Let $A$ be a block upper triangular matrix:

$$A = \begin{pmatrix} A_{1,1}&A_{1,2}\\ 0&A_{2,2} \end{pmatrix}$$

where $A_{1,1} ∈ C^{p \times p}$, $A_{2,2} ∈ C^{(n-p) \times (n-p)}$. 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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    $\begingroup$ Thank you so much for posting this question! You may have just single handedly saved my research!!! $\endgroup$ – Paul Jan 31 '13 at 20:43
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    $\begingroup$ What if $A_{2,1}$ is not $0$ ? $\endgroup$ – Ashutosh Gupta May 12 '16 at 4:40
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    $\begingroup$ @tsiki What is $A_{21}$ is not zero? $\endgroup$ – Babai Jul 19 '16 at 7:44

Let $A$ be the original matrix of size $n \times n$. One way out is to use the identity. (Result from Schur Complement) https://en.wikipedia.org/wiki/Schur_complement

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

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 $\det(A_{22}-\lambda I_{n-k}) = 0$ 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 $\det(A_{22}-\lambda I_{n-k}) = 0$ 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 $\det(A-\lambda I_n) = \det(A_{1,1} - \lambda I_{k}) \times \det(A_{22} - \lambda I_{n-k})$.

  • $\begingroup$ @user17762 Is there a similar statement when $A_{21}$ is not zero matrix? $\endgroup$ – Babai Jul 19 '16 at 7:45

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 $


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

Hence if $\lambda$ is eigenvalue of $A_{1,1}$ then it's also eigenvalue of $A$. 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 $p_2$.

If $\lambda_2$ is also eigenvalue of $A_{1,1}$, we have proved above that it's also eigenvalue of $A$. So, let's assume it's not eigenvalue of $A_{1,1}$ - hence $|A_{1,1} - \lambda_2 I|\ne 0$. Now

$$\left( \begin{matrix} A_{1,1}&A_{1,2} \\ 0 &A_{2,2} \end{matrix} \right) \left( \begin{matrix} x \\ p_2 \end{matrix} \right) = \left( \begin{matrix} A_{1,1} x + A_{1,2} p_2 \\ \lambda_2 p_2 \end{matrix} \right) $$ We can make $ A_{1,1} x + A_{1,2} p_2 = \lambda_2 x$ by choosing $x = - (A_{1,1} - \lambda_2 I)^{-1} A_{1,2} \; p_2$; and so we found an eigenvector for $A$ with $\lambda_2$ as 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 $A_{1,1}$.

  • $\begingroup$ Well, you can't use exactly the same argument for an eigenvector $p_2$ of $A_{2,2}$ since $A (0, p_2)^T = (A_{1,2}p_2, A_{2,2}p_2)^T$. $\endgroup$ – Calle Feb 11 '11 at 1:40
  • $\begingroup$ You are right! Fortunately, I could fix it :-) $\endgroup$ – leonbloy Feb 11 '11 at 2:16
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    $\begingroup$ 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 $A_{11}^T$ $\endgroup$ – user17762 Feb 11 '11 at 2:21
  • $\begingroup$ Nice argument though. $\endgroup$ – user17762 Feb 11 '11 at 5:37
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    $\begingroup$ @Babai If $A_{21}\ne 0$ but $A_{12} = 0$ it's the same thing. If both off diagonal submatrices are non zero, then the thesis is false (eigenvalues of the full matrix are not given by the diagonal submatrices alone). $\endgroup$ – leonbloy Jul 19 '16 at 12:04

For another approach for a proof you can use the Gershgorin disc theorem (sometimes Hirschhorn due to pronounciation differences between alphabets) to prove the disks for the individual matrices are the same as the discs for the large matrix so the sets of possible eigenvalues must be the same. This is because the radial contribution to the disks are 0 all over all entries for the lower left block since $|0| = 0$ and $0+0=0$.


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