Answer to Question #213887 in Linear Algebra for Hetisani Sewela

Question

Suppose T is the element of L(V ) and dim range T = k. Prove that T has at

most k + 1 distinct eigenvalues.

Answer 1

Let $v_1, v_2, v_3....v_n$ $\in V$ are the eigenvectors of T with respect to the eigenvalues $\lambda_1, \lambda_2....\lambda_n$

Which are distinct and non-zero. The vectors $T(v_1), T(v_2).... T(v_n)$ are linearly independent.

In fact, if $c_1, c_2, c_3..... c_n \in K$

Then $c_1 T(v_1)+c_2 T(v_2)+c_3 T(v_3).....c_n T(v_n)=0 v$

Hence, $c_1 \lambda_1v_1 +c_2\lambda_2v_2+....c_n\lambda_nv_n=0$ v

So, we can write it as $c_1 \lambda_1 = c_2\lambda_2 = c_3\lambda3 .....=c_n \lambda_n=0$

From the above, we can conclude that $v_1, v_2, v_3... v_n$ are linearly independent, so $c_1=c_2=c_3..... c_n=0$

But, $\lambda_i$ is non zero, So $n= dim Span(T(v_1),...,T(v_n))≤dim ImageT=k$

So, we can conclude that T has not $k$ distinct non-zero eigenvalues, Hence only possible eigenvalues for T is zero, So we can say that T has $k+1$ distinct eigenvalues.