How to prove the following exercise by using the definition of a determinant?

Question

$\begin{align} \begin{vmatrix} a_{11} & \cdots& a_{1m} & 0 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\ 0 & \cdots & 0 & 1 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \ddots & \cdot \\ 0 & \cdots & 0 & 0 & \cdots & 1 \end{vmatrix} = \begin{vmatrix} a_{11} & \cdots& a_{1m} \\ \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} \\ \end{vmatrix}. \end{align}$

i.e. Definition of a determinant; The determinant of the array A is the number $$ \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots , \lambda_n)} a_{1\lambda_1} ,\cdots, a_{n\lambda_n} $$ where the summation extends over all n! arrangements $(\lambda_1,\cdots ,\lambda_n)$ of $(1, \cdots, n)$. This determinant is denoted by $\begin{align} \begin{vmatrix} a_{11} & a_{21} & \cdots & a_{1m}\\ a_{21} & a_{22} & \cdots & a_{2m}\\ \cdot & \cdot & \cdots & \cdot\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix}, \text{or more briefly, by} \end{align}$ $\begin{align} {\begin{vmatrix} a_{ij}\\ \end{vmatrix}_n}. \end{align}$ $\begin{align} A = \begin{matrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \cdot & \cdot &\cdots & \cdot\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{matrix} \end{align}$

This exercise is taken from the book 'An Introduction to Linear Algebra' by L. Mirsky. Page number 13, exercise 1.3.1.

Answer 1

Using the definition of the determinant you have mentioned, you can show that every non-zero term in the summation for the left side lies in that of the right, and vice versa.

Look at the subpermutation $(\lambda_{m+1}, \ldots, \lambda_n)$ of the permutation on $\{1,\ldots, n\}$. Note that if any for any $m<i\leq n$, $\lambda_i \neq i$, then $a_{i,\lambda_i} = 0$, and hence the corresponding term will be zero.

Hence all the non-zero terms of the left summation have the permutation, when restricted to the last $n-m$ entries, equal to the identity. This also means that the sign of the permutation only depends on the action on the first $m$ terms.

Since it's obvious that every term in the summation of the right side corresponds to the term on the left side with the same permutation on $\{1,\ldots,m\}$ and identity on $\{m+1,\ldots,n\}$, you have that the two summations are equal. Hence the determinants are equal.

Answer 2

Use the fact that the determinant of a block matrix is equal to the product of the determinants of matrix $A$ and $B$.

So $$\begin{vmatrix} A & 0 \\ 0 & B\end{vmatrix} = \det(A) \det(B)$$

Here, $A$ is an $m\times m$ matrix, $B$ is a $(n - m)\times (n - m)$ matrix.

This can be proven by induction.

Answer 3

I strongly believe that the best way to show my indebtedness is to write out the proof. Without (Milind's) assistance I would not have completed it.

$\begin{align} Given; D^{/}= \begin{vmatrix} a_{11} & \cdots& a_{1m} & 0 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\ 0 & \cdots & 0 & 1 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \ddots & \cdot \\ 0 & \cdots & 0 & 0 & \cdots & 1\\ \end{vmatrix}. \end{align}$

$\begin{align} To prove; D^{/}= \begin{vmatrix} a_{11} & \cdots& a_{1m} \\ \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} \\ \end{vmatrix}. \end{align}$

Proof; $\begin{align} {a_{ij}^/} = \begin{cases} a_{ij};\qquad i=j:1\leq\ j\le\ m \\[2ex] a_{ij} = 0; i\neq\ j: m\lt j \le n\\[2ex] a_{ij} = 1; i=j: m+1\le\ j\le\ n; WHY?- (Hypothesis). \end{cases} \end{align}$

The determinant of the array A is the number $$\sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots , \lambda_n)} a_{1\lambda_1} ,\cdots, a_{n\lambda_n}$$ where the summation extends over all n! arrangements $(\lambda_1,\cdots ,\lambda_n)$ of $(1, \cdots, n)$. This determinant is denoted by $\begin{align} \begin{vmatrix} a_{11} & a_{21} & \cdots & a_{1m}\\ a_{21} & a_{22} & \cdots & a_{2m}\\ \cdot & \cdot & \cdots & \cdot\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix}, \text{or more briefly, by} \end{align}$ $\begin{align} {\begin{vmatrix} a_{ij}\\ \end{vmatrix}_n}; (Reason [1]). \end{align}$

$\begin{align} D^{/}= \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots , \lambda_n)} a_{1\lambda_1}^/ ,\cdots, a_{n\lambda_n}^/; WHY?- (Reason [1]).\end{align}$ $\begin{align} D^{/}= \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots ,\lambda_m, \lambda_{m+1}, \cdots, \lambda_n)} a_{1\lambda_1}^/ ,\cdots, a_{m\lambda_m}^/, a_{{m+1}\lambda_{m+1}}^/, \cdots, a_{n\lambda_n}^/ \end{align}$

For any $j$, where $i=j:1\leq\ j\le\ m$, we observe that $a_{ij} = a_{ij}^/$. It follows immediately that $\begin{align} D^{/}= \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots , \lambda_n)} a_{1\lambda_1} ,\cdots, a_{n\lambda_n}; WHY?- (Hypothesis).\end{align}$ $\begin{align} D^{/}= \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots ,\lambda_m, \lambda_{m+1}, \cdots, \lambda_n)} a_{1\lambda_1} ,\cdots, a_{m\lambda_m}, a_{{m+1}\lambda_{m+1}}, \cdots, a_{n\lambda_n} \end{align}$

By looking at the sub-permutation $(\lambda_{m+1}, \cdots, \lambda_n)$ of the longer permutation $(1, \cdots, n)$ we immediately observe that for any $j$, where $i\neq\ j: m\lt j \le n$, $a_{ij} = 0$. Hence the corresponding term will be zero ; WHY?- (Hypothesis).

And in consequence, for any $j$, when $i=j$ and $m+1\le\ j\le\ n$, $a_{ij} = 1$; the non-zero terms which are left in the summation have their permutation, when restricted to the last $n-m$ entries, equal to the identity. This clearly implies that the sign of the permutation is dependent only on the action of the first $m$ terms; WHY?- (Hypothesis).

Hence it follows from the previous results we obtained that $\begin{align}D^{/}= \sum_{(\lambda_1, \cdots , \lambda_m)} {\epsilon (\lambda_1, \cdots , \lambda_m)} a_{1\lambda_1} ,\cdots, a_{m\lambda_m}\end{align}$

$\begin{align} D^{/}= \begin{vmatrix} a_{11} & \cdots& a_{1m} \\ \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} \\ \end{vmatrix} \end{align}$;WHY?- (Reason [1]).

Q.E.D.

This completes the proof of Exercise 1.3.1.

Answer 4

Deduce, by means of multiplication theorem of determinants, that

$\begin{align} \begin{vmatrix} a_{11} & \cdots& a_{1m} & 0 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\ 0 & \cdots & 0 & b_{11} & \cdots & b_{1n} \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ 0 & \cdots & 0 & b_{n1} & \cdots & b_{nn}\\ \end{vmatrix} = \begin{vmatrix} a_{11} & \cdots& a_{1m} \\ \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} \\ \end{vmatrix} \begin{vmatrix} b_{11} & \cdots& a_{1n} \\ \cdot & \cdots & \cdot \\ a_{n1} & \cdots & a_{nn} \\ \end{vmatrix}. \end{align}$

$\begin{align} Given; D^{//} = \begin{vmatrix} a_{11} & \cdots& a_{1m} & 0 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\ 0 & \cdots & 0 & b_{11} & \cdots & b_{1n} \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ 0 & \cdots & 0 & b_{n1} & \cdots & b_{nn}\\ \end{vmatrix}. \end{align}$

$\begin{align} To prove; D^{//} = \begin{vmatrix} a_{11} & \cdots& a_{1m} \\ \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} \\ \end{vmatrix} \begin{vmatrix} b_{11} & \cdots& a_{1n} \\ \cdot & \cdots & \cdot \\ a_{n1} & \cdots & a_{nn} \\ \end{vmatrix}. \end{align}$

Proof; $\begin{align} \begin{vmatrix} a_{11} & \cdots& a_{1m} & 0 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\ 0 & \cdots & 0 & 1 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \ddots & \cdot \\ 0 & \cdots & 0 & 0 & \cdots & 1\\ \end{vmatrix} = \begin{vmatrix} a_{11} & \cdots& a_{1m} \\ \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} \\ \end{vmatrix} \end{align}$; WHY?- (Proof of Exercise 1.3.1)

$\begin{align} \begin{vmatrix} 1 & \cdots& 0 & 0 & \cdots & 0 \\ \cdot & \ddots & \cdot & \cdot & \cdots & \cdot \\ 0 & \cdots & 1 & 0 & \cdots & 0 \\ 0 & \cdots & 0 & b_{11} & \cdots & b_{1n} \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ 0 & \cdots & 0 & b_{n1} & \cdots & b_{nn}\\ \end{vmatrix} = \begin{vmatrix} b_{11} & \cdots& b_{1n} \\ \cdot & \cdots & \cdot \\ b_{n1} & \cdots & b_{nn} \\ \end{vmatrix} \end{align}$; WHY?- (Corollary to Exercise 1.3.1)

$\begin{align} \begin{vmatrix} a_{11} & \cdots& a_{1m} & 0 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\ 0 & \cdots & 0 & 1 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \ddots & \cdot \\ 0 & \cdots & 0 & 0 & \cdots & 1\\ \end{vmatrix} \begin{vmatrix} 1 & \cdots& 0 & 0 & \cdots & 0 \\ \cdot & \ddots & \cdot & \cdot & \cdots & \cdot \\ 0 & \cdots & 1 & 0 & \cdots & 0 \\ 0 & \cdots & 0 & b_{11} & \cdots & b_{1n} \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ 0 & \cdots & 0 & b_{n1} & \cdots & b_{nn}\\ \end{vmatrix}= \begin{vmatrix} a_{11} & \cdots& a_{1m} \\ \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} \\ \end{vmatrix} \begin{vmatrix} b_{11} & \cdots& b_{1n} \\ \cdot & \cdots & \cdot \\ b_{n1} & \cdots & b_{nn} \\ \end{vmatrix}; \end{align}$

WHY? - Reason [2] (Axiom [3]: If equals are multiplied by, or divided by, equals the results are equal).

$\begin{align} Let\, A = {\begin{vmatrix} a_{ij}\\ \end{vmatrix}_n} \end{align}$ and $\begin{align} B = \begin{vmatrix} b_{ij}\\ \end{vmatrix}_n \end{align}$ be given determinants, and write $\begin{align} C = \begin{vmatrix} c_{ij}\\ \end{vmatrix}_n \end{align}$, where $\begin{align} c_{rs} = \sum_{i=1}^{n} a_{ri} b_{is} \quad (r,s = 1,...,n). Then \; AB = C.\end{align}$; WHY ?- Reson [3] (Multiplication theorem for determinants).

Now we are in a position to show that the product of the two determinants on the left hand side is equal to the product of the right hand side. Let the product on the left hand side be equal to $\overline D$, then, if possible, we shall prove that $\begin{align} \overline D = D^{//}.\end{align}$

$\begin{align} \overline D = \begin{vmatrix} a_{11} & \cdots& a_{1m} & 0 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\ 0 & \cdots & 0 & 1 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \ddots & \cdot \\ 0 & \cdots & 0 & 0 & \cdots & 1\\ \end{vmatrix} \begin{vmatrix} 1 & \cdots& 0 & 0 & \cdots & 0 \\ \cdot & \ddots & \cdot & \cdot & \cdots & \cdot \\ 0 & \cdots & 1 & 0 & \cdots & 0 \\ 0 & \cdots & 0 & b_{11} & \cdots & b_{1n} \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ 0 & \cdots & 0 & b_{n1} & \cdots & b_{nn}\\ \end{vmatrix} \end{align}$.

$\begin{align} c_{rs} = \begin{cases} c_{rs} = \sum_{i=1}^{m} a_{ri} b_{is}; \,\qquad (1\leq\ r,s\le\ m)\\[3ex] c_{rs} = \sum_{i=m+1}^{n} a_{ri} b_{is};\,\quad (m+1\leq\ r,s\le\ n)\\[3ex] c_{rs} = \sum_{i=m+1}^{m} a_{ri} b_{is}; \quad (1\leq\ r\le\ m), \, (m+1\leq\ s\le\ n)\\[3ex] c_{rs} = \sum_{i=1}^{m} a_{ri} b_{is}; \qquad (1\leq\ s\le\ m), \, (m+1\leq\ r\le\ n); WHY?- (Reason [3]).\\ \end{cases} \end{align}$

For any ($r,s$), where $1\leq\ r,s\le\ m$, we clearly observe that $\sum_{i=1}^{m} a_{ri} b_{is} = a_{rs}$, or more briefly, $c_{rs} = a_{rs}$; WHY?-(Hypothesis).

Now from the hypothesis and by reason [3] it follows that for any ($r,s$) (1). where $m+1\leq\ r,s\le\ n$, $\sum_{i=m+1}^{n} a_{ri} b_{is}$ = $b_{rs}$, or more briefly, $c_{rs} = b_{rs}$. (2). where $(1\leq\ r\le\ m) \, and \, (m+1\leq\ s\le\ n)$, $\sum_{i=m+1}^{n} a_{ri} b_{is} = 0$: $c_{rs} = 0$. (3). where $(1\leq\ s\le\ m) \, and \, (m+1\leq\ r\le\ n)$, $\sum_{i=m+1}^{n} a_{ri} b_{is} = 0$: $c_{rs} = 0$.

Hence the previous results are sufficient to show that

$\overline D$ = $D^{//}$; WHY?- (Reason [3]).

$\begin{align} \begin{vmatrix} a_{11} & \cdots& a_{1m} & 0 & \cdots & 0 \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\ 0 & \cdots & 0 & b_{11} & \cdots & b_{1n} \\ \cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\ 0 & \cdots & 0 & b_{n1} & \cdots & b_{nn}\\ \end{vmatrix} = \begin{vmatrix} a_{11} & \cdots& a_{1m} \\ \cdot & \cdots & \cdot \\ a_{m1} & \cdots & a_{mm} \\ \end{vmatrix} \begin{vmatrix} b_{11} & \cdots& a_{1n} \\ \cdot & \cdots & \cdot \\ a_{n1} & \cdots & a_{nn} \\ \end{vmatrix} \end{align}$;

WHY?-(Axiom [6]: A magnitiude may be displaced by its equal in any process).

Q.E.D.

This completes the deduction of Exercise 1.3.1.