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]


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?

  • $\begingroup$ 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. $\endgroup$ 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? $\endgroup$
    – Jacob
    Aug 9 '17 at 14:07

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.