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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?

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  • $\begingroup$ No, also in the complex case it may happen that some eigenvalue is not regular. Take the $2 \times 2$ matrix $A$ such that $a_{12}=1$ and all the other entries are $0$. Its only eigenvalue is $0$, so if $A$ were diagonalizable, it would be similar to the zero matrix, absurd. $\endgroup$ – Francesco Polizzi Aug 9 '17 at 14:02
  • $\begingroup$ How do you know you get $n$ linearly independent eigenvectors? Have you considered the case where we have repeated eigenvalues? $\endgroup$ – jacer21 Aug 9 '17 at 14:07
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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.

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