# Help a newbie understand Linear Algebraic terms

I am taking a class in Algebra but I am having a problem grasping exactly what it is I am being asked to do -- I think I am having a problem with the vocabulary being used. I have a couple of questions below and I would appreciate it, if you could take time to explain(in layman's terms or maybe programmer terms, exactly what is being asked. thanks in advance.

1) Let $[e_1; e_2; e_3]$ be the standard basis vectors in $\mathbb{R}^{3}$ and consider the ordered basis: $[e_2; e_1; e_3 + e_1]$ Verify that this is actually a basis and find the coordinates of the vector $(1; 1; 1)^T$ with respect to that basis.

2) Let $T$ be the linear map from $\mathbb{R}^{2}$ to $\mathbb{R}$ defined by: $T ((x, y)^T ) = x-y$ Find its matrix (with respect to the standard bases) and a basis for its kernel

3) Find a spanning vector set for the image of the linear map from $\mathbb{R}^{2}$ to $\mathbb{R}^{3}$ defined by: $T ((x, y)^T) = (2x-y, x + y, y)$

4) Consider the subspace U of $\mathbb{R}^3$ defined by: $U = \{(x; y; z)^T: 2x- y + z = 0; x + y = 0 \}$. Express U as the kernel of an appropriately defined linear map and the matrix of that map with respect to the standard bases of the corresponding $\mathbb{R}^n$'s

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@user10695: Neither the map in (2) nor in (3) is a linear map. Did you copy them correctly, or are you missing a $+$ plus sign somewhere? – Arturo Magidin Jun 22 '11 at 19:36
@user10695: The problem isn't the use of commas versus semicolons. The problem is that mapping $(x,y)^T$ to $xy$ is not linear: the image of $(1,1)^T$ is $1$, but the image of $(1,1)^T+(1,1)^T = (2,2)^T$ is not the sum of the images. – Arturo Magidin Jun 22 '11 at 19:42
^^^ Now the above is part of what I would like to have broken down a bit more. :) . I made changes, does it work now? Could you also please explain yourself a bit more please? – user10695 Jun 22 '11 at 20:00
@user10695: Yes, mapping $(x,y)$ to $x-y$ is linear. The reason the previous one was not linear is that, remember, if $T$ is linear, then $T(\mathbf{v}+\mathbf{w}) = T(\mathbf{v}) + T(\mathbf{w})$ and $T(\alpha\mathbf{v}) = \alpha T(\mathbf{v})$ for all vectors $\mathbf{v}$ and $\mathbf{w}$, and all scalars $\alpha$. To see $T(x,y) = xy$ is not linear, take $\mathbf{x}=(1,1)$, $\mathbf{y}=(1,1)$, and compare $T(\mathbf{x}+\mathbf{y})$ with $T(\mathbf{x}) + T(\mathbf{y})$. You have:$$T((1,1)+(1,1))=T(2,2)=4\neq 1+1 = T(1,1)+T(1,1).$$Not equal, not linear. – Arturo Magidin Jun 22 '11 at 20:24
I think I understand now. Thanks – user10695 Jun 22 '11 at 22:13

All of these terms are in standard use so you should be able to find the precise definitions in any linear algebra text or Wikipedia for the precise definitions. Briefly, however, here are a few definitions that might help you to better understand the terminology.

(A). A subset $S$ of a vector space is said to span the space if every vector in the space, whether in the subset or not, can be written as a linear combination of the elements of $S$.

(B). A basis of a vector space $V$ is a collection of elements of $V$ such that any other vector can be expressed uniquely as a linear combination of those elements. Practically, this means that a basis $B$ both spans the space and is linearly independent. Checking that a given set is a basis then just amounts to verifying that any element $v \in V$ can be expressed as a linear combination of the elements of $B$ and that any linear combination of the elements of $B$ that sums to the $0$ vector necessarily implies that the scalars used in the linear combination are all $0$ - otheerwise the expression for $v$ wouldn't necessarily be unique.

(C). The kernel of a linear transformation $T$ is the set of elements in the domain that are mapped to $0$. So for map $T:\mathbb{R}^2 \rightarrow \mathbb{R}$ the kernel would be given by $ker(T) = \{v \in \mathbb{R}^2 | T(v) = 0 \}$.

Question 1 is basically asking you verify the aspects of the definition elucidated in (B) above to conclude that the given set is, in fact, as basis and to find the scalars that you need to use in order to write $(1,1,1)^T$ as a linear combination of elements in this basis.

For Question 2, you first need to compute the kernel as described in (C) and then determine one set of vectors that satisfy the criteria for a basis as described in (B)

For Question 3, you just need to determine the range of the map, which will turn out in fact to be a subspace, and then set of vectors that will comprise a spanning set as described in (A). This is essentially asking you to solve "half" of the basis problem...

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1) This is just a change-of-basis question. It's asking you to know that the three given vectors are really a basis (i.e. linearly independent). It then asks you to take this vector $[1;1;1]^T$ and express its coordinates in the new basis; that is, what linear combination of the three given vectors produces the same vector as $[1;1;1]^T$ in the original basis.

2) and 3) I'm suspicious about these questions. Neither of the functions given are actually linear maps, which is easily seen in 2) by $T(2{\vec v}) = 4T({\vec v})$. It is a bilinear form (if you treat it as a function of two arguments), but I wouldn't expect those to show up in an intro course. It's also a quadratic form, if that's what you meant. The function in 3) is neither of these, of course. I would please check your transcription? If your transcription is right you're going to have to go ask your teacher what exactly he meant by these questions, because as written they make no sense.

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1) A basis is a set of vectors which can be used to compose any given vector by stretching/compressing/reversing (scaling) and addition/subtraction (combination). Such combinations of vectors are called linear combinations. A basis must fulfill two requirements

(1) each basis vector must be independent (it cannot be formed from a linear combination of the other vectors). (i.e. not too many)

(2) it must be capable of composing any vector in the given space/subspace (it must span the space). (i.e. not too few)

A natural analogue to a basis are prime numbers or primary colors. By combining the components you can compose any whole number or color respectively.

The standard basis is the set of unit vectors which are orthogonal to each other along the coordinate axes. This is not the only choice however, we may form other valid bases by linearly combining the components of the standard basis. This is what is being asked in the first question. Given that a vector is represented as $(1,1,1)$ in the standard basis, what is its representation in the new basis. This is a 3x3 linear system of equations. (Back to the color analogy, if we are given a color in RBG values, how do we represent the same color in terms of HSV or CMYK components).

2) The transformation takes a pair of numbers (2-D vector) and transforms it to a scalar (1-D vector). To find its matrix means to find a matrix such that $T ((x, y)^T ) = \mathbf A [x, y]^T = x-y$. Notice that A is a linear combination of the components $x$ and $y$. The kernel of $\mathbf A$ is the set of vectors $[x, y]^T$ such that $\mathbf A [x, y]^T = 0$. This is commonly referred to as the Null Space.

3) A spanning vector set is similar to a basis except that they do not necessarily have to be linearly independent like a basis does. The image of a transformation is merely the result of a particular transformation of a given vector $[x, y]^T$.

4) You have a subspace defined by two constraints (i.e. two planar equations). Since they are not parallel, they intersect at a line. This is your subspace which is 1-D. The subspace would therefore be represented as a line. You are looking for a 3x3 transformation which maps each point $[x, y, z]^T$ to that line. This is a projection transformation. The basis of the subspace is defined by the cross product of $\mathbf v=[2, -1, 1]^T$ and $\mathbf w=[1, 1, 0]^T$ defined by $\mathbf x = \mathbf v\times \mathbf w$. The projection transformation is $\mathbf x \mathbf x^T / (\mathbf x^T \mathbf x)$. The kernel of the subspace is 2-D and is defined by the plane which is perpendicular to $\mathbf x$ and is spanned by $\mathbf v$ and $\mathbf w$, in fact they form a basis (not orthogonal) for the kernel.

P.S. I highly recommend that you view Gilbert Strang's Introduction to Linear Algebra Course at http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/

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Here is how I taught my students to check if some elements form a basis. You could go by the definition, but it will take longer, and in fact you have to check exactly what I say.

1) The set of vectors $[e_2; e_1; e_3+e_1]$ are the following vectors written in coordinates: $[(0,1,0); (1,0,0);,(1,0,1)]$. Now, to see if they form a basis, calculate the determinant $\left| \begin{matrix} 0 &1&0 \\ 1&0&0 \\ 1&0&1 \end{matrix}\right|$. If the result is not zero, then you found a basis.

To find the coordinates of $(1,1,1)$ in that basis, write $(1,1,1)=a(0,1,0)+b(1,0,0)+c(1,0,1)$ and identify coordinates. You'll get a system of equations from where you should get $a,b,c$, which are the required coordinates.

2) For finding the matrix of a linear map in the standard basis you just calculate the value of $T$ on every element of the basis, and then express every of the found expressions in terms of the second basis(in your case it's standard basis), and put the resulting vectors on the columns of a matrix.

You have $T(1,0)=1,\ T(0,1)=-1$. Therefore the matrix of the application is $(1\ -1)$.

$T(x,y)=0$ implies $x=y$. Therefore $\ker T=\{(a,a) : a \in \Bbb{R}=\{a(1,1): a \in \Bbb{R}\}$. The idea is to solve the kernel as a linear system, and in the end separate the parameters you found in order to find a basis. In your case, $(1,1)$ is a basis for the kernel, since we only have one parameter.

3) In the same way as above, $T(1,0)=(2,1,0)$ and $T(0,1)=(-1,1,1)$, therefore, the matrix of the application is $M=\begin{pmatrix} 2 & -1 \\ 1&1 \\ 0 & 1 \end{pmatrix}$. For a spanning vector set, note that $T(x,y)=x(2,1,0)+y(-1,1,1)$, so a spanning vector set is formed of the vectors $(2,1,0),(-1,1,1)$.

4) $T(x,y,z)=(2x-y+z,x+y)$, $T :\Bbb{R}^3 \to \Bbb{R}^2$. $T(1,0,0)=(2,1),\ T(0,1,0)=(-1,1),\ T(0,0,1)=(1,0)$, and therefore, the matrix is $M=\begin{pmatrix} 2 & -1 & 1 \\ 1 & 1 & 0 \end{pmatrix}$.

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Plugging the values $T(1,0,0)=$ into $(2x-y+z = 0, x+y= 0)$, does not seem to work. Isn't that what is also being required there? – user10695 Jun 22 '11 at 21:49
I was doing that just to illustrate how you find the matrix of the transformation in that case. You can see in the sentence starting with "For a spanning vector set..." what is needed for point 3) – Beni Bogosel Jun 22 '11 at 21:55
I was referring to your response on (4). From what I see, I think I am required to restrict the set of elements in U to only those elements that satisfy the given equations. Is that correct? – user10695 Jun 22 '11 at 22:13
It says there to express the set as the kernel of some linear application. The application is the one mentioned by me. Notice that I took the equations of the kernel and turned them into components of the linear map, without $=0$, of course. – Beni Bogosel Jun 22 '11 at 22:40

1) A basis is a tuple of vectors $e_1, \ldots, e_n \in V$ such that for any vector $x \in V$ you can write $x = x^1 e_1 + \ldots + x^n e_n$ in exactly one way (i.e. the coefficients, or coordinates, $x^1, \ldots, x^n$ are defined uniquely for each vector $x$).

2) A function $f: V \to W$ between two vector spaces over the same field $K$ is called linear if $f(x + y) = f(x) + f(y)$ and $f(\alpha x) = \alpha f(x)$ for any vectors $x, y \in V$ and for any scalar $\alpha$. By picking a basis in V and a basis in W you can represent $f$ by a matrix. Kernel of $f$ is a subset $\operatorname{ker}f \subset V$ of vectors $x \in V$ for which $f(x) = 0$ holds. It is easy to show that this is a subspace (a vector space itself).

3) Likewise, the image of $f$ is the set of all vectors in $W$ that can be its 'outputs'. It is a subspace too. Spanning set is like basis, only you do not require the uniqueness of the coordinates.

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