Determinant of block matrices with non-square blocks

Question

Let $A$ be $m \times n$ matrix, and $B$ be $n \times m$ matrix. Then

1. Show that $$\det\begin{bmatrix}I_{n} & B\\ A & I_{m} \end{bmatrix}=\det\begin{bmatrix}I_{m} & A\\ B & I_{n}\end{bmatrix}$$

2. Show that $$\det(I_{m}-AB)=\det(I_{n}-BA)$$

3. Is it true that $I_{m}-AB$ and $I_{n}-BA$ have the same rank?

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Someone please show me a way to start proving. I tried to use formulas for calculating block matrix with square matrices, but it didn't help me on this one. I really don't know where to start.

Answer 1

Hint: for the first one, note that $$ \begin{bmatrix} 0 & I_m\\ I_n & 0 \end{bmatrix} \begin{bmatrix} I_n & B\\ A & I_m \end{bmatrix} \begin{bmatrix} 0 & I_n\\ I_m & 0 \end{bmatrix} = \begin{bmatrix} I_m & A\\ B & I_n \end{bmatrix} $$

Answer 2

Since $\begin{bmatrix}I_{n} & 0\\ -A & I_{m}\end{bmatrix} \begin{bmatrix}I_{n} & B\\ A & I_{m}\end{bmatrix} = \begin{bmatrix}I_{n} & B\\ 0 & I_{m}-AB \end{bmatrix}$, and $\det \begin{bmatrix}I_{n} & 0\\ -A & I_{m}\end{bmatrix} = 1$, we have

$\det\begin{bmatrix}I_{n} & B\\ A & I_{m}\end{bmatrix} = \det \begin{bmatrix}I_{n} & 0\\ 0 & I_{m}-AB \end{bmatrix} = \det (I_{m}-AB) $.

Similarly, $\begin{bmatrix}I_{n} & -B\\ 0 & I_{m}\end{bmatrix} \begin{bmatrix}I_{n} & B\\ A & I_{m}\end{bmatrix} = \begin{bmatrix}I_{n}-BA & 0\\ A & I_{m} \end{bmatrix}$, and we have $\det\begin{bmatrix}I_{n} & B\\ A & I_{m}\end{bmatrix} = \det \begin{bmatrix}I_{n}-BA & 0\\ 0 & I_{m} \end{bmatrix} = \det (I_{n}-BA ) $.

Hence we have 2.

Swapping the roles of $A,B$ above shows that $\det\begin{bmatrix}I_{m} & A\\ B & I_{n}\end{bmatrix} = \det (I_{n}-BA) $, hence we have 1.

Finally, 3. Let $r_m = \operatorname{rk}(I_{m}-AB)$, $r_n = \operatorname{rk}(I_{n}-BA)$. The identities on Lines 1 and 3 above show that $n+r_m = m+r_n$, hence if $m\le n$, then $r_n = r_m+n-m$.

It follows that $r_n=r_m $ iff $n=m$.

Note: The above argument shows more. Suppose $n\ge m$, and $\lambda \neq 0$. Then \begin{eqnarray} \det (\lambda I -BA) &=& \det (\lambda (I -\frac{1}{\lambda}BA)) \\ &=& \lambda^n \det ( I -\frac{1}{\lambda}AB) \\ &=& \lambda^n \det ( \frac{1}{\lambda} ( \lambda I -AB)) \\ &=& \lambda^{n-m} \det ( \lambda I -AB) \end{eqnarray} This shows that the characteristic polynomials of $AB$ and $BA$ are the same except for $n-m$ roots at $0$.