthonormal}\\ &= [T]_{\beta} \tag{the matrix is real and diagonal} \end{align} Hence, $T^* = T$.

$\endgroup$
0
$\begingroup$

If you assume that T acts on an $n$-dimensional complex inner product space, then $T$ is diagonalizable: There is a basis of orthogonal eigenvectors $v_j$ and in that basis $T$ is represented by a diagonal matrix with the eigenvalues $\lambda_j$ on the diagonal. The adjoint of T is represented in the same basis by a diagonal matrix with entries $\overline{\lambda_j}$, so $T$ is self-adjoint if f $\lambda_j=\overline{\lambda_j}$ for all $j$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .