## Upper-Triangular Matrices

Until further notice, I’ll be assuming that the base field is algebraically closed, like the complex numbers .

What does this assumption buy us? It says that the characteristic polynomial of a linear transformation is — like any polynomial over an algebraically closed field — guaranteed to have a root. Thus any linear transformation has an eigenvalue , as well as a corresponding eigenvector satisfying

So let’s pick an eigenvector and take the subspace it spans. We can take the quotient space and restrict to act on it. Why? Because if we take two representatives of the same vector in the quotient space, then . Then we find

which represents the same vector as .

Now the restriction of to is another linear endomorphism over an algebraically closed field, so *its* characteristic polynomial must have a root, and *it* must have an eigenvalue with associated eigenvector . But let’s be careful. Does this mean that is an eigenvector of ? Not quite. All we know is that

since vectors in the quotient space are only defined up to multiples of .

We can proceed like this, pulling off one vector after another. Each time we find

The image of in the th quotient space is a constant times itself, plus a linear combination of the earlier vectors. Further, each vector is linearly independent of the ones that came before, since if it weren’t, then it would be the zero vector in its quotient space. This procedure only grinds to a halt when the number of vectors equals the dimension of , for then the quotient space is trivial, and the linearly independent collection spans . That is, we’ve come up with a basis.

So, what does look like in this basis? Look at the expansion above. We can set for all . When we set . And in the remaining cases, where , we set . That is, the matrix looks like

Where the star above the diagonal indicates unknown matrix entries, and the zero below the diagonal indicates that all the entries in that region are zero. We call such a matrix “upper-triangular”, since the only nonzero entries in the matrix are on or above the diagonal. What we’ve shown here is that over an algebraically-closed field, any linear transformation has a basis with respect to which the matrix of the transformation is upper-triangular. This is an important first step towards classifying these transformations.

[…] a vector space over an algebraically-closed field has a basis with respect to which its matrix is upper-triangular. That is, it looks […]

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[…] to is the multiplicity of , which is the number of times shows up on the diagonal of an upper-triangular matrix for . Since the total number of diagonal entries is , we see that the dimensions of all the […]

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[…] Upper-Triangular Matrices Over an algebraically closed field we can always find an upper-triangular matrix for any linear endomorphism. Over the real numbers we’re not quite so lucky, but we can come […]

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[…] be a linear map from to itself. Further, let be a basis with respect to which the matrix of is upper-triangular. It turns out that we can also find an orthonormal basis which also gives us an upper-triangular […]

Pingback by Upper-Triangular Matrices and Orthonormal Bases « The Unapologetic Mathematician | May 8, 2009 |

no need of such material

Comment by shasha | May 21, 2009 |

Oh I’m

sosorry that I chose to cover a topic you see as unnecessary. I’ll be sure to run all my future topics by you first.Comment by John Armstrong | May 21, 2009 |

Useful article. Many books just prove this by induction without any explanation

Comment by vish | October 26, 2013 |

Surely in the first equation you meant to write

T(e_1) = lambda_1 * e_1

(instead of T(v) on LHS)

Comment by A Khan | June 17, 2009 |

Yes, sorry. Thanks for catching that.

Comment by John Armstrong | June 17, 2009 |

[…] with a complex transformation we’re done. We can pick a basis so that the matrix for is upper-triangular, and then its determinant is the product of its eigenvalues. Since the eigenvalues are all […]

Pingback by The Determinant of a Positive-Definite Transformation « The Unapologetic Mathematician | August 3, 2009 |

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Pingback by Every Self-Adjoint Transformation has an Eigenvector « The Unapologetic Mathematician | August 12, 2009 |

[…] nilpotent transformation, so all of its eigenvalues are . Specifically, we want those that are also upper-triangular. Thus the matrices we’re talking about have everywhere below the diagonal and all on the […]

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[…] in a given row is to the right of the leftmost nonzero entry in the row above it. For example, an upper-triangular matrix is in row echelon form. We put a matrix into row echelon form by a method called […]

Pingback by Row Echelon Form « The Unapologetic Mathematician | September 1, 2009 |

[…] the difference between and is some scalar multiple of . On the other hand, remember how we found upper-triangular matrices before. This time we peeled off one vector and the remaining transformation was the identity on the […]

Pingback by A Lemma on Reflections « The Unapologetic Mathematician | January 19, 2010 |