# Determinant and Hyper Volume

## Introduction

The determinant is a function that maps a square matrix to a scalar. One of the geometric interpretations of the determinant is that its absolute value represents the hyper volume of the parallelepiped spanned by the row or column vectors of the matrix. This property of the determinant is very useful in many applications, such as Calculus.

In this article, I will first prove that the determinant exists and is unique for any square matrix. Then I will prove that the absolute value of the determinant is the hyper volume of the parallelepiped spanned by the row or column vectors of the matrix.

## The Definition of Determinant

The determinant is a function

$$

\det: \mathbb{R}^{n \times n} \rightarrow \mathbb{R}

$$

where $\mathbb{R}^{n \times n}$ is the set of all $n \times n$ real matrices.

The determinant satisfies the following fundamental properties:

- The determinant of the identity matrix is 1.
- Swapping two rows changes the sign of the determinant.
- Scaling a row by a scalar multiplies the determinant by the same scalar.
- The determinant behaves like a linear function on the rows of the matrix.

There are a few key properties that can be further deduced from the above fundamental properties.

The determinant of a 1x1 matrix is the value of the matrix.

*Proof*Suppose $A = \{ a \}$ is a 1x1 matrix, using the fundamental property one and three, we have

$$

\det(A) = \det(\{ a \times 1 \}) = a \det(\{ 1 \}) = a

$$This concludes the proof. $\square$

If two rows of a matrix are the same, the determinant is 0.

*Proof*Suppose $A$ has two rows that are the same, using the fundamental property two, we can swap the two rows to get a new matrix $A’$, and we have $\det(A) = -\det(A’)$. Because $A$ and $A’$ differ only in the two rows, we also have $\det(A) = \det(A’)$. Therefore, the determinant of $A$ must be 0.

This concludes the proof. $\square$

If one row of a matrix is 0, the determinant is 0.

*Proof*Suppose $A$ has a row that is 0, using the fundamental property three, we can scale the row to get a new matrix $A’$, and we have $\det(A) = 0 \det(A’)$. Because $A$ and $A’$ differ only in the row, we also have $\det(A) = \det(A’)$. Therefore, the determinant of $A$ must be 0.

This concludes the proof. $\square$

Doing a row replacement, i.e., adding a multiple of one row to another row, does not change the determinant.

*Proof*Suppose $A$ and $B$ are exactly the same except that for the $i$-th row. The $i$-th row of $B$ is the the multiple of the $j$-th, where $j \neq i$, row of $A$ (and $B$). Using the fundamental property three and the deduced property two, we know $\det(B) = 0$.

Suppose $C$ is a consequence of row replacement from $A$ and $C_{i,:} = A_{i,:} + B_{i,:}$, using the fundamental property four, we have $\det(C) = \det(A) + \det(B)$. Because $\det(B) = 0$, we have $\det(C) = \det(A)$.

This concludes the proof. $\square$

## The Existence of Determinant

We will first show that the determinant exists by providing a recursive constructive definition of the determinant via cofactor expansion.

Let $A$ be an $n \times n$ matrix.

- The $(i, j)$ minor, denoted $A_{i,j}$, is the $(n-1) \times (n-1)$ matrix obtained from $A$ by deleting the $i$-th row and the $j$-th column.
- The $(i, j)$ cofactor $C_{i,j}$ is defined in terms of the $(i, j)$ minor as

$$

C_{i,j} = (-1)^{i+j} \det(A_{i,j})

$$

We will claim that the determinant of $A$ can be computed recursively via cofactor expansion as

- For any $i = 1, 2, \cdots, n$, we have the cofactor expansion along the $i$-th row as

$$

\begin{align}

\det(A) &= \sum_{j=1}^{n} a_{i,j} C_{i,j} \\

&= \sum_{j=1}^{n} a_{i,j} (-1)^{i+j} \det(A_{i,j})

\end{align}

$$

- For any $j = 1, 2, \cdots, n$, we have the cofactor expansion along the $j$-th column as

$$

\begin{align}

\det(A) &= \sum_{i=1}^{n} a_{i,j} C_{i,j} \\

&= \sum_{i=1}^{n} a_{i,j} (-1)^{i+j} \det(A_{i,j})

\end{align}

$$

Then we will show that such a recursive definition of the determinant actually satisfies the properties of the determinant.

*Proof*

We will only prove the cofactor expansion along the row is the same as the determinant of $A$ via *strong induction*. The proof for the cofactor expansion along the column is similar.

The base case is $n = 1$. In this case, it’s trivial to show that the cofactor expansion along the first row for any $1 \times 1$ matrix satisfies all the properties of the determinant.

Suppose when $n = k$, the cofactor expansion along the $i$-th row, for any $i = 1, 2, \cdots, k$, for any $k \times k$ matrix satisfies all the properties of the determinant.

When $n = k + 1$, we have the cofactor expansion along the the $i$-th row,

$$

\begin{align}

\det(A) &= \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \det(A_{i,j})

\end{align}

$$

To prove the cofactor expansion along the $i$-th row, for any $i = 1, 2, \cdots, k, k + 1$, for any $(k + 1) \times (k + 1)$ matrix has the determinant fundamental property one,

$$

\begin{align}

\det(I) &= \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \det(I_{i,j}) \\

&= 1 (-1)^{i+i} \det(I_{i,j}) \\

&= \det(I_{i,j}) \\

&= 1

\end{align}

$$

Note that we have used the determinant fundamental property one for $k \times k$ matrix so that $\det(I_{i,j}) = 1$.

To prove the cofactor expansion along the $i$-th row, for any $i = 1, 2, \cdots, k, k + 1$, for any $(k + 1) \times (k + 1)$ matrix has the determinant fundamental property three, we assume the row scaling was performed on the $l$-th row by a constant of $c$, resulting in a new matrix $A’$. The matrix $A$ and $A’$ differ only in the $l$-th row.

When $l = i$, it’s straightforward to show that the cofactor expansion along the $i$-th row of $A’$ is the the cofactor expansion along the $i$-th row of $A$ multiplied by $c$, i.e., $\det(A’) = c \det(A)$.

$$

\begin{align}

\det(A’) &= \sum_{j=1}^{k+1} a’_{i,j} (-1)^{i+j} \det(A’_{i,j}) \\

&= \sum_{j=1}^{k+1} ca_{i,j} (-1)^{i+j} \det(A_{i,j}) \\

&= c \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \det(A_{i,j}) \\

&= c \det(A)

\end{align}

$$

When $l \neq i$, the cofactor expansion along the $i$-th row of $A’$ can be computed as

$$

\begin{align}

\det(A’)

&= \sum_{j=1}^{k+1} a’_{i,j} (-1)^{i+j} \det(A’_{i,j}) \\

&= \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \left( c \det(A_{i,j}) \right) \\

&= c \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \det(A_{i,j}) \\

&= c \det(A)

\end{align}

$$

Note we have used the determinant fundamental property three for $k \times k$ matrix so that $\det(A’_{i,j}) = c \det(A_{i,j})$.

To prove the cofactor expansion along the $i$-th row, for any $i = 1, 2, \cdots, k, k + 1$, for any $(k + 1) \times (k + 1)$ matrix has the determinant fundamental property four, suppose matrix $A$, $B$, and $C$ are exactly the same except that for the $l$-th row. The $l$-th row of $C$ is sum of the $i$-th row of $A$ and the $i$-th row of $B$.

When $l = i$, it’s straightforward to show that the cofactor expansion along the $i$-th row of $C$ is the sum of the the cofactor expansion along the $i$-th row of $A$ and the cofactor expansion along the $i$-th row of $B$, i.e., $\det(C) = \det(A) + \det(B)$.

$$

\begin{align}

\det(C) &= \sum_{j=1}^{k+1} c_{i,j} (-1)^{i+j} \det(C_{i,j}) \\

&= \sum_{j=1}^{k+1} \left( a_{i,j} + b_{i,j} \right) (-1)^{i+j} \det(C_{i,j}) \\

&= \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \det(C_{i,j}) + \sum_{j=1}^{k+1} b_{i,j} (-1)^{i+j} \det(C_{i,j}) \\

&= \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \det(A_{i,j}) + \sum_{j=1}^{k+1} b_{i,j} (-1)^{i+j} \det(B_{i,j}) \\

&= \det(A) + \det(B)

\end{align}

$$

When $l \neq i$, the cofactor expansion along the $i$-th row of $C$ can be computed as

$$

\begin{align}

\det(C) &= \sum_{j=1}^{k+1} c_{i,j} (-1)^{i+j} \det(C_{i,j}) \\

&= \sum_{j=1}^{k+1} c_{i,j} (-1)^{i+j} \left( \det(A_{i,j}) + \det(B_{i,j}) \right) \\

&= \sum_{j=1}^{k+1} c_{i,j} (-1)^{i+j} \det(A_{i,j}) + \sum_{j=1}^{k+1} c_{i,j} (-1)^{i+j} \det(B_{i,j}) \\

&= \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \det(A_{i,j}) + \sum_{j=1}^{k+1} b_{i,j} (-1)^{i+j} \det(B_{i,j}) \\

&= \det(A) + \det(B)

\end{align}

$$

Note we have used the determinant fundamental property four for $k \times k$ matrix so that $\det(C_{i,j}) = \det(A_{i,j}) + \det(B_{i,j})$.

To prove the cofactor expansion along the $i$-th row, for any $i = 1, 2, \cdots, k, k + 1$, for any $(k + 1) \times (k + 1)$ matrix has the determinant fundamental property two, suppose the two exchanging rows are the $h$-th and the $l$-th row, resulting in a new matrix $A’$. The matrix $A$ and $A’$ differ only in the $h$-th and the $l$-th row.

When $l \neq i$ and $h \neq i$, the cofactor expansion along the $i$-th row of $A’$ is the same as the cofactor expansion along the $i$-th row of $A$, i.e., $\det(A’) = -\det(A)$.

$$

\begin{align}

\det(A’) &= \sum_{j=1}^{k+1} a’_{i,j} (-1)^{i+j} \det(A’_{i,j}) \\

&= \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \left( - \det(A’_{i,j}) \right) \\

&= - \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \det(A’_{i,j}) \\

&= - \det(A)

\end{align}

$$

Note we have used the determinant fundamental property two for $k \times k$ matrix so that $\det(A’_{i,j}) = - \det(A_{i,j})$.

When $l = i$, the situation becomes more complicated.

$$

\begin{align}

\det(A) &= \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \det(A_{i,j}) \\

&= \sum_{j=1}^{k+1} a_{l,j} (-1)^{l+j} \det(A_{l,j}) \\

\end{align}

$$

The $\det(A_{l,j})$ can also be computed using the cofactor expansion along any row $A_{l,j}$. This row is designated as the $i’$-th row in the original matrix $A$. The $i’$-th row, however, is not necessarily the $h$-th row in the original matrix $A$. But we could always swap the $h$-th row in the original matrix $A$ for $| i’ -h |$ times to get a new matrix $A’_{l,j}$ such that the $i’$-th row in the original matrix $A$ is the $h$-th row in the new matrix $A’_{l,j}$. According to the fundamental property two, for $k \times k$ matrix $A_{l,j}$ and $A’_{l,j}$, we have $\det(A_{l,j}) = (-1)^{| i’ - h |} \det(A’_{l,j})$.

We further define $A_{\{i, i^{\prime}\},\{j, j^{\prime}\}}$ as the $(n - 2) \times (n - 2)$ matrix obtained from $A$ by deleting the $i$-th row, the $i^{\prime}$-th row, the $j$-th column, and the $j^{\prime}$-th column.

$$

\begin{align}

\det(A) &= \sum_{j=1}^{k+1} a_{i,j} (-1)^{i+j} \det(A_{i,j}) \\

&= \sum_{j=1}^{k+1} a_{l,j} (-1)^{l+j} \det(A_{l,j}) \\

&= \sum_{j=1}^{k+1} a_{l,j} (-1)^{l+j} (-1)^{| i’ - h |} \det(A’_{l,j}) \\

&= (-1)^{| i’ - h |} \sum_{j=1}^{k+1} a_{l,j} (-1)^{l+j} \det(A’_{l,j}) \\

&= (-1)^{| i’ - h |} \sum_{j=1}^{k+1} \sum_{j’=1, j’ \neq j}^{k+1} a_{l,j} (-1)^{l+j} a_{h,j’} (-1)^{h+j’ - b_1 - b_2} \det(A_{\{l, h\},\{j, j’\}}) \\

&= (-1)^{| i’ - h |} \sum_{j=1}^{k+1} \sum_{j’=1, j’ \neq j}^{k+1} a_{l,j} a_{h,j’} (-1)^{l+j + h+j’ - b_1 - b_2} \det(A_{\{l, h\},\{j, j’\}}) \\

\end{align}

$$

where $b_1 = 1$ if $h < l$ and $b_1 = 0$ if $h > l$, and $b_2 = 1$ if $j < j’$ and $b_2 = 0$ if $j > j’$.

The coefficient for $\det(A_{\{l, h\},\{j, j’\}})$ must be $a_{l,j} a_{h,j’} (-1)^{l+j + h+j’ - b_1 - b_2} + a_{l,j’} a_{h,j} (-1)^{l+j’ + h+j - b’_1 - b’_2}$.

Because $b_1 = b’_1$ and $b_2 \neq b’_2$, we must have

$$

(-1)^{l+j + h+j’ - b_1 - b_2} = (-1) (-1)^{l+j’ + h+j - b’_1 - b’_2}

$$

Thus, in $\det(A)$, the coefficient for $\det(A_{\{l, h\},\{j, j’\}})$ must be either $a_{l,j} a_{h,j’} - a_{l,j’} a_{h,j}$ or $a_{l,j’} a_{h,j} - a_{l,j} a_{h,j’}$.

In $\det(A’)$, the coefficient for $\det(A_{\{l, h\},\{j, j’\}})$ must be $a_{l,j’} a_{h,j} - a_{l,j} a_{h,j’}$ when the coefficient for $\det(A_{\{l, h\},\{j, j’\}})$ in $\det(A)$ is $a_{l,j} a_{h,j’} - a_{l,j’} a_{h,j}$ and $a_{l,j} a_{h,j’} - a_{l,j’} a_{h,j}$ when the coefficient for $\det(A_{\{l, h\},\{j, j’\}})$ in $\det(A)$ is $a_{l,j’} a_{h,j} - a_{l,j} a_{h,j’}$.

Therefore, we must have

$$

\det(A) + \det(A’) = 0

$$

and

$$

\det(A) = -\det(A’)

$$

Taken together, we have shown that the cofactor expansion along the $i$-th row, for any $i = 1, 2, \cdots, k, k + 1$, for any $(k + 1) \times (k + 1)$ matrix satisfies all the fundamental properties of the determinant. The determinant must exist.

This concludes the proof. $\square$

## The Uniqueness of Determinant

With the cofactor expansion, we have shown that the determinant exists. But since there are many ways to do cofactor expansion for a matrix, the determinant might not be unique.

Because we have shown that the determinant exists, if we can show that the determinant is unique if it exists, then the determinant for any square matrix must exist and it is unique.

In this section, we will prove that the determinant is unique if it exists.

*Proof*

Using the fundamental property four of the determinant, the determinant of every $n \times n$ matrix can be uniquely described as the sum of the determinants of $n$ $n \times n$ matrices.

$$

\det(A) = \sum_{j=1}^{n} \det(A_{i,j})

$$

where $A_{i,j}$ is the $n \times n$ matrix that is almost the same as $A$ except that for the $i$-th row. The $i$-th row of $A_{i,j}$ consists of zeros except that the $j$-th element is the same as the $j$-th element of the $i$-th row of $A$.

The fundamental property four of the determinant can be applied recursively to each $\det(A_{i,j})$ until each line of the matrix contains at most one non-zero element. Using the fundamental property two and three of the determinant, the determinant of the matrix can be uniquely computed as a multiple of determinant of an identity matrix.

Therefore, the determinant is unique if it exists.

This concludes the proof. $\square$

## The Properties of Hyper Volume

Let $v_1$, $v_2$, $\cdots$, $v_n$ be $n$ vectors in $\mathbb{R}^n$. The hyper volume of the parallelepiped $P$ spanned by $v_1$, $v_2$, $\cdots$, $v_n$, can be uniquely determined from $v_1$, $v_2$, $\cdots$, $v_n$.

Let matrix $A$ be a matrix whose rows are $v_1$, $v_2$, $\cdots$, $\text{vol}(\{v_1, v_2, \cdots, v_n\}) = \text{vol}(A)$.

The hyper volume function $\text{vol}(A)$ has the following fundamental properties:

- The hyper volume of the identity matrix is 1.
- Swapping two rows does not the hyper volume.
- Scaling a row by a scalar multiplies the determinant by the absolute value of the same scalar.
- Doing a row replacement, i.e., adding a multiple of one row to another row, does not change the determinant.

## Determinant and Hyper Volume

We have shown that the determinant is the only function that satisfies the four fundamental properties and some other deduced properties of the determinant.

Similarly, we will show that the absolute value of the determinant is the only function that satisfies the four fundamental properties of the hyper volume.

*Proof*

The hyper volume function $\text{vol}(A)$ satisfies the four fundamental properties of the hyper volume must exist. The absolute value of the determinant is such a function.

The hyper volume function $\text{vol}(A)$ must also be unique if it exist. Similar to the proof of the uniqueness of the determinant, we could reduce the matrix $A$ to a diagonal matrix using the fundamental property four of the hyper volume. The hyper volume of the matrix can be uniquely computed as a multiple of the hyper volume of an identity matrix.

Therefore, the absolute value of the determinant is the only function that satisfies the four fundamental properties of the hyper volume.

$$

\text{vol}(A) = |\det(A)|

$$

This concludes the proof. $\square$

## FAQ

### The Definition of Determinant Variants

The four fundamental properties that defines determinant on Wikipedia and Interactive Linear Algebra are slightly different.

- The determinant of the identity matrix is 1.
- Swapping two rows changes the sign of the determinant.
- Scaling a row by a scalar multiplies the determinant by the same scalar.
- Doing a row replacement, i.e., adding a multiple of one row to another row, does not change the determinant.

Note that the fundamental property four becomes the deduced property four in my definition of the determinant and the original fundamental property three is gone. This definition of the determinant is actually equivalent to the definition of the determinant that I have provided in the beginning of this article.

Although I am not going into the details of the proof, during the induction proof of the existence of the determinant, once the fundamental property two has been proved for $(k+1) \times (k+1)$ matrix, the deduced property two can be used for $(k+1) \times (k+1)$ matrix. Then proving the fundamental property three for $(k+1) \times (k+1)$ matrix becomes very straightforward. The proof of the uniqueness of the determinant is also very similar to the proof of the uniqueness of the determinant that I have provided.

## References

Determinant and Hyper Volume