# Proving that the linear transformation $T$ is diagonizable

Let $$\ T: \mathbf R^n \rightarrow \mathbf R^n$$ be a linear transformation such that there is a vector $$\ u \in \mathbf R^n$$, $$\ T^2(u) \not = 0$$ and $$\ \dim\ker T = n-1$$

Prove that $$\operatorname{Im} T = \operatorname{span}\{T(u)\}$$ and that $$T$$ can be diagonalised.

My solution so far is that $$T^2(u) \not =0 \rightarrow T(T(u)) \neq 0 \rightarrow T(u) \notin \ker T \rightarrow T(u) \in \operatorname{Im} T$$

and because

$$\dim \ker T = n-1 \Rightarrow \dim\operatorname{Im} T = 1 = \dim(\operatorname{span}\{T(u)\}$$

and so $$\dim(\operatorname{span}\{T(u)\}) \in \operatorname{Im} T$$ and therefore $$\operatorname{Im} T = \operatorname{span}\{T(u)\}$$

But how do I prove that $$T$$ is diagonalizable?

My first guess is that because $$\dim \ker T = n-1$$ then $$0$$ is an eigenvalue with $$n-1$$ eigenvectors but what can I conclude about the last vector?

• $T^2u\in\operatorname{image}T=\operatorname{span}\{Tu\}$ and is nonzero, so $Tu$ is an eigenvector with nonzero eigenvalue. – user10354138 Nov 5 '18 at 10:21

If we denote by $$\;V_\lambda\;$$ the eigenspace corresponding to eigenvector $$\;\lambda\;$$ , then we simply have that $$\;\ker T= T_0\;$$ so $$\;\dim T_0=n-1\;$$, and taking $$\;n-1\;$$ lin. independent vectors here together with any non-zero vector mapping onto Im$$\;T\;$$ automatically gives us a basis of $$\;\Bbb R^n\;$$ (why?), and this basis' elements are all of them eigenvectors of $$\;T\iff T\;$$ is diagonalizable. Fill in details now.
Take any basis of ker($$T$$) and include $$Tu$$ . This will give you a basis of $$\mathbf{R}^n$$. Check that with respect to this basis matrix of $$T$$ is diagonal.
Since $$\dim \ker T=n-1$$ consider a basis $$e_1,...,e_n$$, where $$e_1,...,e_{n-1}\in \ker T$$, write the matrix of $$T$$ in this basis, you have $$T(e_n)=ae_n+a_1e_1+..+a_{n-1}e_{n-1}$$ and the polynomial characteristic of $$T$$ is $$(x-a)x^{n-1}$$, $$a\neq 0$$ otherwise $$Im T\subset \ker T$$ and $$T^2=0$$, let $$u$$ be an eigenvector associated to $$a$$, $$T$$ is diagonalizable in $$(e_1,..,_{n-1},u)$$.