Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It's been a while since I've studied linear algebra, and I wanted to follow up on something I read on MathOverflow.

In this answer, KConrad mentions you can write any invertible $2\times2$ matrix as a conjugate of an upper triangular matrix. How does this work again?

If I wanted to more formally pose my question:

Suppose $A=\begin{bmatrix} a & b \\ c & d\end{bmatrix}\in GL_2(\mathbb{C})$. How could I determine matrices $B$ and $C$ such that $A=CBC^{-1}$ with $B$ upper triangular?

share|cite|improve this question
Does this help? – Billy Sep 2 '11 at 6:55
up vote 2 down vote accepted

You have two options, depending on whether you are working in exact or inexact arithmetic.

For exact arithmetic, Billy has mentioned the Jordan decomposition: in the decomposition $\mathbf A=\mathbf C\mathbf B\mathbf C^{-1}$, $\mathbf B$ can be diagonal if $\mathbf A$ is not defective; i.e. $\mathbf A$ has a complete eigenvector set (and thus the Jordan decomposition and eigendecomposition are equivalent in this case). If $\mathbf A$ is defective, $\mathbf B$ will have the form of a Jordan block:

$$\mathbf B=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$$

where $\lambda$ is the sole eigenvalue of $\mathbf A$.

In inexact arithmetic, the computation of the Jordan decomposition can be unstable. (See this article for a discussion of the difficulties.) The appropriate decomposition here is the Schur decomposition: here, in the decomposition $\mathbf A=\mathbf C\mathbf B\mathbf C^{-1}$, $\mathbf C$ is chosen to be unitary, $\mathbf C^{-1}=\mathbf C^{\ast}$. $\mathbf B$ becomes triangular.

If $\mathbf A$ is real and the eigenvalues of $\mathbf A$ are all real, then this decomposition can be done with an orthogonal matrix $\mathbf C$ for real $\mathbf A$. But if the two eigenvalues are complex conjugates, one must necessarily use a unitary $\mathbf C$ for decomposing $\mathbf A$. The best one can do with an orthogonal $\mathbf C$ in this case is to have $\mathbf B$ in the form

$$\mathbf B=\begin{pmatrix}\lambda&u\\v&\lambda\end{pmatrix}$$

where $\lambda\pm i\mu$ are the eigenvalues of $\mathbf A$, and $uv=-\mu$.

share|cite|improve this answer
Thanks for answering. – Gotye Sep 2 '11 at 21:15

First calculate the characteristic polynomial, i.e. $\det(A-xI)$, where $x$ is an indeterminate. Then get the roots, if they are different your matrix $B$ is $$\begin{bmatrix}\lambda_1 &0\\0 &\lambda_2\end{bmatrix}.$$ Otherwise your matrix $B$ is either $$\begin{bmatrix}\lambda_1 & 0\\\ 0&\lambda_1\end{bmatrix} \, \text{or}\, \begin{bmatrix}\lambda_1 & 1\\0&\lambda_1\end{bmatrix}.$$ For computing the matrix $C$ (and $B$) you have to compute the spaces $\ker(A-\lambda_1 I)$ and $\ker(A-\lambda_2 I)$ and $\ker (A-\lambda_1 I)^2$. Then take a basis of $\ker(A-\lambda_1 I)$ and complement it to a basis of $\ker(A-\lambda_1I)^2$. After rescaling your matrix $A$ will look like one in the second line with respect to this basis. The matrix $C$ is the base change matrix.

EDIT: Of course this is just the normal form explained in this concrete example as it was said in the comment.

share|cite|improve this answer
Thanks for explaining. – Gotye Sep 2 '11 at 21:15

Actually any complex square matrix (invertible or not and regardless of size) can be unitarily triangularized. The theoretical way to illustrate this fact is quite standand. For a $2\times 2$ matrix $A$, let $u=(x,y)^T$ be an eigenvector of $A$ corresponding to some eigenvalue $\lambda$. Normalize $u$ (i.e. $u\leftarrow u/(u^\ast u)$) to make it a unit vector. Let $U=\begin{pmatrix} x & -\bar{y} \\ y & \bar{x}\end{pmatrix}$. Then $U$ is a unitary matrix with $U^{-1}=U^\ast$ (the asterisk means conjugate transpose, i.e. $X^\ast=\bar{X}^T$) and hence the $(2,1)$-th entry of $U^{-1}AU$ is given by $$ \begin{eqnarray} &&(\textrm{2nd row of } U^{-1}) (A)(\textrm{1st col. of } U)\\ &=&(\textrm{2nd row of } U^\ast)\ (A)\ (\textrm{1st col. of } U)\\ &=&\begin{pmatrix} -y & x\end{pmatrix} A \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} -y & x\end{pmatrix} \lambda \begin{pmatrix} x\\y\end{pmatrix} =0. \end{eqnarray} $$ Since the $(2,1)$-th entry is zero, $B=U^{-1}AU$ is upper triangular.

share|cite|improve this answer
Thanks user1551. – Gotye Sep 2 '11 at 21:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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