Properties of Parallelograms Practice with Our Examples

While sitting in a determinant class, our Professor, while describing determinants, coined an observation which made me wonder. He stated that the determinant of a square matrix can be represented as a function in the following way -

Let M be the set of all square matrices, then a function, say X, is defined from $M\to <!--\; \to \; -->\mathbb{R}$ which returns the determinant of any square matrix $C\in <!--\; \in \; -->M$ i.e.
$X:M\to <!--\; \to \; -->\mathbb{R}$
which returns |M|
As soon as he said this, I began to wonder about the injectivity, surjectivity and bijectivity of this function.
I was told that for a given $C\in <!--\; \in \; -->M$, there exists only one unique determinant $|M|\in <!--\; \in \; -->\mathbb{R}$. But the converse is not true according to me. This can be proved by the following argument -

Let's say that
$\left|\begin{array}{cc}x& 5\\ x& x\end{array}\right|=-<!--\; -\; -->6$
On solving for x, we obtain two values i.e 2 and 3, therefore the matrix is not unique for the given determinant i.e. for one image (−6), there exist multiple preimages. Hence, according to me, the function so defined is not injective, and consequently not bijective as well.
Now checking for surjectivity, we'll have to look for the equality of the codomain and the range of the function. In my opinion, for infinite square matrices, there will be infinite unique determinants belonging to $\mathbb{R}$. Hence the function should be surjective, but only for real entries into the matrix.
For complex entries, I don't think the function remains surjective, as the range will not coincide with the codomain (i.e. $\mathbb{R}$). However, in both cases, I feel that the function is neither injective not bijective.

ABCD is a parallelogram, P is any point on AC. Through P, MN is drawn parallel to BA cutting BC in M and AD in N. SR is drawn parallel to BC cutting BA in S and CD in R. Show that [ASN]+[AMR]=[ABD] (where $[.]$ denotes the area of the rectilinear figure).

My attempt : used base division method to find the area of ASN but I get extra variables which is tough...

When using the definition and properties of the inner product, we get the parallelogram law:
$$\begin{gathered} \Vert <!--\; \Vert \; -->x+y{\Vert <!--\; \Vert \; -->}^2=⟨<!--\; ⟨\; -->x+y,x+y⟩<!--\; ⟩\; -->=⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->y,x⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->y,y⟩<!--\; ⟩\; -->, \\ \Vert <!--\; \Vert \; -->x-<!--\; -\; -->y{\Vert <!--\; \Vert \; -->}^2=⟨<!--\; ⟨\; -->x-<!--\; -\; -->y,x-<!--\; -\; -->y⟩<!--\; ⟩\; -->=⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->-<!--\; -\; -->⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->-<!--\; -\; -->⟨<!--\; ⟨\; -->y,x⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->y,y⟩<!--\; ⟩\; --> \end{gathered}$$
Adding these two expressions:
$\Vert <!--\; \Vert \; -->x+y{\Vert <!--\; \Vert \; -->}^2+\Vert <!--\; \Vert \; -->x-<!--\; -\; -->y{\Vert <!--\; \Vert \; -->}^2=2⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->+2⟨<!--\; ⟨\; -->y,y⟩<!--\; ⟩\; -->=2\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}^2+2\Vert <!--\; \Vert \; -->y{\Vert <!--\; \Vert \; -->}^2$
as required. But the above does not simplify to Pythagoras' theorem $\Vert <!--\; \Vert \; -->x+y{\Vert <!--\; \Vert \; -->}^2=\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}^2+\Vert <!--\; \Vert \; -->y{\Vert <!--\; \Vert \; -->}^2$if $x$ and $y$ are orthogonal.

How can we get Pythagoras from the parallelogram law?

Given a family of parallelograms such that the corresponding edges of all members are parallel and any two members of this family intersect. Can we conclude that any three members of this family intersect?
Intuitively I thought the answer should be positive. But I am having difficulty to write a rigorous proof (or find a counter example). Can anyone provide help me with this? Thanks

How to solve the following problem:

Let $f_n,f\in <!--\; \in \; -->L^2({\mathbb{R}}^d)$ for all $n\ge <!--\; \ge \; -->1$ be such that $\Vert <!--\; \Vert \; -->f_n{\Vert <!--\; \Vert \; -->}_2\to <!--\; \to \; -->\Vert <!--\; \Vert \; -->f{\Vert <!--\; \Vert \; -->}_2$ as $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$. Suppose, moreover, that
$\int <!--\; \int \; -->f_{n}g\to <!--\; \to \; -->\int <!--\; \int \; -->fg$
for all $g\in <!--\; \in \; -->L^2({\mathbb{R}}^d)$. Then $f_n$ converges to $f$ in $L^2$-norm.

Let say I have 2 vectors (1,0,0) and (0,2,0), and I want to find a third vector that is orthogonal to both of them. I can do a cross product and get (0,0,2). However, I know there are infinite vector in the following form $(0,0,x)$ where $x$$\in <!--\; \in \; -->$$R$ that are orthogonal to the other two. My question is what is the difference between the orthogonal vector results from cross product and any other orthogonal vector? Why the cross product gives only 1 specific orthogonal vector? What is the significance of this vector? Thank you!

Let ${\mathbf{R}}^{\ast <!--\; \ast \; -->}$ be group of non-zero real numbers where the operation is multiplication. The group is 2x2 matrices with non-zero determinant.Let $f:G\to <!--\; \to \; -->{\mathbf{R}}^{\ast <!--\; \ast \; -->}$ be a function given by $f(A)=det(A)$.

My proof to show f is homomorphorism is:

We must show $f(ab)=f(a)f(b)$. Here we can see

if $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $B=\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)$

Then $f(AB)=f(C)$ where $C=AB=\left(\begin{array}{cc}ae+bg& af+bh\\ ce+dg& cf+dh\end{array}\right)$

Note $det(C)$
$=f(C)$$=(ae+bg)(cf+dh)-<!--\; -\; -->(af+bh)(ce+dg)$
$=(aecf+bgcf)+(aedh+bgdh)-<!--\; -\; -->(afce+bhce)-<!--\; -\; -->(afdg+bhdg)$
$=bgcf+aedh-<!--\; -\; -->bhce-<!--\; -\; -->afdg$
Also note that $f(A)=ad-<!--\; -\; -->bc$ and $f(B)=eh-<!--\; -\; -->fg$

then $f(A)f(B)=(ad-<!--\; -\; -->bc)(eh-<!--\; -\; -->fg)=adeh-<!--\; -\; -->bceh+adfg-<!--\; -\; -->bcfg$

Thus $f(AB)=f(A)f(B)$ and as such $f$ is a homomorphism.

Now since we showed $f$ is surjective, we need to show $f$ is injective to prove $f$ is a bijective mapping and thus isomorphic.

One aspect of $2\times <!--\; \times \; -->2$ determinant we can use is that it produce the area of parallelogram. If we think of the properties of rotation matrices, we know their determinants all produce 1. Thus $f$ is not injective and bijective. Meaning $f$ is not isomorphic.

Is this the correct way to go about this proof or am I making wild mis-understandings?

The property I'm talking about is:
There is some partition of the plane figure P into $n$ congruent figures for any $n$.
Is it true that only discs, sectors of discs, annuli, sectors of annuli and parallelograms have this property?

I tried figuring out what's the geometrical meaning of this identity in vectors. Proving it isn't a problem, however I'd like to know a more geometry oriented explanation to it.
I'd appreciate if someone could explain it to me. Thank you.
$\Vert <!--\; \Vert \; -->a+b{\Vert <!--\; \Vert \; -->}^2+\Vert <!--\; \Vert \; -->a-<!--\; -\; -->b{\Vert <!--\; \Vert \; -->}^2=2(\Vert <!--\; \Vert \; -->a{\Vert <!--\; \Vert \; -->}^2+\Vert <!--\; \Vert \; -->b{\Vert <!--\; \Vert \; -->}^2)$

How to rigorously prove that the diagonals of a parallelogram are never parallel? It is intuitively obvious, but since it is not an axiom, it is a proposition that needs to be proved. I would like to see a proof without using analytic geometry, but only the old methods of Euclidean synthetic geometry.

The midpoints of the sides of $\mathrm{△<!--\; △\; -->}ABC$ along with any of the vertices as the fourth point make a parallelogram of area equal to what? The answer is, obviously, $\frac{1}{2}\mathrm{area}<!--\; \; -->(\mathrm{△<!--\; △\; -->}ABC)$

In the method, I took $\mathrm{△<!--\; △\; -->}ABC$ with $D$, $E$, and $F$ as midpoints of $AB$, $AC$, and $BC$, respectively; and I joined $DE$ and $EF$ so that I get a parallelogram $\mathrm{◻<!--\; ◻\; -->}DEFB$.
I know what the answer is because one can easily make that out. Also, those four triangles (four because the parallelogram can still be divided into two triangles and the rest two triangles add up to four) so it's simple that the area of $\mathrm{◻<!--\; ◻\; -->}DEBF$ will be 1/4 of $\mathrm{△<!--\; △\; -->}ABC$, but how?
Can anyone explain me this?

A circle is circumscribed by a parallelogram. prove by using tangent property that the parallelogram is a rhombus. I tried to prove that the adjacent sides of the parallelogram are equal but I lack idea to apply tangent property

I had a question about a problem that I was working on for my pre-calculus class.

Here's the problem statement:

The area of the parallelogram with vertices 0, $м$, $w$, and $v+w$ is 34. Find the area of the parallelogram with vertices 0, $Av$, $Aw$, and $Av+Aw$, where
$A=\left(\begin{array}{cc}3& -<!--\; -\; -->5\\ -<!--\; -\; -->1& -<!--\; -\; -->3\end{array}\right).$
I got the answer by doing something very tedious. I set $v=(\genfrac{}{}{0ex}{}{17}{0})$ and $w=(\genfrac{}{}{0ex}{}{0}{2})$, and did some really crazy matrix multiplication and a lot of plotting points of GeoGebra to get the answer of: $\boxed{{\displaystyle 476}}$.

Now, I'm 100% sure that was not the fastest way, can someone tell me the non-bash way to do the problem?

The problem is: Given parallelogram ABCD with diagonals $\stackrel{¯<!--\; ¯\; -->}{\mathrm{A}\mathrm{C}}$ and $\stackrel{¯<!--\; ¯\; -->}{\mathrm{B}\mathrm{D}}$ intersecting at E. If m$\mathrm{\angle <!--\; \angle \; -->}AEB$=60 degrees, m$\mathrm{\angle <!--\; \angle \; -->}CAB$=30 degrees, and AB=18, find AD.
This problem seems simple, and I presume that you use the properties of special right triangles; however, I can't seem to get the right answer. The answer is 6 $\sqrt{}21$. Thank you!

I'm curious about is there any geometric property relative to negative value for determinant of matrix.
$det(A)<0$
I knew about some of determinant of matrix properties as following, but it seems to me that it is nothing relative to negative value of determinant of matrix
$det(AB)=det(A)det(B)(Multiplicative)$
$det(A)=\mathit{\text{0}}⟺<!--\; ⟺\; -->\mathit{\text{A is singular}}$
$M_{2,2}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
$|<!--\; |\; -->det(M_{2\times <!--\; \times \; -->2})|<!--\; |\; -->=|<!--\; |\; -->ad-<!--\; -\; -->bd|<!--\; |\; -->=\mathit{\text{volumn of parallelogram}}$
$|<!--\; |\; -->det(M_{n\times <!--\; \times \; -->n})|<!--\; |\; -->=\underset{j=1}{\overset{n}{\prod <!--\; \prod \; -->}}{a}_{i,j}(-<!--\; -\; -->1{)}^{i+j}det(M_{i,j})\mathit{\text{expansion of determinant alone the }}{\mathit{\text{i}}}^{th}\mathit{\text{ row}}$

Is it true that:

-an inner product satisfies the properties of a norm if and only if the norm satisfies the parallelogram equality

-a norm can be induced by a metric if and only if the metric satisfies $d(x+a,y+a)=d(x,y)$ and $d(ax,ay)=ad(x,y)$

or are the implications one way?

I've looked all over and I can't find a good proof of why the diagonals of a rhombus should intersect at right angles. I can intuitively see its true, just by drawing rhombuses, but I'm trying to prove that the slopes of the diagonals are negative reciprocals and its not working out.

I'm defining my rhombus as follows: $[(0,0),(a,0),(b,c),(a+b,c)]$

I've managed to figure out that $c=\sqrt{a^2-<!--\; -\; -->b^2}$ and that the slopes of the diagonals are $\frac{\sqrt{a^2-<!--\; -\; -->b^2}}{a+b}$ and $\frac{-<!--\; -\; -->\sqrt{a^2-<!--\; -\; -->b^2}}{a-<!--\; -\; -->b}$

What I can't figure out is how they can be negative reciprocals of one another.

I mean to say that I could not find the algebraic proof. I've seen and understand the geometric proof, but I needed help translating it into coordinate form.

I'm trying to prove some properties of sequence spaces. I already know that the space $l^{\mathrm{\infty <!--\; \infty \; -->}}$ of all bounded sequences isn't an inner product space, isn't separable but it is complete with sup norm.
The space c of all convergent sequences is also complete, as a closed subspace of $l^{\mathrm{\infty <!--\; \infty \; -->}}$, and so is $c_0$ - the space off all sequences with entries equal to zero from some point on.
I also know that both $c$ and $c_0$ are separable.
My question is - are $c$ and $c_0$ inner product spaces. Should I look for an example of sequences which don't satisfy the parallelogram law.

Prove $|x+y{|}^2+|x-<!--\; -\; -->y{|}^2=2|x{|}^2+2|y{|}^2$ if $x,y\in <!--\; \in \; -->{\mathbb{R}}^k$.
Interpret this geometrically, as a statement about parallelograms.
I've shown that the expression given equates to $x\cdot <!--\; \cdot \; -->x+2x\cdot <!--\; \cdot \; -->y+y\cdot <!--\; \cdot \; -->y+x\cdot <!--\; \cdot \; -->x-<!--\; -\; -->2x\cdot <!--\; \cdot \; -->y+y\cdot <!--\; \cdot \; -->y$. But what does this have to do with parallelograms?

How can we use vectors and dot products to show that the diagonals of a parallelogram intersect at ${90}^{\circ <!--\; \circ \; -->}$ if and only if the figure is a rhombus?
I did the proof, but I realized my final answer would be a rectangle. (I know a rhombus is a type of rectangle, too). But I only want to prove the two diagonals are orthogonal.