If $W_j$ is the subspace associated to $\lambda_j$ for $j\in{1,...,k}$ prove that $W_1,...,W_k$ are independent

Asked May 04 '19 at 13:13

Active May 04 '19 at 13:29

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Let $V$ be finite-dimensional vector space such that $dim(V)=n$ and consider $T \in L(V)$ and $\lambda_1,...,\lambda_k$ all the different eigenvalues.

a) If $W_j$ is the subspace associated to $\lambda_j$ for $j\in{1,...,k}$ prove that $W_1,...,W_k$ are independent

What I know here is that we have to prove that $w_1+...+w_k=0, w_i \in W_i$

If T is diagonalizable, prove that $V=W_1 \oplus...\oplus W_k$

If T is diagonalizable $\exists\beta=${$v_1,...,v_n$} such that $v_i$ is a characteristic vector $\forall i \in${$1,...,n$} how do I prove $V=W_1 \oplus...\oplus W_k$?

edited Jun 12 '20 at 10:38

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asked May 04 '19 at 13:13

Juju9708

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$W_1+\cdots+W_k=0$ is nonsense. You must mean something else. – Gerry Myerson May 04 '19 at 13:15
Equation edited, but still nonsense. – Gerry Myerson May 05 '19 at 03:39

If $W_j$ is the subspace associated to $\lambda_j$ for $j\in{1,...,k}$ prove that $W_1,...,W_k$ are independent

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