$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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For $v_1, \dots, v_n\in\mathbb{R}^n$, let the determinant of $v_1, \dots, v_n$ be defined as the signed volume of a parallelpiped formed with $v_1, \dots, v_n$ ie. $\text{det}(v_1, \dots, v_n)=\text{signed volume of }P(v_1, \dots, v_n)$, where $P(v_1, \dots, v_n)=\{\alpha_1 v_1 + \dots + \alpha_n v_n: \alpha_1\in [0, 1]\}$. The convension is that for $v_1$ to $v_2$ where we go clockwise, the volume is positive and negative volume otherwise.
For a matrix $A\in M_n(P_n)$ with columns $v_1, \dots, v_n\in\mathbb{R}^n$, we define $\text{det} A=\text{det}(v_1, \dots, v_n)$

Concepts

In two dimension, two vectors can determine a parallelogram. In general for $n$ dimensions, $n$ vectors can determine a parallelpiped.
For the determinant, we are interested in the volume of the parallelpiped.
There are four basic properties of a determinant we want to cover.
  1. The determinant detects linear independence. That is, the determinant of a set is nonzero if and only if the set is linearly independent
  2. Antisymmetry. That is, $\text{det}(v_1, \dots, v_n)=-\text{det}(v_n, \dots, v_1)$
  3. Multilinearity. That is, if we fix all variables except the first, we can express the determinant as a sum of two separate determinants with each of the first terms and so $\text{det}(v_1+v_1', \dots, v_n)=\text{det}(v_1, \dots, v_n)+\text{det}(v_1', \dots, v_n)$.
  4. Normalized. That is, for unit vectors $e_1$, we have $\text{det}(e_1, \dots, e_n)=1$
For calculating the determinant, let $$\begin{align*} A=\begin{bmatrix}a_1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & \dots & a_n \end{bmatrix}. \end{align*}$$ Then $$\begin{align*} \text{det}(A)&=a_1\begin{bmatrix}1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_n\end{bmatrix}\\ &=a_1a_2\begin{bmatrix}1 & 0 & \dots & 0\\ 0 & 1 & \dots & 0 \\ \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & \dots & a_n \end{bmatrix}\\ &=a_1\dots a_n\begin{bmatrix}1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 1 \end{bmatrix} \end{align*}$$ Note that for an upper triangular matrix if $a_{ii}=0$ for some $i$, then $A$ is not invertible, so $\text{det }A=0$
Recall that for elementary column operations, if we multiple one column by $\alpha$, then the determinant gets multiplied by $\alpha$. If we swap two columns, the determinant gets multiplied by $-1$, and adding scalar multiples of columns does not affect the determinant. Using this, we can create a formula to calculate the determinant.
Another method to calculate determinants is using cofactor expansion.

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For $v_1, \dots, v_n\in\mathbb{R}^n$, let the determinant of $v_1, \dots, v_n$ be defined as the signed volume of a parallelpiped formed with $v_1, \dots, v_n$ ie. $\text{det}(v_1, \dots, v_n)=\text{signed volume of }P(v_1, \dots, v_n)$, where $P(v_1, \dots, v_n)=\{\alpha_1 v_1 + \dots + \alpha_n v_n: \alpha_1\in [0, 1]\}$. The convension is that for $v_1$ to $v_2$ where we go clockwise, the volume is positive and negative volume otherwise.
For a matrix $A\in M_n(P_n)$ with columns $v_1, \dots, v_n\in\mathbb{R}^n$, we define $\text{det} A=\text{det}(v_1, \dots, v_n)$

Concepts

In two dimension, two vectors can determine a parallelogram. In general for $n$ dimensions, $n$ vectors can determine a parallelpiped.
For the determinant, we are interested in the volume of the parallelpiped.
There are four basic properties of a determinant we want to cover.
  1. The determinant detects linear independence. That is, the determinant of a set is nonzero if and only if the set is linearly independent
  2. Antisymmetry. That is, $\text{det}(v_1, \dots, v_n)=-\text{det}(v_n, \dots, v_1)$
  3. Multilinearity. That is, if we fix all variables except the first, we can express the determinant as a sum of two separate determinants with each of the first terms and so $\text{det}(v_1+v_1', \dots, v_n)=\text{det}(v_1, \dots, v_n)+\text{det}(v_1', \dots, v_n)$.
  4. Normalized. That is, for unit vectors $e_1$, we have $\text{det}(e_1, \dots, e_n)=1$
For calculating the determinant, let $$\begin{align*} A=\begin{bmatrix}a_1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & \dots & a_n \end{bmatrix}. \end{align*}$$ Then $$\begin{align*} \text{det}(A)&=a_1\begin{bmatrix}1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_n\end{bmatrix}\\ &=a_1a_2\begin{bmatrix}1 & 0 & \dots & 0\\ 0 & 1 & \dots & 0 \\ \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & \dots & a_n \end{bmatrix}\\ &=a_1\dots a_n\begin{bmatrix}1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 1 \end{bmatrix} \end{align*}$$ Note that for an upper triangular matrix if $a_{ii}=0$ for some $i$, then $A$ is not invertible, so $\text{det }A=0$
Recall that for elementary column operations, if we multiple one column by $\alpha$, then the determinant gets multiplied by $\alpha$. If we swap two columns, the determinant gets multiplied by $-1$, and adding scalar multiples of columns does not affect the determinant. Using this, we can create a formula to calculate the determinant.
Another method to calculate determinants is using cofactor expansion.

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definition
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FullPage
definition
concepts
used in
hypothesis
results
For $v_1, \dots, v_n\in\mathbb{R}^n$, let the determinant of $v_1, \dots, v_n$ be defined as the signed volume of a parallelpiped formed with $v_1, \dots, v_n$ ie. $\text{det}(v_1, \dots, v_n)=\text{signed volume of }P(v_1, \dots, v_n)$, where $P(v_1, \dots, v_n)=\{\alpha_1 v_1 + \dots + \alpha_n v_n: \alpha_1\in [0, 1]\}$. The convension is that for $v_1$ to $v_2$ where we go clockwise, the volume is positive and negative volume otherwise.
For a matrix $A\in M_n(P_n)$ with columns $v_1, \dots, v_n\in\mathbb{R}^n$, we define $\text{det} A=\text{det}(v_1, \dots, v_n)$

Concepts

In two dimension, two vectors can determine a parallelogram. In general for $n$ dimensions, $n$ vectors can determine a parallelpiped.
For the determinant, we are interested in the volume of the parallelpiped.
There are four basic properties of a determinant we want to cover.
  1. The determinant detects linear independence. That is, the determinant of a set is nonzero if and only if the set is linearly independent
  2. Antisymmetry. That is, $\text{det}(v_1, \dots, v_n)=-\text{det}(v_n, \dots, v_1)$
  3. Multilinearity. That is, if we fix all variables except the first, we can express the determinant as a sum of two separate determinants with each of the first terms and so $\text{det}(v_1+v_1', \dots, v_n)=\text{det}(v_1, \dots, v_n)+\text{det}(v_1', \dots, v_n)$.
  4. Normalized. That is, for unit vectors $e_1$, we have $\text{det}(e_1, \dots, e_n)=1$
For calculating the determinant, let $$\begin{align*} A=\begin{bmatrix}a_1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & \dots & a_n \end{bmatrix}. \end{align*}$$ Then $$\begin{align*} \text{det}(A)&=a_1\begin{bmatrix}1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_n\end{bmatrix}\\ &=a_1a_2\begin{bmatrix}1 & 0 & \dots & 0\\ 0 & 1 & \dots & 0 \\ \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & \dots & a_n \end{bmatrix}\\ &=a_1\dots a_n\begin{bmatrix}1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 1 \end{bmatrix} \end{align*}$$ Note that for an upper triangular matrix if $a_{ii}=0$ for some $i$, then $A$ is not invertible, so $\text{det }A=0$
Recall that for elementary column operations, if we multiple one column by $\alpha$, then the determinant gets multiplied by $\alpha$. If we swap two columns, the determinant gets multiplied by $-1$, and adding scalar multiples of columns does not affect the determinant. Using this, we can create a formula to calculate the determinant.
Another method to calculate determinants is using cofactor expansion.

Used In

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Hypothesis

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For $v_1, \dots, v_n\in\mathbb{R}^n$, let the determinant of $v_1, \dots, v_n$ be defined as the signed volume of a parallelpiped formed with $v_1, \dots, v_n$ ie. $\text{det}(v_1, \dots, v_n)=\text{signed volume of }P(v_1, \dots, v_n)$, where $P(v_1, \dots, v_n)=\{\alpha_1 v_1 + \dots + \alpha_n v_n: \alpha_1\in [0, 1]\}$. The convension is that for $v_1$ to $v_2$ where we go clockwise, the volume is positive and negative volume otherwise.
For a matrix $A\in M_n(P_n)$ with columns $v_1, \dots, v_n\in\mathbb{R}^n$, we define $\text{det} A=\text{det}(v_1, \dots, v_n)$

Concepts

In two dimension, two vectors can determine a parallelogram. In general for $n$ dimensions, $n$ vectors can determine a parallelpiped.
For the determinant, we are interested in the volume of the parallelpiped.
There are four basic properties of a determinant we want to cover.
  1. The determinant detects linear independence. That is, the determinant of a set is nonzero if and only if the set is linearly independent
  2. Antisymmetry. That is, $\text{det}(v_1, \dots, v_n)=-\text{det}(v_n, \dots, v_1)$
  3. Multilinearity. That is, if we fix all variables except the first, we can express the determinant as a sum of two separate determinants with each of the first terms and so $\text{det}(v_1+v_1', \dots, v_n)=\text{det}(v_1, \dots, v_n)+\text{det}(v_1', \dots, v_n)$.
  4. Normalized. That is, for unit vectors $e_1$, we have $\text{det}(e_1, \dots, e_n)=1$
For calculating the determinant, let $$\begin{align*} A=\begin{bmatrix}a_1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & \dots & a_n \end{bmatrix}. \end{align*}$$ Then $$\begin{align*} \text{det}(A)&=a_1\begin{bmatrix}1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_n\end{bmatrix}\\ &=a_1a_2\begin{bmatrix}1 & 0 & \dots & 0\\ 0 & 1 & \dots & 0 \\ \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & \dots & a_n \end{bmatrix}\\ &=a_1\dots a_n\begin{bmatrix}1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 1 \end{bmatrix} \end{align*}$$ Note that for an upper triangular matrix if $a_{ii}=0$ for some $i$, then $A$ is not invertible, so $\text{det }A=0$
Recall that for elementary column operations, if we multiple one column by $\alpha$, then the determinant gets multiplied by $\alpha$. If we swap two columns, the determinant gets multiplied by $-1$, and adding scalar multiples of columns does not affect the determinant. Using this, we can create a formula to calculate the determinant.
Another method to calculate determinants is using cofactor expansion.

Used In

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Hypothesis

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Results

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FullPage
definition
concepts
used in
hypothesis
results