Show that $B$ $\in$ $GL_{n-1}($K$)$ $\Rightarrow$ $A$ $\in$ $GL_n$($K$)

Question

Let $K$ be a field and A = $ \begin{bmatrix} -1 & A_{12} \\ 0 & B \\ \end{bmatrix} $ $\in$ $K^{n,n}$ and $B$ $\in$ $K^{n-1,n-1}$. Show that $A$ $\in$ $GL_n$($K$) applies if $B$ $\in$ $GL_{n-1}($K$)$.

My thoughts and questions: $B$ is obviously a (n - 1) $\times$ (n - 1) matrix in the field $\mathbb{K}$. For example: For n = 4 is $B$ a $\underbrace{3}_{4-1}$ $\times$ $\underbrace{3}_{4-1}$ matrix in the field $\mathbb{K}$.

But generally I don't understand how $A$ $\in$ $GL_n$($K$) applies if $B$ $\in$ $GL_{n-1}($K$)$.

I thought about using this lemma: Let $A$ $\in$ $K^{n,m}$ with $A$ = $ \begin{bmatrix} A_{1} \\ A_{2} \\ \end{bmatrix}$ which means that $A_{1}$ $\in$ $K^{l,m}$, $A_{2}$ $\in$ $K^{n-l,m}$.

Therefore we have that Rank(A) $\le$ Rank($A_{1}$) + Rank($A_{2}$). But it is made for $A$ $\in$ $K^{n,m}$ and not for $A$ $\in$ $K^{n,n}$.

How do I beginn this proof? and how does $A$ $\in$ $GL_n$($K$) applies if $B$ $\in$ $GL_{n-1}($K$)$? (I really don't why this correct). Any hints guiding me to the right direction I much appreciate.

Answer 1

We don't really need determinants to solve this one; we can work directly from the definition of $GL_m(K)$.

We need to show that

$B \in GL_{n - 1}(K) \Longrightarrow A = \begin{bmatrix} -1 & A_{12} \\ 0 & B \end{bmatrix} \in GL_n(K); \tag 1$

that is, $A$ is invertible if $B$ is. We recall that invertible matrices are characterized by vanishing kernel; therefore we need to show

$\ker B = \{0\} \Longrightarrow \ker A = \{0\}; \tag 2$

so let

$\vec a \in K^n; \tag 3$

we may write

$\vec a = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix}; \tag 4$

if

$\vec a \in \ker A, \tag 5$

then we have

$A \vec a = 0, \tag 6$

or

$\begin{bmatrix} -1 & A_{12} \\ 0 & B \end{bmatrix} \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} = 0; \tag 7$

we may also set

$\vec b = \begin{pmatrix} a_2 \\ a_3 \\ \vdots \\ a_n \end{pmatrix} \in K^{n - 1}; \tag 8$

then (7) may be written

$\begin{bmatrix} -1 & A_{12} \\ 0 & B \end{bmatrix} \begin{pmatrix} a_1 \\ \vec b \end{pmatrix} = 0; \tag 9$

thus, performing the matrix-vector multiplication indicated in this equation,

$\begin{pmatrix} -a_1 + A_{12} \vec b \\ B\vec b \end{pmatrix} = 0, \tag{10}$

whence

$-a_1 + A_{12} \vec b = 0, \tag{11}$

and

$B \vec b = 0; \tag{12}$

since we assume $B \in GL_{n - 1}(K)$ we see that (12) implies

$\vec b = 0, \tag{13}$

from which (11) yields

$-a_1 = -a_1 + A_{12} \vec b = 0; \tag{14}$

we thus see from (4), (8), (11), (12), (13) and (14) that

$\vec a = 0 \Longrightarrow \ker A = 0 \Longrightarrow A \in GL_n(K). \tag{15}$

It is worth noting that the above demonstration may be "walked backwards" to show that

$A \in GL_n(K) \Longrightarrow B \in GL_{n - 1}(K) \tag{16}$

as well; we need show that

$\ker A = \{0\} \Longrightarrow \ker B = \{0\}; \tag{17}$

now if

$\vec b \in \ker B, \tag{18}$

that is,

$B \vec b = 0, \tag{19}$

then with

$a_1 = A_{12} \vec b \tag{20}$

we have

$\vec a = \begin{pmatrix} a_1 \\ \vec b \end{pmatrix} \tag{21}$

satisfying

$A \vec a = 0; \tag{22}$

therefore

$\vec a \in \ker A \Longrightarrow \vec a = 0 \Longrightarrow \vec b = 0 \Longrightarrow \ker B = \{0\}, \tag{23}$

whence

$B \in GL_{n - 1}(K). \tag{24}$

Answer 2

HINT:

$$ \det A=-\det B\neq0 $$

Answer 3

Hint:

The matrix $A$ is block-triangular (with square blocks). Hence $\det A$ is the product of the determinants of the blocks.

Answer 4

It is enough to prove that $A$ is surjective as a linear map.

Let $y \in K^n$. We need to find $x \in K^n$ such that $Ax^T=y^T$.

Since $B$ is surjective, there is $(x_2,\dots,x_n) \in K^{n-1}$ such that $B(x_2,\dots,x_n)^T=(y_2,\dots,y_n)^T$. Let $x_1 = A_{12} \cdot (x_2,\dots,x_n)-y_1$. Then $x=(x_1,x_2,\dots,x_n)$ satisfies $Ax^T=y^T$.

Answer 5

Using Leibniz formula : $$\det(A)=\sum_{\sigma \in \mathfrak S_n}\varepsilon(\sigma )\prod_{i=1}^n a_{i\sigma (i)}=-\sum_{\sigma \in \mathfrak S_n: \sigma (1)=1}\varepsilon(\sigma )\prod_{i=2}^na_{i\sigma (i)}=-\det(B).$$

Answer 6

You can find an inverse of $A$, which directly proves that $A$ is invertible:

The matrix $A$ is in upper triangular block form, so the inverse ought to be of the form $$ \begin{bmatrix} -1 & ? \\ 0 & B^{-1} \end{bmatrix}. $$ By considering the product $$ \begin{bmatrix} -1 & A_{12} \\ 0 & B \end{bmatrix} \begin{bmatrix} -1 & ? \\ 0 & B^{-1} \end{bmatrix} = \begin{bmatrix} 1 & -? + A_{12} B^{-1} \\ 0 & I_{n-1} \end{bmatrix}, $$ so we see that the unknown entry $?$ should be $A_{12} B^{-1}$.

We have guessed the right matrix because $$ \begin{bmatrix} -1 & A_{12} \\ 0 & B \end{bmatrix} \begin{bmatrix} -1 & A_{12} B^{-1} \\ 0 & B^{-1} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & I_{n-1} \end{bmatrix} = I_n $$ and $$ \begin{bmatrix} -1 & A_{12} B^{-1} \\ 0 & B^{-1} \end{bmatrix} \begin{bmatrix} -1 & A_{12} \\ 0 & B \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & I_{n-1} \end{bmatrix} = I_n. $$ This shows that $A$ is invertible with $$ \begin{bmatrix} -1 & A_{12} \\ 0 & B \end{bmatrix}^{-1} = \begin{bmatrix} -1 & A_{12} B^{-1} \\ 0 & B^{-1} \end{bmatrix}. $$