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