Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is problem 4 from page 258 of Curtis's Linear Algebra: An Introductory Approach. I seem to be having trouble understanding something needed to solve the problem, which reads

Suppose A and B are matrices in triangular form, with zeros above the diagonal. Show that A $\times$ B has the same property, and hence that every eigenvalue of A $\times$ B can be expressed in the form $\alpha\beta$, where $\alpha$ is an eigenvalue on A and $\beta$ is an eigenvalue of B.

In the actual problem, there's a little dot above the cross in the "matrix product" A $\times$ B to signify that the resulting matrix represents a transformation on the vector space of tensors $V \otimes W$. Because this matrix has entries given by

$$\pmatrix{ \alpha_{11}\mathbf{B} & \alpha_{12}\mathbf{B} & ... & \alpha_{1n}\mathbf{B} \\ ... \\ \alpha_{n1}\mathbf{B} & \alpha_{n2}\mathbf{B} & ... & \alpha_{nn}\mathbf{B} }$$

when represented against the basis $\lbrace v_1 \otimes w_1 , ... , v_1 \otimes w_m , v_2 \otimes w_1 , ... , v_2 \otimes w_m , ... , v_n \otimes w_m \rbrace$, I can understand the first claim in the problem (i.e. that the matrix A $\times$ B is also triangular). I'm having trouble proving that every eigenvector of the above matrix can be represented as a product of eigenvectors of A and B.

Because of how linear transformations on tensor products are defined, we know that

$$ (\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{a} \otimes \mathbf{b}) = (\mathbf{Aa} \otimes \mathbf{Bb}) $$

Also, if $\gamma$ is an eigenvalue of A $\times$ B, then

$$ (\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{a} \otimes \mathbf{b}) = \gamma (\mathbf{a} \otimes \mathbf{b}). $$

But at this point, I'm stuck...

share|improve this question
2  
The eigenvalues of a triangular matrix are its diagonal entries. –  Qiaochu Yuan Jun 21 '12 at 23:52
    
@QiaochuYuan Thanks! –  Andrew Jun 21 '12 at 23:56
    
@QiaochuYuan Would you be able to post your comment as an answer? Then I can mark it as a solution to the problem. I realize I got tripped up on something really minor, but you did figure out where I got stuck:) –  Andrew Jun 22 '12 at 0:35

1 Answer 1

up vote 2 down vote accepted

The eigenvalues of a triangular matrix are its diagonal values.

You can prove this by computing the characteristic polynomial, but you can also without too much effort write down the eigenvectors inductively.

share|improve this answer

Your Answer

 
discard

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.