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[SOLVED] Statements with determinant

mathmari

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Apr 14, 2013
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Hey!! :eek:

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?

(Wondering)
 

HallsofIvy

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MHB Math Helper
Jan 29, 2012
1,151
[tex]v_1= Ae_1= (a_1 a_2)[/tex] and [tex]v_2= Ae_2= (b_2 b_2)[/tex]. The area of a parallelogram having those vectors as sides is [tex]|v_1||v_2|sin(\theta)[/tex] where [tex]\theta[/tex] is the angle between the two vectors. The lengths of the two vectors are [tex]|v_1|= \sqrt{a_1^2+ a_2^2}[/tex] and [tex]|v_2|= \sqrt{b_1^2+ b_2^2}[/tex]. What is the angle between them?
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,711
Hey!! :eek:

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? (Wondering)

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

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? (Wondering)
 

mathmari

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Apr 14, 2013
4,028
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? (Wondering)
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,711
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? (Wondering)
Basically we consider both. (Emo)

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? (Thinking)

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? (Wondering)
 
Last edited:

mathmari

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Apr 14, 2013
4,028
Basically we consider both. (Emo)

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? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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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? (Wondering)
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. (Thinking)

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. (Nerd)
 

mathmari

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Apr 14, 2013
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A map is injective if every point in the co-domain has at most 1 original.
Is that the case? (Wondering)
How do we check that if we know that the determinant is equal to $0$ ? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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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? (Wondering)
 

mathmari

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Apr 14, 2013
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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? (Wondering)
 

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? (Wondering)
 

mathmari

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Apr 14, 2013
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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? (Wondering)
Ah yes! So the vectors must be linear dependent, or not? (Wondering)
 

Klaas van Aarsen

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mathmari

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Apr 14, 2013
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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. (Nerd)
 

mathmari

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Apr 14, 2013
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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. (Nerd)
I see!! Thanks a lot!! (Star)