Show that T is diagonalizable if nullity$(T) + m = n.$

Question

$V$ is a vector space of dimension $n > 0$, let $T: V \to V$ be a linear operator and $ { 0, \lambda _{1}, ..., \lambda_{m} } $ be a set of the distinct eigenvalues of T . Show that T is diagonalizable if nullity$(T) + m = n$.

My Try:I think we can directly claim that $T$ is diagonalizable since it splits into distinct linear factor.Thank you

Answer 1

Remember that a linear map (or a matrix...) is diagonalizable iff for each and every eigenvalue, it algebraic and geometric multiplicities are equal, and this is the same as saying that the sum of all the geometric multiplicities of all the eigenvalues equals the space's dimension, i.e. $\;n\;$

For the eigenvalue zero: its geometric multiplicity equals the dimension of the corresponding eigenspace, which of course is $\;\ker T\;$ : $\;\dim\ker T=\text{null.}\,T\;$ , but then we get the above mentioned condition for diagonability as we're givenb

$$\overbrace{null.\,T+m}^{\text{sum of geom. mult.}}=n$$

Answer 2

Let $u_1, \dots,u_k$ be a basis of $\ker T$. Let $v_1, \dots,v_m$ be eigenvectors for $\lambda_1, \dots,\lambda_m$ respectively.

Then $u_1, \dots,u_k, v_1, \dots,v_m$ are a basis for $V$ because eigenvectors of distinct eigenvalues are linearly independent and $k+m=n$.

With respect to this basis, $T$ has a diagonal matrix: $diag(0,\dots,0,\lambda_1, \dots,\lambda_m)$.

Answer 3

Let $\delta_0,\delta_1,\dots,\delta_m$ be the geometric multiplicities for the eigenvalues $0,\lambda_1,\dots,\lambda_m$. Then $\delta_0=\operatorname{nullity}(T)$ by definition. Let $\mu_0,\mu_1,\dots,\mu_m$ be the algebraic multiplicities. You know that $1\le\delta_i\le\mu_i$, for $i=0,1,\dots,m$.

In particular $$ \delta_0+\delta_1+\dots+\delta_m\ge\delta_0+m $$ and so $$ n=\mu_0+\mu_1+\dots+\mu_n\ge \delta_0+\delta_1+\dots+\delta_m\ge\delta_0+m=n $$ the rightmost relation holding by assumption.

Can you show this forces $\delta_i=\mu_i$ for $i=0,1,\dots,m$?

By the way, what can you say about $\mu_i$ for $i=1,\dots,m$? (This is however not relevant for diagonalizability.)