0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\lambda$).

$\endgroup$

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.