Triangular Similar Matrix question

[Alupsaiu]

Hi,

Is every matrix similar to a triangular matrix? If it is, anyone have an idea how to prove it?
Thanks

 

[Petr Mugver]

Hi,

Is every matrix similar to a triangular matrix? If it is, anyone have an idea how to prove it?
Thanks

The proof can be done by induction. (Idea): In the field of complex numbers any n x n matrix has at least an eigenvector v. Complete v to get a basis of C^n. With respect to this basis, the first column of A has only the first element (let's call it a) eventually non zero, so A is similar to a block-triangular matrix, with a being the first 1 x 1 block and a certain (n-1)x(n-1) matrix B being the second block. Apply the inductive hypotesis to B and you get the result.

 

[HallsofIvy]

Every matrix is similar to a diagonal matrix or to a "Jordan Normal Form" both of which are upper triangular.

 

[td21]

schur triangle theorem and it can be unitary similar

 

[blue_raver22]

for your question. The answer is no, not every matrix is similar to a triangular matrix. In fact, there are many matrices that cannot be transformed into a triangular matrix through similarity transformations. One way to prove this is by considering the Jordan canonical form of a matrix, which is a triangular matrix that represents the same linear transformation as the original matrix. It is known that not all matrices have a Jordan canonical form, which means they cannot be similar to a triangular matrix. Another way to prove this is by considering the eigenvalues of a matrix. If a matrix has distinct eigenvalues, it can be diagonalized and therefore similar to a diagonal matrix, which is a special case of a triangular matrix. However, if a matrix has repeated eigenvalues, it may not be diagonalizable and therefore not similar to a triangular matrix. I hope this helps clarify things.