# Upper triangularising a matrix when $\lambda$ is a root of the characteristic polynomial of multiplicty r

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.

## 1 Answer

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 $$\lambda$$).