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.

Let $A$ be a block upper triangular matrix:

$A = \left( \begin{matrix} A_{1,1}&A_{1,2}\\ 0&A_{2,2} \end{matrix} \right)$ where $A_{1,1} ∈ C^{p×p}$, $A_{2,2} ∈ C^{(n-p)×(n-p)}$

Show that the eigenvalues of $A$ are the combined eigenvalues of $A_{1,1}$ and $A_{2,2}$

I've been pretty much stuck looking at this for a good hour and a half, so any help would be much appreciated. Thanks.

share|improve this question
    
Thank you so much for posting this question! You may have just single handedly saved my research!!! –  Paul Jan 31 '13 at 20:43

2 Answers 2

up vote 7 down vote accepted

Let $A$ be the original matrix of size $n \times n$. One way out is to use the identity.

$\det \left( \begin{matrix} B_{1,1}&B_{1,2}\\ B_{2,1 }&B_{2,2} \end{matrix} \right) = \det(B_{1,1}) \times \det(B_{22} - B_{21}B_{11}^{-1}B_{12})$.

We know that $\lambda$ is a number such that $Ax = \lambda x$. From which we get $\det(A-\lambda I_n) = 0$.

In your case, the matrix $A_{21}$ is a zero matrix and hence, we get $\det(A-\lambda I_n) = \det \left( \left( \begin{matrix} A_{1,1}&A_{1,2}\\ 0&A_{2,2} \end{matrix}\right) - \lambda I_n \right) = \det \left( \begin{matrix} A_{1,1} - \lambda I_{k}&A_{1,2}\\ 0&A_{2,2} - \lambda I_{n-k} \end{matrix}\right)$

Hence $\det(A-\lambda I_n) = \det(A_{1,1} - \lambda I_{k}) \times \det(A_{22} - \lambda I_{n-k})$.

So we get that if $\lambda$ is an eigen value of $A_{11}$ or $A_{22}$, then either $\det(A_{11}-\lambda I_k) = 0$ or $\det(A_{11}-\lambda I_{n-k}) = 0$ and hence $\det(A-\lambda I_n) = 0$ and hence $\lambda$ is an eigenvalue of $A$.

Similarly, if $\lambda$ is an eigenvalue of $A$, then $\det(A-\lambda I_n) = 0$, then either $\det(A_{11}-\lambda I_k) = 0$ or $\det(A_{11}-\lambda I_{n-k}) = 0$ and hence $\lambda$ is an eigen value of $A_{11}$ or $A_{22}$.

Edit

There is actually a small error in the above argument.

You might wonder that if $\lambda$ is an eigenvalue of $A_{11}$, then $A_{11} - \lambda I_k$ is not invertible and hence the identity $\det \left( \begin{matrix} B_{1,1}&B_{1,2}\\ B_{2,1 }&B_{2,2} \end{matrix} \right) = \det(B_{1,1}) \times \det(B_{22} - B_{21}B_{11}^{-1}B_{12})$ is false since $B_{11}$ is not invertible.

However, there is an another identity $\det \left( \begin{matrix} B_{1,1}&B_{1,2}\\ 0&B_{2,2} \end{matrix} \right) = \det(B_{1,1}) \times \det(B_{22})$ which is always true. (Prove both the identites as an exercise).

We can make use of this identity to get $\det(A-\lambda I_n) = \det(A_{1,1} - \lambda I_{k}) \times \det(A_{22} - \lambda I_{n-k})$.

share|improve this answer

A simpler way is from the definition. Is is easy to show that if $\lambda_1$ is an eigenvalue of the upper diagonal block $A_{1,1}$, with eigenvector $p_1$, (size $n_1$) then it's also an eigenvalue of the full matrix, with the same eigenvector augmented with zeros.

$A_{1,1} \; p_1 = \lambda_1 p_1$ with $p_1 \ne 0 $

So

$ \left( \begin{matrix} A_{1,1}&A_{1,2} \\ 0 &A_{2,2} \end{matrix} \right) \left( \begin{matrix} p_1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} A_{1,1} \; p_1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} \lambda_1 p_1 \\ 0 \end{matrix} \right) = \lambda_1 \left( \begin{matrix} p_1 \\ 0 \end{matrix} \right) $

There are $n_1$ (counting multiplicity) such eigenvalues. The same applies to the lower diagonal block $A_{2,2}$. So we have found the $n_1$ + $n_2 = n$ eigenvalues of the full matrix. (Wrong! This only applied to block diagonal matrix - Fixed below)

Suposse now that $\lambda_2$ is eigenvalue of $A_{2,2}$ with eigenvector $p_2$

Then

$\left( \begin{matrix} A_{1,1}&A_{1,2} \\ 0 &A_{2,2} \end{matrix} \right) \left( \begin{matrix} x \\ p_2 \end{matrix} \right) = \left( \begin{matrix} A_{1,1} x + A_{1,2} p_2 \\ \lambda_2 p_2 \end{matrix} \right) $

If $\lambda_2$ is not an eigenvalue of $A_{1,1}$ , then we can make $ A_{1,1} x + A_{1,2} p_2 = \lambda x$ by choosing $x = - (A_{1,1} - \lambda_2 I)^{-1} A_{1,2} \; p_2$ and then we found an eigenvector for $A$ with $\lambda_2$ eigenvalue.

It this way, we showed that if $\lambda$ is eigenvalue of $A_{1,1}$ or $A_{2,2}$, then it's an eigenvalue of $A$.

To complete the proof, one should show the other way round: that if $\lambda$ is eigenvalue of $A$ then it's eigenvalue of $A_{1,1}$ or $A_{2,2}$. But that's easy:

$\left( \begin{matrix} A_{1,1}&A_{1,2} \\ 0 &A_{2,2} \end{matrix} \right) \left( \begin{matrix} x_1 \\ x_2 \end{matrix} \right) = \left( \begin{matrix} A_{1,1} \; x_1 + A_{1,2} \; x_2 \\ A_{2,2} \; x_2 \end{matrix} \right) = \left( \begin{matrix} \lambda \; x_1 \\ \lambda \; x_2 \end{matrix} \right) $

Now, either $x_2 = 0$ or not. If not, then $\lambda$ is eigenvalue of $A_{2,2}$. If yes, it's eigenvalue of $A_{1,1}$.

share|improve this answer
    
Well, you can't use exactly the same argument for an eigenvector $p_2$ of $A_{2,2}$ since $A (0, p_2)^T = (A_{1,2}p_2, A_{2,2}p_2)^T$. –  Calle Feb 11 '11 at 1:40
    
You are right! Fortunately, I could fix it :-) –  leonbloy Feb 11 '11 at 2:16
1  
You could probably save some writing by noting that $A$ and $A^T$ have the same eigen values. So your initial argument could be carried on to $A_{11}$ by transposing $A$ and doing the same argument for $A_{11}^T$ –  user17762 Feb 11 '11 at 2:21
    
Nice argument though. –  user17762 Feb 11 '11 at 5:37

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.