Result: Suppose $V$ is a finite dimensional complex vector space and $T:V \to V$ is a linear map. Then $T$ has an upper triangular matrix with respect to some basis of $V.$ [For the proof see: Linear algebra done right by sheldon axler third edition, p.149]
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My claim is: Suppose $V$ is a finite dimensional complex vector space and $T:V \to V$ is a linear map. Then $T$ has a diagonal with respect to some basis of $V.$
Here is my proof: Assume dim $V=n.$ Since $V$ is complex vector space, $T$ has an eigenvalue, say $\lambda_1$ (This is known fact). There for there is nonzero vector $v_1\in V$
such that $$Tv_1= \lambda_1 v_1$$.
Put $U_1=\text{span} (v_1).$ Note that $U_1$ is a subspace of $V.$ Then there is subspace $W\subset V$ such that
$$V= U_1\oplus W$$
and dim $W=n-1.$
Again, we may say that $T$ is linear map on a complex vector space $W,$ and therefore it has a eigenvalue, say $\lambda_2$, and the corresponding eigenvector $0\neq v_2\in W$ and $Tv_2= \lambda_2 v_2.$
We may continue this process, and I think, we get $v_1, v_2,…, v_n$ linearly independent eigenvector of $T$. And therefore the matrix of $T$ with respec to this basis is diagonal.
My Question: Is my claim is correct? If not, where is the mistake in proof?
Best Answer
Your claim is not correct. An error in your proof is that $T$ is not a linear map on $W$. Of course you can restrict $T$ to $W$, but for $w \in W$, there is no reason for $T(w)$ to be in $W$, so you have a map from $W$ to $V$.
For a counterxample, consider the map on $\mathbb{C}^2$ given by $(z_1,z_2)\mapsto (z_2,0)$.
For more information on what is true, look up Jordan Canonical Form.