# Show that $\lambda$ is an eigenvalue of $T$ if and only if $\bar{\lambda}$ is an eigenvalue of $T^*$.

Let $T:V\to V$ be a linear transformation on a finite-dimensional inner product space $V$.Show that $\lambda$ is an eigenvalue of $T$ if and only if $\bar{\lambda}$ is an eigenvalue of $T^*$.

Proof given:

Let O be an orthonormal basis for V. Then $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is an eigenvalue of $[T]_O$ if and only if rank($[T]_O - \lambda I) \geq 1$ if and only if rank($([T]_O - \lambda I)^*) \geq 1$ if and only if rank($[T^*]_O - \bar{\lambda} I) \geq 1$ if and only if $\bar{\lambda}$ is an eigenvalue of $[T^*]_O$ if and only if $\bar{\lambda}$ is an eigenvalue of $T^*$.

My question is why $\lambda$ is an eigenvalue of a matrix, say A if and only if rank$(A- \lambda I) \geq 1$?

• Explain your notation: what is $\;[T]_0\;$ ? And $\;\lambda\;$ is eigenvalue of an operator (or matrix) $\;T\;$ iff $\;\det (T-\lambda I)=0\iff \text{rank}\,(T-\lambda I)\;$ isn't full. What does the rank being greater than or equal one means? Oh, I see that's your question...perhaps they mean that rank is the dimension of the kernel... Dec 1, 2014 at 3:52
• I think that instead of rank, you should have "null" or "dim ker". Otherwise, the statements make no sense. Dec 1, 2014 at 3:54
• $[T]_O$ is the matrix representing T in terms of basis O. Dec 1, 2014 at 3:54
• Consider $A = 0$ on $\mathbb{R}$. You have $\lambda = 0$ is an eigenvalue and $rank(A) = 0$. Your statement is wrong. Dec 1, 2014 at 3:55
• So it should be kernel >= 1 instead? Dec 1, 2014 at 3:55

Perhaps for them $\;rank A=\dim\ker A\;$ , and then everything follows:
$$\lambda\;\;\text{eigenvalue of}\;\;A\iff \det (A-\lambda I)=0\iff \ker (A-\lambda I)\neq 0\iff$$
$$\dim\ker (A-\lambda I)>0\iff \text{rank} (A-\lambda I)\ge 1\,$$