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

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?

• It’s just the definition of linear independence: $c_1x_1+…+c_nx_n=0$ implies all $c_i=0$. Here, just write $u_i=c_i v_i$ where $Tv_i=\lambda_iv_i$ Commented Aug 17, 2023 at 23:27

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 $$1 \le k \le n$$.