$T :V\rightarrow\ V$ be a linear transformation with dimension of Range of $T = k$.

Question

Let $V$ be a finite dimensional vector space over $\mathbb R$ and $T :V\rightarrow\ V$ be a linear transformation with dimension of Range of $T = k$.Then how to show that $T$ can have at most $k + 1$ distinct eigenvalues? somebody please help me.thanks for your kind help.

Answer 1

If there are $k+1$ distinct eigenvalues $\lambda_1,\dots,\lambda_{k+1}$, then there are $k+1$ (non-zero) eigenvectors $v_1,\dots,v_{k+1}$ corresponding to $\lambda_1,\dots,\lambda_{k+1}$, respectively.

Q. What is the dimension of the image of the span of $v_1,\dots,v_{k+1}$ under $T$?

Hope this hint helps!

Answer 2

Let V be $R^n$ Now If we denote matrix of T by A, then A is NxN matrix with characteristic ploynomial of degree at most n.And since dimension of range is k , this should imply that there are k linearly indep vectors in A .

this imply, there are k e-values of multiplicity one and 1 e-value of multiplicity n-k. so total distint e-values are k+1.