# What do determinants and eigenvalues ​​actually mean?

## Determinant

The determinant (also called the determinant function in mathematical terms) assigns a number \$ a \ in \ mathbb {R} \$ to every square matrix \$ {A} \ in \ mathbb {R} ^ {n \ times n} \$:
\ begin {align *}
& {f}: \ mathbb {R} ^ {n {x} n} \ rightarrow \ mathbb {R}, \ textbf {A} \ rightarrow \ text {det} (\ textbf {A}) \
\ text {or} \ quad & {f}: \ mathbb {R} ^ {n {x} n} \ rightarrow \ mathbb {R}, \ textbf {A} \ rightarrow {| \ textbf {A} |}
\ end {align *}

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### What does the determinant indicate?

The determinant of a matrix \$ {A} \$ (\$ \ det ({A}) \$ or \$ | {A} | \$) specifies how the volume of a geometry composed of vertices scales when it passes through the matrix \$ A \$ is mapped. If the determinant is negative, the orientation of the corner points also changes.

• This means that the corner points of a geometry are mapped into a sub-vector space which is of a lower dimension than the one from which the column vectors of the matrix originate, e.g. if a 3 \$ x \$ 3 matrix points from the space (\$ \ dim (\ mathbb {R} ^ 3) \$ = 3) on a two-dimensional plane (or on a 1-dimensional straight line).
• This then means that the column vectors of the matrix do not form a basis.
• From this it can in turn be concluded that the column vectors are linearly dependent.
• So the matrix is ​​not invertible \$ \ ldots \$ etc.

\ begin {align}
\ text {det} (A) \ begin {cases} \ neq 0 & \ Rightarrow \ {A} ^ {- 1} \ text {exists.} \
= 0 & \ Rightarrow \ {A} ^ {- 1} \ text {does not exist.} \ End {cases}
\ end {align}

Here again a basic video on the determinant

### Calculation rules for determinants

matrix A. is square (\$ \ textbf {A} \ in \ mathbb {R} ^ {n x n} \$). Therefore, the following laws of calculation apply:
\ begin {align *}
& \ text {det} (\ textbf {A} \ cdot \ textbf {B}) && = \ text {det} (\ textbf {A}) \ cdot \ text {det} (\ textbf {B}) \ [2mm]
& \ text {det} (\ textbf {A} ^ T) && = \ text {det} (\ textbf {A}) \ [2mm]
& \ text {det} (\ textbf {A} ^ {- 1}) && = \ frac {1} {\ text {det} (\ textbf {A})} \ [2mm]
& \ text {det} (m \ cdot \ textbf {A}) && = m ^ n \ cdot \ text {det} (\ textbf {A}) \ quad \ text {with} \ quad m \ in \ mathbb { R}
\ end {align *}

If \$ \ text {det} (\ textbf {A}) = a \$, then:

1. Multiplication of a row (column) of A. with a number \$ \ lambda \$ \$ \ Rightarrow \ text {det} (\ textbf {A}) = \ lambda \ cdot a \$
2. Swap 2 rows (2 columns) of A. \$ \ Rightarrow \ det (\ widetilde {\ textbf {A}}) = - a \$
3. Addition of a multiple of a line to another line \$ \ Rightarrow \ det (\ widetilde {\ textbf {A}}) = a \$
4. Addition of a multiple of a column to another line \$ \ Rightarrow \ det (\ widetilde {\ textbf {A}}) = a \$

“… To another line (column)” ‚. We have to change this row (column) in the changed matrix and not the row (column) from which we have taken a multiple for the calculation!

..with \$ \ widetilde {\ textbf {A}} \$ as matrix \$ \ textbf {A} \$ after the respective manipulation

example

To 1) First line multiplied by 3
\ begin {align *}
\ text {det} (\ textbf {A}) = \ text {det}
\ left (
\ begin {matrix}% 3x3
3 & -1 & 2 \\
0 & 3 & 4 \\
1 & -2 & 0
\ end {matrix} \ right) = - 14, \ qquad \ text {det} (\ widetilde {\ textbf {A}}) = \ text {det}
\ left (
\ begin {matrix}% 3x3
9 & -3 & 6 \\
0 & 3 & 4 \\
1 & -2 & 0
\ end {matrix} \ right) = 3 \ cdot -14 = -42
\ end {align *}

To 2) Second and third columns swapped.
\ begin {align *}
\ text {det} (\ textbf {A}) = \ text {det}
\ left (
\ begin {matrix}% 3x3
3 & -1 & 2 \\
0 & 3 & 4 \\
1 & -2 & 0
\ end {matrix}
\ right) = - 14, \ qquad \ text {det} (\ widetilde {\ textbf {A}}) = \ text {det}
\ left (
\ begin {matrix}% 3x3
3 & 2 & -1 \\
0 & 4 & 3 \\
1 & 0 & -2
\ end {matrix}
\ right) = 14
\ end {align *}

To 3) \$ 3 \ cdot \ text {"third line"} - \ text {"first line"} \ Rightarrow \$ becomes the new first line
\ begin {align *}
\ text {det} (\ textbf {A}) = \ text {det}
\ left (
\ begin {matrix}% 3x3
3 & -1 & 2 \\
0 & 3 & 4 \\
1 & -2 & 0
\ end {matrix}
\ right) = - 14, \ qquad \ text {det} (\ widetilde {\ textbf {A}}) = \ text {det}
\ left (
\ begin {matrix}% 3x3
0 & -5 & -2 \\
0 & 3 & 4 \\
1 & -2 & 0
\ end {matrix}
\ right) = - 14
\ end {align *}

### Calculation of determinants

The general method for computing determinants of a \$ n \$ x \$ n \$ matrix is ​​Laplace's expansion theorem. The Sarrus rule exists especially for a 3 x 3 matrix and you should also remember the arithmetic scheme for a 2 x 2 matrix apart from the development theorem.

1 x 1 matrix

The determinant of a 1 x 1 matrix corresponds to its single element.

2 x 2 matrix
\ begin {align}
& \ text {det}
\ left (
\ begin {matrix}
a & b \
c & d
\ end {matrix} \ right) = a \ cdot d \ - \ c \ cdot b
\ end {align}

3 x 3 matrix - rule of Sarrus
\ begin {align}
& \ text {det}
\ left (
\ begin {matrix}
a & b & c \
d & e & f \
g & h & i \
\ end {matrix} \ right) = aei + bfg + cdh-ceg-bdi-afh
\ end {align}

n x n matrix - Laplace's development theorem (general procedure)
\ begin {align}%
\ text {Either} \ quad \ text {det} (\ textbf {A}) & = \ sum_ {j = 1} ^ n \ left [a_ {ij} \ cdot (-1) ^ {i + j} \ cdot D_ {ij} \ right] \ quad \ text {Development after \$ i \$ -th line} \
\ text {or} \ quad \ text {det} (\ textbf {A}) & = \ sum_ {i = 1} ^ n \ left [a_ {ij} \ cdot (-1) ^ {i + j} \ cdot D_ {ij} \ right] \ quad \ text {Development according to \$ j \$ -th column}
\ end {align}

What do the formulas mean and how are they to be read?

• It doesn't matter which row or column you develop according to (if possible, according to the row / column with the most zeros)
• Term \$ a_ {ij} \$: The element \$ a_ {ij} \$ from the matrix A.
• Term \$ (- 1) ^ {i + j} \$: "Checkerboard rule for signs".
Make a note of the matrix with the sign scheme
\ begin {align *}
\ left (
\ begin {matrix}
+ & - & + & - & \ cdots \
- & + & - & + & \ cdots \
+ & - & + & - & \ cdots \
- & + & - & + & \ cdots \
\ vdots & \ vdots & \ vdots & \ vdots & \ ddots
\ end {matrix}
\ right)
\ end {align *}
• Term \$ D_ {ij} \$: The so-called sub-determinant (or the minor) of the matrix A.if the \$ i \$ -th row and \$ j \$ -th column of A. is crossed out
\ begin {align *}
\ textbf {A} =
\ left (
\ begin {matrix}% 3x3
1 & 3 & 8 \\
-2 & 5 & 0 \\
4 & 6 & 7
\ end {matrix} \ right), \ quad D_ {32} =
\ left |
\ begin {matrix}% 2x2
1 & 8 \\
-2 & 0
\ end {matrix} \ right |, \ quad D_ {13} =
\ left |
\ begin {matrix}% 2x2
-2 & 5 \\
4 & 6
\ end {matrix} \ right |, \ quad D_ {22} =
\ left |
\ begin {matrix}% 2x2
1 & 8 \\
4 & 7
\ end {matrix}
\ right |
\ end {align *}
• (Terms \$ (- 1) ^ {i + j} \ cdot D_ {ij} \$ are called cofactors of the matrix A..)

Another example

\$ \ text {det} (\ textbf {A}) = \ text {det}
\ left (\ begin {matrix}% 4x4
-4 & 0 & \ pmb {-1} & -1 \
2 & 2 & \ pmb {0} & 2 \
-2 & 3 & \ pmb {1} & -5 \
-1 & 1 & \ pmb {0} & -2
\ end {matrix} \ right)
\$ \$ \ stackrel {\ text {Sarrus}} {=} - 1 \ cdot (-12 + 10-4 - (- 6) - (8) - (- 10)) + 1 \ cdot (16 + 0 + (- 2) - (2) - (0) - (- 8)) \
\ stackrel {\ phantom {Sarrus}} {=} - (2) + (20) = 18 \$

### Application of determinants

Why should we calculate determinants?

It is a "mathematical tool" with which we can easily check that:

• A (square) matrix is ​​invertible / regular
• A mapping is an automorphism (and thus also iso-, mono- and epimorphism)
• A given EZS is a base
• ... etc ...

example
For which parameters \$ a \ in \ mathbb {R} \$ is the mapping \$ f_A \$ an isomorphism?
\ begin {align *}
f_A: {\ mathbb {R} ^ 3} \ rightarrow \ mathbb {R} ^ 3, \ overrightarrow {x} \ rightarrow \ textbf {A} \ overrightarrow {x} \ quad \ text {with} \ quad \ textbf { A} =
\ left (
\ begin {matrix}% 3x3
-1 & 2 & 9 + a \
1 + a & 4-a & -3 \
a & 2 & 4
\ end {matrix}
\ right)
\ end {align *}

Smart approach: If \$ f_A \$ is an isomorphism, it is also an automorphism (\$ f_A \$ is already an endomorphism, since \$ \ mathbb {R} ^ 3 \$ is mapped to \$ \ mathbb {R} ^ 3 \$). If so, \$ {A} ^ {- 1} \$ exists and thus det (A.) = 0.

det (A.) calculate: In matrix A. is not an entry 0. We can, for example, take line \$ 3 \ cdot (-1) \$ and add it to the first line. The determinant does not change, element \$ a_ {12} \$ becomes zero and we have saved calculation steps (not a must!). Then development according to 2nd column: \ begin {align *}
& = + (4-a) \ Bigl ((- 1-a) \ cdot 4-a (5 + a) \ Bigr) -2 \ Bigl ((- 1-a) (- 3) - (1 + a ) (5 + a) \ Bigr) \
& = (4-a) \ Bigl (-4-4a-5a-a ^ 2 \ Bigr) -2 \ Bigl (3 + 3a-5-6a-a ^ 2 \ Bigr) \
& = (4-a) \ Bigl (-a ^ 2-9a-4 \ Bigr) -2 \ Bigl (-a ^ 2-3a-2 \ Bigr) = (a-4) \ Bigl (a ^ 2 + 9a + 4 \ Bigr) +2 \ Bigl (a ^ 2 + 3a + 2 \ Bigr) \
& = a ^ 3 + 9a ^ 2 + 4a-4a ^ 2-36a-16 + 2a ^ 2 + 6a + 4 \
& = a ^ 3 + 7a ^ 2-26a-12
\ end {align *}

Well det (A.) = 0 to calculate the \$ a \$ that must be excluded afterwards!

det (A.) is a third degree polynomial. If there is a "nice" solution for \$ a \$ (\$ a \ in \ mathbb {Z} \$), then this is a divisor of the absolute term (-12). So \$ a \ in \ left \ {- 12, -6, -4, -3, -2, -1,1,2,3,4,6,12 \ right \} \$. By trying it out (recommendation: calculate the function value using the Horner scheme) we find \$ a = 3 \$ as a solution. Separate polynomial using Horner's scheme: \$ \ Rightarrow a ^ 3 + 7a ^ 2-26a-12 = \ underbrace {(a-3)} _ {a_1 \, = \, 3} \ underbrace {(a ^ 2 + 10a + 4)} _ {a \, = \ ,?} = 0 \ stackrel {pq} {\ quad} a_ {2,3} = - 5 \ pm \ sqrt {21} \$

So \$ f_A \$ is an isomorphism if \$ a \ in \ mathbb {R} \ setminus \ left \ {3, -5- \ sqrt {21}, - 5+ \ sqrt {21} \ right \} \$