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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...

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    $\begingroup$ The eigenvalues of a triangular matrix are its diagonal entries. $\endgroup$ – Qiaochu Yuan Jun 21 '12 at 23:52
  • $\begingroup$ @QiaochuYuan Thanks! $\endgroup$ – Andrew Jun 21 '12 at 23:56
  • $\begingroup$ @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:) $\endgroup$ – Andrew Jun 22 '12 at 0:35
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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.

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