####
**Introducing a Fellow Slogger:**

Please note that this is not a problem solving episode (see Week 9). This final week, the instructors introduce

**simple induction**. A fellow slogger gives a nice induction post under Week 12 here.

You can also find my comments to the following slog under Week 3:

What is really cool about the induction post is that the slogger talks about how induction is used in a proof of the following statement: the determinant of any $n \times n$ identity matrix is $1$. In fact, I am so excited that a fellow slogger uses CSC165 concepts for greater and better things that I feel compelled to follow the slogger's footsteps. I will also use

**CSC165 concepts**as a guide for a problem that is outside the scope of this course:

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**The Fun Problem:**

**Problem**: Is the following statement true: the determinant of any upper triangular matrix is

the product of its diagonal entries?

The truth is I have no clue. In my Problem Solving Episode under Week 9
(under the heading Step 1: Understand the Problem), I work out examples of an
idea that I do not fully understand to gain intuition. We should work out examples for this problem
too.

First we know that an

**upper triangular matrix**is a $n \times n$ matrix $A$ that is defined by:
$A_{ij} = \begin{cases} a_{ij}
& i \leq j \\ 0 & i > j \end{cases}$.

Then the following matrices satisfy the definition nicely:

$\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}, \begin{bmatrix}
4 & 5 & 6 \\ 0 & 7 & 8 \\ 0 & 0 & 9 \end{bmatrix}$.

Secondly, let $A$ be a $n \times n$ matrix and $F$ be a field. Then the

**determinant**is a function $det: M_{n \times n}(F) \to F$ that is defined by:
$det(A) = \begin{cases}
A_{ij} & n = 1 \\ \sum\limits_{j= 1}^n (-1)^{i + j} A_{ij} det(\bar{A_{ij}})
& n \geq 2 \end{cases}$.

There is some fancy notation here but it is not difficult to
understand. For simplicity sakes, think
of $F$ as the set of real numbers. Also,
the strange bar on top of $A_{ij}$ just means $A$ with both row $i$ and column $j$ removed. We see that the determinant is
recursively defined. This may be useful
later.

To find the
determinant of the above matrices, use the definition. For the $2 \times 2$ matrix, we have:

$det(\begin{bmatrix}
1 & 2 \\ 0 & 3 \end{bmatrix}) = (-1)^{1 + 1}(1)(3) + (-1)^{1 + 2}(0) =
3$.

For the $3 \times 3$ matrix, we have:

$det(\begin{bmatrix} 4
& 5 & 6 \\ 0 & 7 & 8 \\ 0 & 0 & 9\end{bmatrix}) = $

$(-1)^{1 + 1}(4)det(\begin{bmatrix}
7 & 8 \\ 0 & 9 \end{bmatrix}) + (-1)^{1 + 2}(5)det(\begin{bmatrix}
0 & 8 \\ 0 & 9 \end{bmatrix}) + (-1)^{1 + 3}(6)det(\begin{bmatrix}
0 & 7 \\ 0 & 0 \end{bmatrix}) = $

$(4)(-1)^{1 +
1}(7)(8) + (-1)(5)(-1)^{1 + 1}(0)(9) + (5)(-1)^{1 + 1}(0)(0) =$

$(4)(7)(8) + 0 + 0 =
224$.

It seems like the statement is true and that we should try
proving it. Something interesting in the calculation is that the determinant of a matrix with a column of $0$s is $0$. Again, this may be useful later.

####

How should we prove the statement? A direct proof comes to mind but remember that the determinant is recursively defined. This makes it difficult to work with unless we know the determinant of a previous recursive case. Proof by simple induction sounds attractive here. The induction hypothesis may give us the determinant of a previous recursive case. Let us try it.

####
**The Attempt at a Solution:**

How should we prove the statement? A direct proof comes to mind but remember that the determinant is recursively defined. This makes it difficult to work with unless we know the determinant of a previous recursive case. Proof by simple induction sounds attractive here. The induction hypothesis may give us the determinant of a previous recursive case. Let us try it.

Since a $0
\times 0$ matrix is undefined, then we show that $\forall n \in N, n \geq 1
\implies$ the determinant of an upper triangular $n \times n$ matrix is
the product of its diagonal entries. By

**simple induction**on $n$, we show that this statement is true. The CSC165 proof structure for induction is our guide.
For the

**base case**, assume that $n = 1$. Let $A$ be a $1 \times 1$ upper triangular matrix. By the definition of the determinant, then $det(A) = A_{ij}$. Since $A_{ij}$ is the only diagonal entry, then statement is true for $n = 1$.
For the

**induction step**, assume that $n \in N, n \geq 1$ and that the statement is true for $n$. That will be our**induction hypothesis**. We need to show that the statement is true for $n + 1$. Let $A$ be a $(n + 1) \times (n + 1)$ upper triangular matrix. By the definition of the determinant, we have that
$det(A) = \sum\limits_{j = 1}^{n + 1} (-1)^{1 + j}A_{1j}
det(\bar{A_{1j}})$.

Remember
this observation: the determinant of a matrix with a column of $0$s is
$0$? Turns out, that is because a square
matrix with a column of $0$s is not invertible and some fancy theorem tells us
that the determinant of such a matrix is $0$.

Luckily, we
make the following observation:

$det(A) = \sum\limits_{j
= 1}^{n + 1} (-1)^{1 + j} A_{1j} det(\bar{A_{1j}}) = (-1) ^{1 + 1}A_{11} det(\bar{A_{11}}) + \sum\limits_{j = 2}^{n + 1} (-1)^{1 +
j}A_{1j} det(\bar{A_{1j}})$.

Notice that the matrix $\bar{A_{1j}}$ is $A$ without row $1$
and without a column $j \geq 2$. Minus the first element, column $1$ of
$A$ is a column of $0$s So the matrix $\bar{A_{1j}}$
also has a column of $0$s. Thus, $det(\bar{A_{1j}})
= 0$ and

$det(A) = (-1) ^{1 +
1}A_{11} det(\bar{A_{11}}) + \sum\limits_{j
= 2}^{n + 1} (-1)^{1 + j}A_{1j} det(\bar{A_{1j}}) = (-1) ^{1 + 1}A_{11} det(\bar{A_{11}})$.

The matrix $\bar{A_{11}}$
has both row $1$ and column $1$ removed but it is still an upper triangular $n \times n$ matrix. By the induction hypothesis, then $det(\bar{A_{11}})$
is the product of its diagonal entries. Notice
that the diagonal entries of $det(\bar{A_{11}})$ are the diagonal entries of $A$ without
$A_{11}$. Since

$det(A) = (-1) ^{1 +
1}A_{11} det(\bar{A_{11}}) = A_{11} det(\bar{A_{11}})$,

then we have $det(A)$ as the product of its diagonal entries. Since $A$ is a $(n + 1)
\times (n + 1)$ upper triangular matrix, then the statement is true for $n +
1$. By simple induction, then $\forall n
\in N, n \geq 1 \implies$ the statement is true for $n$. So that is my attempt is at this fun problem.