# [SOLVED]Statements with determinant

#### mathmari

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Hey!! We have the matrix $A=\begin{pmatrix}a_1 & b_1 \\ a_2 & b_2\end{pmatrix}$.

We consider the vectors $\vec{v}:=A\vec{e}_1$ and $\vec{w}:=A\vec{e}_2$.
1. Justify geometrically, why the area of the parallelogram spanned by $\vec{v}$ and $\vec{w}$ is equal to $\det A$.
2. Calculate the determinant of the matrix that we get from $A$ if we swap the two columns. (This is related to whether the orientation of the parallelogram is equal to the orientation of $\mathbb{R}^2$ which is given $\vec{e}_1$ and $\vec{e}_2$.)
3. Give the interpretation of the case $\det A=0$. What can we say in this case for the map $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$, $\vec{x}\mapsto A\vec{x}$ ? Is $f$ injectiv?

Could you give me a hint for 1. ?

At 2. we just have to calculate the determinant. The statement of the parentheses is related to the answer of 1., or not?

At 3. if the determinant is equal to $0$, then it follows that the map is injective. But how could we explain that? #### HallsofIvy

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$$v_1= Ae_1= (a_1 a_2)$$ and $$v_2= Ae_2= (b_2 b_2)$$. The area of a parallelogram having those vectors as sides is $$|v_1||v_2|sin(\theta)$$ where $$\theta$$ is the angle between the two vectors. The lengths of the two vectors are $$|v_1|= \sqrt{a_1^2+ a_2^2}$$ and $$|v_2|= \sqrt{b_1^2+ b_2^2}$$. What is the angle between them?

#### Klaas van Aarsen

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Hey!! We have the matrix $A=\begin{pmatrix}a_1 & b_1 \\ a_2 & b_2\end{pmatrix}$.

We consider the vectors $\vec{v}:=A\vec{e}_1$ and $\vec{w}:=A\vec{e}_2$.
1. Justify geometrically, why the area of the parallelogram spanned by $\vec{v}$ and $\vec{w}$ is equal to $\det A$.
2. Calculate the determinant of the matrix that we get from $A$ if we swap the two columns. (This is related to whether the orientation of the parallelogram is equal to the orientation of $\mathbb{R}^2$ which is given $\vec{e}_1$ and $\vec{e}_2$.)
3. Give the interpretation of the case $\det A=0$. What can we say in this case for the map $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$, $\vec{x}\mapsto A\vec{x}$ ? Is $f$ injectiv?

Could you give me a hint for 1. ?
Hey mathmari !!

Remember the cross product ($\times$)? How can it be used to find the area of a parallellogram? At 2. we just have to calculate the determinant. The statement of the parentheses is related to the answer of 1., or not?
Yes. At 3. if the determinant is equal to $0$, then it follows that the map is injective. But how could we explain that?
A map is injective if every point in the co-domain has at most 1 original.
Is that the case? #### mathmari

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We consider a parallelogram that has as vectors the vectors $\vec{v}$ and $\vec{w}$.
Do we consider then the half of the parallelogram, which is a triangle or do we take the formula of the area using the cross product? #### Klaas van Aarsen

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We consider a parallelogram that has as vectors the vectors $\vec{v}$ and $\vec{w}$.
Do we consider then the half of the parallelogram, which is a triangle or do we take the formula of the area using the cross product? Basically we consider both. Suppose $\vec a$ and $\vec b$ are vectors in 3-dimensional space.

If we consider half of the parallelogram between $\vec a$ and $\vec b$, which is a triangle, then its area is:
$$\text{area triangle} = \frac 12 \cdot\text{base}\cdot\text{height} = \frac 12 \cdot\|\vec a\| \cdot \|\vec b\| \sin\angle(\vec a,\vec b)$$
isn't it? So the area of the parallelogram is $\|\vec a\| \cdot \|\vec b\| \sin\angle(\vec a,\vec b)$.

Now consider the cross product, for which we have:
$$\|\vec a\times \vec b\| = \|\vec a\|\cdot \|\vec b\| \sin\angle(\vec a,\vec b) = \left\|\begin{pmatrix}a_2b_3-a_3b_2 \\ a_3b_1 - a_1b_3\\ a_1b_2 - a_2b_1\end{pmatrix}\right\|$$
What do we get if we pick $a_3=b_3=0$? Does it look like the determinant in the OP? Last edited:

#### mathmari

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Basically we consider both. Suppose $\vec a$ and $\vec b$ are vectors in 3-dimensional space.
Why do we have to consider vectors in 3-dimensional space? We are given a $2\times 2$-matrix and wouldn't that mean that we consider a 2-dimensional space? #### Klaas van Aarsen

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Why do we have to consider vectors in 3-dimensional space? We are given a $2\times 2$-matrix and wouldn't that mean that we consider a 2-dimensional space? Because the cross product is only defined for 3-dimensional vectors.
And we can treat 2-dimensional vectors as 3-dimensional vectors if we set the 3rd coordinate to 0. Consequently the cross product will be a vector along the 3rd axis.
It will be either along the positive axis or the negative axis.
This direction identifies the orientation of the vectors with respect to the parallelogram. #### mathmari

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A map is injective if every point in the co-domain has at most 1 original.
Is that the case? How do we check that if we know that the determinant is equal to $0$ ? #### Klaas van Aarsen

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How do we check that if we know that the determinant is equal to $0$ ?
In question 1 it is established that the determinant is equal to the area of the parallelogram between the vectors in the matrix.
So a determinant of 0 implies that the area of the parallelogram is 0.
What does that imply for the vectors $\vec v$ and $\vec w$?
What does it imply for the image of the domain into the co-domain? #### mathmari

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In question 1 it is established that the determinant is equal to the area of the parallelogram between the vectors in the matrix.
So a determinant of 0 implies that the area of the parallelogram is 0.
What does that imply for the vectors $\vec v$ and $\vec w$?
The length of vectors must be $0$, and so they represent points, or not? #### Klaas van Aarsen

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The length of vectors must be $0$, and so they represent points, or not?
That is one case, but the vectors do not have to be 0.
Don't we get an area of 0 size if the vectors are on the same line? #### mathmari

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That is one case, but the vectors do not have to be 0.
Don't we get an area of 0 size if the vectors are on the same line? Ah yes! So the vectors must be linear dependent, or not? #### Klaas van Aarsen

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Ah yes! So the vectors must be linear dependent, or not?
Yep. #### mathmari

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Yep. And what do we get from that? How is this related to $f$ ? I got stuck right now. Last edited:

#### Klaas van Aarsen

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And what do we get from that? How is this related to $f$ ? I got stuck right now.
The range of $f$ consists of linear combinations of $\vec v$ and $\vec w$.
If they are linearly dependent, then there is more than one point that maps to the same point.
That is, there will for instance be $x,y$, that are not both zero, such that $A\binom x0=A\binom 0y$.
Consequently there are points in the co-domain that have multiple originals. In the example we have $f(x,0)=f(0,y)$.
Therefore $f$ is not injective. #### mathmari

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The range of $f$ consists of linear combinations of $\vec v$ and $\vec w$.
If they are linearly dependent, then there is more than one point that maps to the same point.
That is, there will for instance be $x,y$, that are not both zero, such that $A\binom x0=A\binom 0y$.
Consequently there are points in the co-domain that have multiple originals. In the example we have $f(x,0)=f(0,y)$.
Therefore $f$ is not injective. I see!! Thanks a lot!! 