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Finding the Image and Kernel


New member
May 15, 2013
Consider d map f:R^4 into R^2 defines by f(x,y,z,w)=(2x+y+z+w,x+z-w). find the image and the kernel, please include explanations...

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
Consider d map f:R^4 into R^2 defines by f(x,y,z,w)=(2x+y+z+w,x+z-w). find the image and the kernel, pls include explanations pls..
Welcome to MHB, Jerome! :)

Do you have definitions handy for image and kernel?
If so, we can see how we can apply them.

Let's start with the image.
In my book the image is the set of all vectors that can be "reached" by f.
Can for instance (1,0) be "reached"?
Or put otherwise, can you find an (x,y,z,w) in $\mathbb R^4$ that has (1,0) as its image?

And how about (0,1)?
If both can be reached, the image is $\mathbb R^2$, since (1,0) and (0,1) "span" $\mathbb R^2$.


Well-known member
MHB Math Helper
Jan 29, 2012
The kernel is the set of vectors that the linear tranformation maps to the 0 vector. (Note that if the linear transformation is from vector space U to vector space V, then the kernel is a subspace of U and the image is a subspace of V)
Since f maps (x, y, z, w) to (2x+y+z+w,x+z-w) so the kernel must have 2x+ y+ z+ w= 0, x+ z- w= 0. If we add the two equations the "w"s cancel and 3x+ y+ 2z= 0 so that y= -3x-2z. Clearly w= x+ z. So (x, y, z, w)= (x, -3x- 2z, z, x+ z)- (x, -3x, 0, x)+ (0, -2z, z, z)= x(1, -3, 0, 1)+ z(0, -2, 1, 1) which tells you immediately what the dimension of the kernel and a basis for the kernel is.

Notice that this satifies the "rank nullity" theorem: if a linear transformation is from a vector space of dimension n to a vector space of dimension m, the rank (dimension of the image) and nullity (dimension of the kernel) add to n.
Here, that is 2+ 2= 4.


Well-known member
MHB Math Scholar
Feb 15, 2012
Given an arbitrary linear mapping between two vector spaces of finite dimension, my first impulse is to choose bases and create a matrix for the linear map. My choice of bases is nothing special, I pick:

\(\displaystyle B_1 = \{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}\)
\(\displaystyle B_2 = \{(1,0),(0,1)\}\)

The 2x4 matrix I obtain relative to these two bases is:

\(\displaystyle \begin{bmatrix}2&1&1&1\\1&0&1&-1 \end{bmatrix}\)

It should be clear immediately that this matrix has rank 2 (the two rows are linearly independent), so the image of f is of dimension 2, and the ONLY 2-dimensional subspace of \(\displaystyle \Bbb R^2\) is, of course, \(\displaystyle \Bbb R^2\) itself.

This settles the image question (f is surjective), but only tells us (via the rank-nullity theorem) that the kernel has dimension 2. So we actually need to FIND a basis for the kernel (or: equivalently, the null space of our matrix above). How do we do that?

We need to solve the following HOMOGENEOUS system of equations:

2x + y + z + w = 0
x + z - w = 0

Putting our 2x4 matrix into rref form, we get the matrix:

\(\displaystyle \begin{bmatrix}1&0&1&-1\\0&1&-1&3 \end{bmatrix}\)

This is equivalent to the system:

x + z - w = 0
y - z + 3w = 0

If we pick z = 1, w = 0, we get:

x + 1 = 0
y - 1 = 0, leading to the vector (-1,1,1,0).

If we pick z = 0, w = 1, we get:

x - 1 = 0
y + 3 = 0, leading to the vector (1,-3,0,1). These two vectors are clearly linearly independent. Are they elements of the kernel? We check:

f(-1,1,1,0) = (-2+1+1+0,-1+1+0) = (0,0)
f(1,-3,0,1) = (2-3+0+1, 1+0-1) = (0,0)

so {(-1,1,1,0),(1,-3,0,1)} is a basis for the kernel.

Does this agree with Hall's answer? If so, we should be able to express both:

(1,-3,0,1) and (0,-2,1,1) as a linear combination of our two vectors (remember, bases are not UNIQUE). And indeed:

(1,-3,0,1) = (0)(-1,1,1,0) + (1)(1,-3,0,1)
(0,-2,1,1) = (1)(-1,1,1,0) + (1)(1,-3,0,1)