What is the geometric meaning of the cross product of two vectors?
Consider two vectors in the plane of the screen you are looking at now. When you cross two vectors you end up with another vector that is perpendicular to the two vectors. The way this vector points is based off of the right hand rule. You point the fingers of your right hand in the direction of the first vector. You then curl your fingers in the direction of the second vector you are crossing. Whatever way your thumb is pointing is the direction the crossed vector will be. EX: If your first vector pointed along the length of your monitor and the second vector pointed along the width of the monitor, the cross product would result in a vector pointing out of the screen. Also, A X B = ABsin(theta), Where theta is the angle between the two vectors. If the two vectors are parallel, the angle between them is zero, thus theta is zero. Sin(zero) is zero, thus A X B equals zero.
Give the geometric description of Span {v1,v2} for the vectors?
I'm a bit confused because the y term is 0. If it wasn't 0, I would have been able to solve this. I can actually solve this, but I prefer to use the old school: x = 3s - 2t y = 0 z = 2s + 3t From here on you can eliminate parameters, but for some reason I can't do it when the y term is 0. Let's try to get the equation of the line by using the cross product. If u =
Is this correct in giving a geometric description of Span[math]\{\vec{v}_{1},\vec{v}_{2}\}[/math] for [math]\vec{v}_{1}=(8,2,-6),\vec{v}_{2}=(12,3,-9)[/math]?
The span of [math]\vec{v_{1}}[/math] and [math]\vec{v_{2}}[/math] is the set of all possible linear combinations of [math]\vec{v_{1}}[/math] and [math]\vec{v_{2}}[/math] of the form [math]a\vec{v_{1}}+b\vec{v_{2}}[/math], where a,b are scalars. Now [math]\vec{v_{1}}=\frac{2}{3}\vec{v_{2}}[/math], which means that a linear combination is of the form [math](\frac{2}{3}a+b)\vec{v_{2}}=c\vec{v_{2}}[/math], c being another scalar. So all linear combinations of the two vectors are of the form [math]c\vec{v_{2}}[/math] which means that the vector space is one-dimensional along the line through the origin whose direction numbers are given by [math]\vec{v_{2}}[/math].
Geometrically how do you understand polynomial vector space?
Start with the subspace of nth degree polynomials, those of the form [math]a_0+a_1x+\dots+a_nx^n.[/math] This is an n+1 dimensional vector space with orthonormal basis [math]\{1,x,x^2,\dots,x^n\}[/math]. One beautiful result in linear algebra is that all finite dimensional vector spaces are isomorphic. So you can geometrically understand the space of all degree 2 polynomials as just being the same space as [math]\mathbb{R}^3.[/math] Any two linearly independent polynomials span a "plane" in the space. You can give the polynomials vector coordinates. For example you can write [math]1+x^2-x^3-2x^4[/math] as the vector [math](1,0,1,-1,-2)[/math]. You can think of two polynomials being orthogonal to each other if their dot product is zero. So [math]x^3-x+1[/math] is orthogonal to [math]x^2+x-1[/math] since [math]1-x+x^3\cdot -1+x+x^2=(1,-1,0,1)\cdot(-1,1,1,0)=0.[/math] Working in this way, you can extend all of your intuition and knowledge about [math]\mathbb{R}^{n+1}[/math] to the space of polynomials of degree n. You can express linear transformations of polynomials as matrices too. Basically, the fact that they are polynomials doesn't really matter. The x doesn't do much except change what the vectors and linear operators look like. The space of all polynomials on the other hand is something you may not be as familiar with: an infinite dimensional vector space. It has an orthonormal basis [math]\{1,x,x^2,x^3,\dots\}[/math] consisting of infinitely many vectors. Almost everything you have learned about finite dimensional spaces works for this space, but there are a few little differences, such as the fact that not every subspace is closed (if you know what a closed set is), and not every infinite linearly independent set of vectors forms a basis. You can still use orthonormality and all that.
We define the geometric transformation of translation by a vector v = (v1, v2)^T in the plane as?
No that's not linear. That's part of the "affine" class, which contains all functions of the form x->Mx+v. In your case M=Id and the function is simply a translation. There are sometimes ambiguities because in some contexts we say "linear" when we actually mean "affine". A linear function is a function f such that f(ax+by)=af(x)+bf(y) In R the only linear functions are the multiplications by some constant. In R^n (or any dimension n space) any linear function f can be represented with a matrix M such that f(x)=Mx And vice-versa: any x->Mx function has to be linear. So your translation cannot be expressed under that form.
Describe geometrically the span of the set and give a simplified vector equation which describes it?
Hi, the span of a set of vectors is defined as the smallest vector space containing them. Equivalently it can be defined as the vector space generated by the given vectors. Now, usually what one makes in order to determine the span of a given set of vectors is this: - find the maximum number of linearly independent vectors in the given set and isolate them: they will be the base of the span; - write the general vector of the span as their linear combination; - try to figure out some sort of "nice" equation describing it. In your case you see that the two vectors in S are linearly dependent (their sum is (0, 0, 0)). Then the span will have a 1-vector base (for example the base containing just (1, 1, -3)), it will be a 1-dimensional vector space, better known as line (this is its geometric description, a line through the origin). Now all you have to find is a vector equation for it. The span will contain only those vectors that are multiples of the vector of the base (all of them actually), so it will contain all the vectors of the form t(1, 1, -3). Allowing t to vary in |R you get v(t) = t(1, 1, -3) = (t, t, -3t) which is the vector equation describing the span of S. I hope this helped you. Bye!
The span of the sum of two vectors...?
Yes. if u ∈ Span{v1, v2, . . . , vk} then u = c1v1 + c2v2 + ... + ckvk (definition of span) and if w ∈ Span{v1, v2, . . . , vk} then w = p1v1 + p2v2 + ... + pkvk (definition of span) u+w = (c1+p1)v1 + .... + (ck+pk)vk = a1v1 + a2v2 + .. + akvk. Since u + w can be written as a linear combination of vectors in the set {v1, v2, . . . , vk}, we have u + w ∈ Span{v1, v2, . . . , vk}
Geometrically, how can three vectors span a plane in [math]\mathbb{R}^3[/math]?
What makes you think the three vectors are linearly independent? Indeed, if they were, it wouldn't be a plane. In the top picture, for instance, u+v is clearly not independent from {u, v} and that is why it lies in that plane. In your second picture, it's unclear what [math]a_1, a_2, a_3[/math] are. I assume they are linearly dependent too.Basically, two linearly independent vectors span a plane, 3 linearly independent vectors span a 3-dimensional subspace, and so on.
Linear Algebra: Geometric Forms of Matrices?
The question equates to whether the dimension of the subspace they generate is 0, 1, 2, or 3. (Why? Make sure you know, as that's the key point.) After that -- well, v1 and v2 are obviously linearly independent of each other, so the dimension is at least 2. So the answer is C if v3 can be written as a linear combination of v1 and v2, D if it cannot be. (Why?) Proceed from there. :)
Algebraically and geometrically, how do you identify a span of vectors a line, a point, or a plane?
To span n dimensions takes n vectors, minimum. For instance a plane requires 2 vectors. Suppose we have 3 vectors, a, b, and c. Vector a has 3 coordinates (ax, ay, az), and so on.Algebraically:If they span only:point : they all must be zero (0,0,0)line: they must be multiples of each other. For instance (1,1,1), (2,2,2) and (3,3,3) are all on the same line.plane: If they’re not all multiples, then they span a plane if there are 3 numbers A, B, and C, not all zero, such that A*a+B*b+C*c = 0. Two of them might be multiples, or else two of them combined equals the third.volume : If they don’t meet the conditions to span a plane we say they're independent. For instance (1,0,0), (0,1,0), (0,0,1).BTW you can also compute this algebraic result by taking the determinant of the matrix formed by the 3 vectors. If it’s nonzero, they span 3-d space.Geometrically they look like:point: all 0, just a point at the originline: they're all on the same line, called collinear, or dependent. Simplest case, they're all on the x-axis.plane: if two of them aren't on the same line, they must span a plane. For instance the x-axis and y-axis.volume: if the third one doesn't lie on the plane defined by the first two, but comes off at an angle, they span the whole 3-d space.BTW this all generalizes to n dimensions very easily.