I am reviewing Upper Triangular matrices and am confused on this problem.

We have that $T$ is a linear transformation on a finite dimensional complex vector space. We are to suppose that $\lambda$ is a root of the characteristic polynomial, $c(x)$, or multiplicity $r$. From here, show that there is a basis for $T$ with respect to which the matrix for $T$ has for all $i$ and for $1\leq j\leq r$, $a_{ij}= \lambda$ if $i=j$, and $0$ if $i>j$. That is, the top left hand corner block has dimension r, and the eigenvalue $\lambda$ across the diagonal, and this block is upper triangular.

I'm supposing here that I want to tweak the argument to show that a matrix that's characteristic polynomial is the product of linear factors is upper triangular, so I want some kind of induction.However, I am confused about where to start, and how I can relate the algebraic multiplicity to the size of this block.

Any help in seeing and constructing this argument really appreciated.


Take a look at the Schur Decomposition Lemma. It is exactly what you need. The basic idea is to continuously construct find unitary matrices $U_i$ such that $A = U_1 U_2 \cdots U_k T U_k^{-1} \cdots U_{2}^{-1} U_{1}^{-1}$ has $A_{ij} = 0$ for all $i > j$ and $i \leq k$. After iteration $n$, we will have $A_{ij} = 0$ for all $i > j$, and the matrix you seek will be upper triangular, with respect to the basis represented in the matrix $U_1 U_{2}U_3 \cdots U_n$. Note that since all of the $U_i$ are unitary, so is their product. Letting $W$ be the product of all the $U_i$, we see that $T = W^{-1} A W$, for upper triangular $A$. Because the eigenvalues of an upper triangular matrix are simply the diagonal elements, and $T$ is similar to $A$, they must have the same eigenvalues (i.e. roots of the characteristic polynomial $\lambda$).


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