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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.

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

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 $A_{22}$.


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

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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 $


$ \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 $p_2$


$\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 $A_{1,1}$.

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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
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
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