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Manfred Weis
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This question is a more precise version of this question.

Let's assume we have the matrix

$$\left( \begin{array}{ccccc} 0 & a & 0 & 0 & 0 \\ f & 0 & b & 0 & 0 \\ 0 & e & 0 & c & 0 \\ 0 & 0 & d & 0 & r \\ 0 & 0 & 0 & g & 0 \\ \end{array} \right)$$

If we square it, we get the matrix

$$\left( \begin{array}{ccccc} a f & 0 & a b & 0 & 0 \\ 0 & a f+b e & 0 & b c & 0 \\ e f & 0 & b e+c d & 0 & c r \\ 0 & d e & 0 & c d+g r & 0 \\ 0 & 0 & d g & 0 & g r \\ \end{array} \right)$$

We see that this matrix decomposes into two submatrices

$$C_1:=\left(\begin{array}{cccc} a f & a b & 0 \\ e f & b e+c d & c r \\ 0 & d g & g r \\ \end{array}\right)$$ and $$C_2:=\left(\begin{array}{cc} a f+b e & b c \\ d e & c d+g r \\ \end{array}\right)$$ Now, one can check explicitly that the two submatrices are isospectral apart apart from from one eigenvalue zero. I wonder if there is an abstract argument why this is so?

It would for instance follow if we can write the matrices as $C_1 = AB$ and $C_2= BA$, but I don't see how such a decomposition could work. In particular, this does not seem to be restricted to 5x5 matrices but holds for arbitrary matrices of the above form.

This question is a more precise version of this question.

Let's assume we have the matrix

$$\left( \begin{array}{ccccc} 0 & a & 0 & 0 & 0 \\ f & 0 & b & 0 & 0 \\ 0 & e & 0 & c & 0 \\ 0 & 0 & d & 0 & r \\ 0 & 0 & 0 & g & 0 \\ \end{array} \right)$$

If we square it, we get the matrix

$$\left( \begin{array}{ccccc} a f & 0 & a b & 0 & 0 \\ 0 & a f+b e & 0 & b c & 0 \\ e f & 0 & b e+c d & 0 & c r \\ 0 & d e & 0 & c d+g r & 0 \\ 0 & 0 & d g & 0 & g r \\ \end{array} \right)$$

We see that this matrix decomposes into two submatrices

$$C_1:=\left(\begin{array}{cccc} a f & a b & 0 \\ e f & b e+c d & c r \\ 0 & d g & g r \\ \end{array}\right)$$ and $$C_2:=\left(\begin{array}{cc} a f+b e & b c \\ d e & c d+g r \\ \end{array}\right)$$ Now, one can check explicitly that the two submatrices are isospectral apart apart from one eigenvalue zero. I wonder if there is an abstract argument why this is so?

It would for instance follow if we can write the matrices as $C_1 = AB$ and $C_2= BA$, but I don't see how such a decomposition could work. In particular, this does not seem to be restricted to 5x5 matrices but holds for arbitrary matrices of the above form.

This question is a more precise version of this question.

Let's assume we have the matrix

$$\left( \begin{array}{ccccc} 0 & a & 0 & 0 & 0 \\ f & 0 & b & 0 & 0 \\ 0 & e & 0 & c & 0 \\ 0 & 0 & d & 0 & r \\ 0 & 0 & 0 & g & 0 \\ \end{array} \right)$$

If we square it, we get the matrix

$$\left( \begin{array}{ccccc} a f & 0 & a b & 0 & 0 \\ 0 & a f+b e & 0 & b c & 0 \\ e f & 0 & b e+c d & 0 & c r \\ 0 & d e & 0 & c d+g r & 0 \\ 0 & 0 & d g & 0 & g r \\ \end{array} \right)$$

We see that this matrix decomposes into two submatrices

$$C_1:=\left(\begin{array}{cccc} a f & a b & 0 \\ e f & b e+c d & c r \\ 0 & d g & g r \\ \end{array}\right)$$ and $$C_2:=\left(\begin{array}{cc} a f+b e & b c \\ d e & c d+g r \\ \end{array}\right)$$ Now, one can check explicitly that the two submatrices are isospectral apart from one eigenvalue zero. I wonder if there is an abstract argument why this is so?

It would for instance follow if we can write the matrices as $C_1 = AB$ and $C_2= BA$, but I don't see how such a decomposition could work. In particular, this does not seem to be restricted to 5x5 matrices but holds for arbitrary matrices of the above form.

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Matrices with same eigenvalues

This question is a more precise version of this question.

Let's assume we have the matrix

$$\left( \begin{array}{ccccc} 0 & a & 0 & 0 & 0 \\ f & 0 & b & 0 & 0 \\ 0 & e & 0 & c & 0 \\ 0 & 0 & d & 0 & r \\ 0 & 0 & 0 & g & 0 \\ \end{array} \right)$$

If we square it, we get the matrix

$$\left( \begin{array}{ccccc} a f & 0 & a b & 0 & 0 \\ 0 & a f+b e & 0 & b c & 0 \\ e f & 0 & b e+c d & 0 & c r \\ 0 & d e & 0 & c d+g r & 0 \\ 0 & 0 & d g & 0 & g r \\ \end{array} \right)$$

We see that this matrix decomposes into two submatrices

$$C_1:=\left(\begin{array}{cccc} a f & a b & 0 \\ e f & b e+c d & c r \\ 0 & d g & g r \\ \end{array}\right)$$ and $$C_2:=\left(\begin{array}{cc} a f+b e & b c \\ d e & c d+g r \\ \end{array}\right)$$ Now, one can check explicitly that the two submatrices are isospectral apart apart from one eigenvalue zero. I wonder if there is an abstract argument why this is so?

It would for instance follow if we can write the matrices as $C_1 = AB$ and $C_2= BA$, but I don't see how such a decomposition could work. In particular, this does not seem to be restricted to 5x5 matrices but holds for arbitrary matrices of the above form.