Anki Deck Changes

Note 1: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Classic
GUID: Br>vM#El56

added

New Note

Front

For the co-factor formula for the determinant what's the pattern of signs to multiply by?

Back

For the co-factor formula for the determinant what's the pattern of signs to multiply by?

\(\begin{bmatrix} + & - & + & - & + & \dots \\ - & + & - & + & - & \dots \\ + & - & + & - & + & \dots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \end{bmatrix}\)

Field-by-field Comparison

Field	Before	After
Front		For the <b>co-factor</b> formula for the determinant what's the pattern of signs to multiply by?
Back		\(\begin{bmatrix} + & - & + & - & + & \dots \\ - & + & - & + & - & \dots \\ + & - & + & - & + & \dots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \end{bmatrix}\)

Tags: ETH::1._Semester::LinAlg::7._The_determinant::3._Cofactors,_Cramer's_Rule_and_Beyond

Note 2: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: B~w={<:u.n

added

New Note

Front

For \(A \in \mathbb{R}^{n \times n}\) and \(\lambda \in \mathbb{R}\) we have \(\det(\lambda B) = \lambda^n \det(B) \).

Back

For \(A \in \mathbb{R}^{n \times n}\) and \(\lambda \in \mathbb{R}\) we have \(\det(\lambda B) = \lambda^n \det(B) \).

Each row is scaled by \(\lambda\) and by multi-linearity we have to take it out of each one (n times)

Field-by-field Comparison

Field	Before	After
Text		For \(A \in \mathbb{R}^{n \times n}\) and \(\lambda \in \mathbb{R}\) we have \(\det(\lambda B) = {{c1:: \lambda^n \det(B) }}\).
Extra		Each row is scaled by \(\lambda\) and by multi-linearity we have to take it out of each one (n times)

Tags: ETH::1._Semester::LinAlg::7._The_determinant::2._The_general_case::1._Properties

Note 3: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: HnqZA9P|@G

added

New Note

Front

Given a triangular (either upper or lower) matrix \(T \in \mathbb{R}^{n \times n}\), we have \[ \det(T) = {{c1:: \prod_{k = 1}^n T_{kk} }}\]

Back

Given a triangular (either upper or lower) matrix \(T \in \mathbb{R}^{n \times n}\), we have \[ \det(T) = {{c1:: \prod_{k = 1}^n T_{kk} }}\]

For a triangular matrix, if we choose an element off the diagonal, we are then forced to choose one in the \(0\)s thus making that factor \(0\). The only valid permutation is thus the \(\text{id}\), which means we just multiply the diagonals.

Field-by-field Comparison

Field	Before	After
Text		Given a <b>triangular</b> (either upper or lower) matrix \(T \in \mathbb{R}^{n \times n}\), we have \[ \det(T) = {{c1:: \prod_{k = 1}^n T_{kk} }}\]
Extra		For a triangular matrix, if we choose an element off the diagonal, we are then forced to choose one in the \(0\)s thus making that factor \(0\). The only valid permutation is thus the \(\text{id}\), which means we just multiply the diagonals.

Tags: ETH::1._Semester::LinAlg::7._The_determinant::2._The_general_case::1._Properties

Note 4: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: I5WrW!E]G<

added

New Note

Front

The determinant expressed in terms of co-factors is: \[\det(A) = {{c1:: \sum_{j = 1}^n A_{ij}C_{ij} }}\]

Back

The determinant expressed in terms of co-factors is: \[\det(A) = {{c1:: \sum_{j = 1}^n A_{ij}C_{ij} }}\]

in which we multiply the cofactor of every element by the element itself, as is clear in the example for a 3x3.

Field-by-field Comparison

Field	Before	After
Text		The determinant expressed in terms of <b>co-factors</b> is: \[\det(A) = {{c1:: \sum_{j = 1}^n A_{ij}C_{ij} }}\]<br>
Extra		in which we multiply the cofactor of every element by the element itself, as is clear in the example for a 3x3.<br><img src="paste-5b306ce2f1c5340a372c470f868d00a247f2c566.jpg">

Tags: ETH::1._Semester::LinAlg::7._The_determinant::3._Cofactors,_Cramer's_Rule_and_Beyond

Note 5: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Classic
GUID: JAn~&+e&|!

added

New Note

Front

The determinant is linear in each row (or each column). In other words for any \(a_0, a_1, a_2, \dots, a_n \in \mathbb{R}^n\) and \(\alpha_0, \alpha_1 \in \mathbb{R}\) we have: (Two linearity properties)

Back

The determinant is linear in each row (or each column). In other words for any \(a_0, a_1, a_2, \dots, a_n \in \mathbb{R}^n\) and \(\alpha_0, \alpha_1 \in \mathbb{R}\) we have: (Two linearity properties)

Field-by-field Comparison

Field	Before	After
Front		The determinant is linear in each row (or each <i>column</i>). In other words for any \(a_0, a_1, a_2, \dots, a_n \in \mathbb{R}^n\) and \(\alpha_0, \alpha_1 \in \mathbb{R}\) we have: (<i>Two linearity properties)</i>
Back		<img src="paste-b0314843c81b23252762fd0a50059644aa1dfffe.jpg">

Tags: ETH::1._Semester::LinAlg::7._The_determinant::3._Cofactors,_Cramer's_Rule_and_Beyond

Note 6: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: NJVswNKOd~

added

New Note

Front

For a permutation \(\sigma\), if \(\sigma(i) \neq i\) then there exists a \(j\) such that \(\sigma(j) \neq j\).

Back

For a permutation \(\sigma\), if \(\sigma(i) \neq i\) then there exists a \(j\) such that \(\sigma(j) \neq j\).

We're going to have to venture off the diagonal for at least one other element.

If we have a matrix \(A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\), the only permutation that doesn't produce a \(0\) product is the \(\text{id}\) permutation.

Field-by-field Comparison

Field	Before	After
Text		For a permutation \(\sigma\), if \(\sigma(i) \neq i\) then {{c1:: there exists a \(j\) such that \(\sigma(j) \neq j\)}}.
Extra		We're going to have to venture off the diagonal for at least one other element.<br><br>If we have a matrix \(A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\), the only permutation that doesn't produce a \(0\) product is the \(\text{id}\) permutation.

Tags: ETH::1._Semester::LinAlg::7._The_determinant::2._The_general_case::2._Permutations

Note 7: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: O?@!`_xk3T

added

New Note

Front

In a triangular matrix, if one of the diagonals is zero, the determinant is \(0\).

Back

In a triangular matrix, if one of the diagonals is zero, the determinant is \(0\).

Field-by-field Comparison

Field	Before	After
Text		In a <b>triangular matrix</b>, if {{c2::one of the diagonals is zero}}, the determinant is {{c1::\(0\)}}.

Tags: ETH::1._Semester::LinAlg::7._The_determinant::2._The_general_case::1._Properties

Note 8: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: Oow<}IKdC,

added

New Note

Front

Given a matrix \(A \in \mathbb{R}^{n \times n}\), then \[ \det(A) = \det(A^\top) \]

Back

Given a matrix \(A \in \mathbb{R}^{n \times n}\), then \[ \det(A) = \det(A^\top) \]

This follows from the fact that the inverse of a permutation has the same sign, and transposing is the same as doing the inverse permutation.

Field-by-field Comparison

Field	Before	After
Text		Given a matrix \(A \in \mathbb{R}^{n \times n}\), then \[ {{c1::\det(A)}} = \det(A^\top) \]
Extra		This follows from the fact that the inverse of a permutation has the same sign, and transposing is the same as doing the inverse permutation.

Tags: ETH::1._Semester::LinAlg::7._The_determinant::2._The_general_case::1._Properties

Note 9: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: P0)~~JgL|z

added

New Note

Front

Given matrices \(A, B \in \mathbb{R}^{n \times n}\), we have \[ \det(AB) = \det(A) \cdot \det(B) \]

Back

Given matrices \(A, B \in \mathbb{R}^{n \times n}\), we have \[ \det(AB) = \det(A) \cdot \det(B) \]

If we multiply first by \(A\) then \(B\) the unit cube will be stretched the same way as if we did both at once.

Field-by-field Comparison

Field	Before	After
Text		Given matrices \(A, B \in \mathbb{R}^{n \times n}\), we have \[ \det(AB) = {{c1:: \det(A) \cdot \det(B) }}\]
Extra		If we multiply first by \(A\) then \(B\) the unit cube will be stretched the same way as if we did both at once.

Tags: ETH::1._Semester::LinAlg::7._The_determinant::2._The_general_case::1._Properties

Note 10: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: i{D+8us!vH

added

New Note

Front

Given a matrix \(A\) such that \(\det(A) \neq 0\) then \(A\) is invertible and \[ \det(A^{-1}) = {{c1:: \frac{1}{\det(A)} }}\]

Back

Given a matrix \(A\) such that \(\det(A) \neq 0\) then \(A\) is invertible and \[ \det(A^{-1}) = {{c1:: \frac{1}{\det(A)} }}\]

If \(A\) shrinks the unit cube and \(A^{-1}\) inflates it back to the unit dimensions then the ratio of the changes is \(1\).

Field-by-field Comparison

Field	Before	After
Text		Given a matrix \(A\) such that \(\det(A) \neq 0\) then \(A\) is <b>invertible</b> and \[ \det(A^{-1}) = {{c1:: \frac{1}{\det(A)} }}\]
Extra		If \(A\) shrinks the unit cube and \(A^{-1}\) inflates it back to the unit dimensions then the ratio of the changes is \(1\).

Tags: ETH::1._Semester::LinAlg::7._The_determinant::2._The_general_case::1._Properties

Note 11: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: oKlE5w/*RH

added

New Note

Front

If \(Q \in \mathbb{R}^{n \times n}\) is an orthogonal matrix then \(\det(Q) = \) \(1\) or \(-1\).

Back

If \(Q \in \mathbb{R}^{n \times n}\) is an orthogonal matrix then \(\det(Q) = \) \(1\) or \(-1\).

As \(Q\) is orthogonal, we don't scale (preserves \(\top\)) thus the unit cube is just turned, not scaled.

Field-by-field Comparison

Field	Before	After
Text		If \(Q \in \mathbb{R}^{n \times n}\) is an <i>orthogonal</i> matrix then  \(\det(Q) = \) {{c1:: \(1\) or \(-1\)}}.
Extra		As \(Q\) is orthogonal, we don't scale (preserves \(\top\)) thus the unit cube is just turned, not scaled.

Tags: ETH::1._Semester::LinAlg::7._The_determinant::2._The_general_case::1._Properties

Note 12: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: pdv(<_+/|h

added

New Note

Front

Given a permutation matrix \(P \in \mathbb{R}^{n \times n}\) corresponding to a permutation \(\sigma\), then \(\det(P) = {{c1::\text{sgn}(\sigma)}}\)

Back

Given a permutation matrix \(P \in \mathbb{R}^{n \times n}\) corresponding to a permutation \(\sigma\), then \(\det(P) = {{c1::\text{sgn}(\sigma)}}\)

(this is as \(P\) is also an orthogonal matrix, see 3.). We sometimes write \(\text{sgn}(P)\).

For the permutation matrix, each row contains only one entry: a \(1\). Thus the only permutation \(\sigma\) in the product that doesn't have a \(0\) factor is the permutation corresponding to the matrix \(P\) itself. The product is \(1 \cdot 1 \dots \cdot 1\) thus we get \(\text{sgn}(\sigma) = \text{sgn}(P)\).

Field-by-field Comparison

Field	Before	After
Text		Given a permutation matrix \(P \in \mathbb{R}^{n \times n}\) corresponding to a permutation \(\sigma\), then \(\det(P) = {{c1::\text{sgn}(\sigma)}}\)
Extra		(this is as \(P\) is also an orthogonal matrix, see 3.). We sometimes write \(\text{sgn}(P)\).<br><br>For the permutation matrix, each row contains only one entry: a \(1\). Thus the only permutation \(\sigma\) in the product that doesn't have a \(0\) factor is the permutation corresponding to the matrix \(P\) itself. The product is \(1 \cdot 1 \dots \cdot 1\) thus we get \(\text{sgn}(\sigma) = \text{sgn}(P)\).

Tags: ETH::1._Semester::LinAlg::7._The_determinant::2._The_general_case::1._Properties

Note 13: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: sH/[5Mikuh

added

New Note

Front

If \(A\) is a matrix and \(P\) is a permutation that swaps two elements (i.e. \(\text{sgn}(P) = -1\)): \[\det(PA) = - \det(A) \]

Back

If \(A\) is a matrix and \(P\) is a permutation that swaps two elements (i.e. \(\text{sgn}(P) = -1\)): \[\det(PA) = - \det(A) \]

\(PA\) corresponds to swapping two rows of \(A\)

Field-by-field Comparison

Field	Before	After
Text		If \(A\) is a matrix and \(P\) is a permutation that <i>swaps two elements</i> (i.e. \(\text{sgn}(P) = -1\)): \[\det(PA) = {{c1:: - \det(A) }}\]<br>
Extra		\(PA\) corresponds to swapping two rows of \(A\)

Tags: ETH::1._Semester::LinAlg::7._The_determinant::3._Cofactors,_Cramer's_Rule_and_Beyond

Note 14: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: u:|nk(8Sda

added

New Note

Front

The Co-Factors of \(A\) are \( C_{ij} = {{c1:: (-1)^{i + j} \det(\mathcal{A}_{ij}) }}\) where \(\mathcal{A}_{ij}\) is the \((n - 1)\times (n-1)\) matrix without row \(i\) and column \(j\).

Back

The Co-Factors of \(A\) are \( C_{ij} = {{c1:: (-1)^{i + j} \det(\mathcal{A}_{ij}) }}\) where \(\mathcal{A}_{ij}\) is the \((n - 1)\times (n-1)\) matrix without row \(i\) and column \(j\).

Field-by-field Comparison

Field	Before	After
Text		The <b>Co-Factors</b> of \(A\) are \( C_{ij} = {{c1:: (-1)^{i + j} \det(\mathcal{A}_{ij}) }}\) where \(\mathcal{A}_{ij}\) is the \((n - 1)\times (n-1)\) matrix without row \(i\) and column \(j\).
Extra		<img src="paste-5380635095a8a7665119ccf8d988a2ee74964ad3.jpg">

Tags: ETH::1._Semester::LinAlg::7._The_determinant::3._Cofactors,_Cramer's_Rule_and_Beyond

Note 15: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: zz.w6i^0.V

added

New Note

Front

A matrix \(A \in \mathbb{R}^{n \times n}\) is invertible if and only if \[ \det(A) \neq 0 \]

Back

A matrix \(A \in \mathbb{R}^{n \times n}\) is invertible if and only if \[ \det(A) \neq 0 \]

If the unit cube collapses to have 0 volume (i.e. \(\det(A) = 0\)) then we lost a dimension and \(A\) cannot be invertible.

Field-by-field Comparison

Field	Before	After
Text		A matrix \(A \in \mathbb{R}^{n \times n}\)  is <i>invertible</i> if and only if \[ \det(A) {{c1:: \neq 0 }}\]
Extra		If the unit cube collapses to have 0 volume (i.e. \(\det(A) = 0\)) then we lost a dimension and \(A\) cannot be invertible.

Tags: ETH::1._Semester::LinAlg::7._The_determinant::2._The_general_case::1._Properties

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