5.38 Sum of eigenspaces is a direct sum Linear Algebra Done Right

Question

In the book Linear Algebra Done Right by Sheldon Axler there is the next theorem:

Theorem. Suppose $V$ is finite-dimensional and $T\in\mathcal{L}(V)$, where $\mathcal{L}(V)$ is the set of all linear operator on $V$. Suppose also that $\lambda_1,...,\lambda_n$ are distinct eigenvalues of $T$. Then $$E(T,\lambda_1)+...+E(T,\lambda_n)$$ is a direct sum.

Here the proof he gives

I can't understand the implication which tell why each $u_j$ is equals $0$. As far as I understand $$u_1+...+u_n=0$$ implies $u_1,...,u_n$ are linearly independent, and therefore there existes $u_j$ such that $u_j$ is not an eigenvector corresponding to $\lambda$, which implies that $u_j=0$ because we know by how we define $E(T,\lambda_j)$ that $T(x_j)=\lambda_j x_j$. So we have one $u_j$ is zero, but it doesn't imply all are, maybe by induction it could, but I want to see if that is possible without induction as it seems the book's proof suggest.

Can any one help me please?

Answer 1

If you suppose that $$ u_1+u_2+u_3+\cdots+u_n = 0, $$ then you can apply $\prod_{j=1,j\ne k}^{n}(T-\lambda_jI)$ to the above in order to obtain $$ \prod_{j=1,j\ne k}^{n}(\lambda_k-\lambda_j)u_k=0, $$ which implies that $u_k=0$. It follows that $u_k=0$ for all $1 \le k \le n$.