Linear Algebra: Image, Kernel and Basis

[tangibleLime]

Homework Statement

I've been browsing the Internet but can't find a straightforward explanation for a procedure on how to find the image and kernel of a matrix.

Question: Find a basis of the image of A, and a basis of the kernel of A.

$$\[ A = \left[ {\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \\ \end{array} } \right] \]$$

The Attempt at a Solution

I think that the image (Googling doesn't help much for some reason... is there another, more popular term for 'image'?) is just the span of vectors of A that are linearly independent; that is, no two vectors can be combined to create another vector in the span (or is it only the vectors that came before it?).

So just by eyeballing it, I can tell that the image of A is [1,1,1] and [1,2,3] since [2,2,2] can be expressed as 2*[1,1,1]. But what if this matrix was more complicated and I couldn't eyeball it? From reading the definition, I think a basis of the image is just the entire span of the image... since by definition, the image already contains no redundant vectors.

I think that the kernel (or is it null-space? Are those terms synonymous?) is the span of vectors that, when multiplied by the matrix, result in the zero vector. Therefore, the kernel would always be PART of the image, but not necessarily the entire span of the image... correct?

So to find the kernel of A, I took A and put it into RREF, eventually resulting in:

$$\[\left[ {\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} } \right] \]$$

Translating this to a set of equations gives me:

$$x_{1} + 2x_{2} = 0$$
$$x_{3} = 0$$

But I'm not sure what to do from there. There are no free variables like there usually are when I solve a problem like this. I tried to do this, but now it looks weird and I know it's wrong:

$$x_{1} = -2x_{2}$$
$$x_{2} = \frac{1}{2}x_{1}$$
$$x_{3} = 0$$

I do know (thanks to a calculator) that the kernel of A is [-2,1,0], but I don't know how to get that manually.

Anyways, assuming I do know how to get the kernel of that matrix, I need to find the basis of the kernel. My understanding of basis is that it is simply the span of a vector space with linearly independent vectors. Since kernel is a vector space (right?) the basis of it would simply be the span of linearly independent vectors that make up the kernel. In this case, there is only one vector... so that is the basis?

Any help would be greatly appreciated.

 

[Mark44]

Homework Statement

I've been browsing the Internet but can't find a straightforward explanation for a procedure on how to find the image and kernel of a matrix.

Question: Find a basis of the image of A, and a basis of the kernel of A.

$$\[ A = \left[ {\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \\ \end{array} } \right] \]$$

The Attempt at a Solution

I think that the image (Googling doesn't help much for some reason... is there another, more popular term for 'image'?) is just the span of vectors of A that are linearly independent; that is, no two vectors can be combined to create another vector in the span (or is it only the vectors that came before it?).

The image of A is the set of vectors y such that Ax = y. IOW, the image is the set of outputs.

So just by eyeballing it, I can tell that the image of A is [1,1,1] and [1,2,3] since [2,2,2] can be expressed as 2*[1,1,1]. But what if this matrix was more complicated and I couldn't eyeball it? From reading the definition, I think a basis of the image is just the entire span of the image... since by definition, the image already contains no redundant vectors.

I think that the kernel (or is it null-space? Are those terms synonymous?) is the span of vectors that, when multiplied by the matrix, result in the zero vector.

Yes, that's right. And the terms nullspace and kernel are synonyms.

Therefore, the kernel would always be PART of the image, but not necessarily the entire span of the image... correct?

Right. Also, the kernel and image are subspaces of, in this case, R³. Further, dim(ker) + dim(image) = dim(R³) = 3.

So to find the kernel of A, I took A and put it into RREF, eventually resulting in:

$$\[\left[ {\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} } \right] \]$$

Translating this to a set of equations gives me:

$$x_{1} + 2x_{2} = 0$$
$$x_{3} = 0$$

But I'm not sure what to do from there. There are no free variables like there usually are when I solve a problem like this. I tried to do this, but now it looks weird and I know it's wrong:

$$x_{1} = -2x_{2}$$
$$x_{2} = \frac{1}{2}x_{1}$$
$$x_{3} = 0$$

A good start.
$$x_{1} = -2x_{2}$$
$$x_{2} = x_2$$
$$x_{3} = 0$$

So there is one free variable, x₂. Another way to write these equations is more vector-like.
$$\left[ \begin{array}{c} x_1 \\ x_2 \\x_3 \end{array} \right] = x_2\left[ \begin{array}{c} -2 \\ 1 \\0 \end{array} \right]$$

This expresses any vector in the kernel of A as a multiple of the vector shown.

I do know (thanks to a calculator) that the kernel of A is [-2,1,0], but I don't know how to get that manually.

Anyways, assuming I do know how to get the kernel of that matrix, I need to find the basis of the kernel. My understanding of basis is that it is simply the span of a vector space with linearly independent vectors. Since kernel is a vector space (right?) the basis of it would simply be the span of linearly independent vectors that make up the kernel. In this case, there is only one vector... so that is the basis?

Any help would be greatly appreciated.